Solutions to Axler’s Measure, Integration & Real Analysis

Let \(P\) be the partition \(a = x_0 < x_1 < \cdots < x_n = b\). By the definition of the lower and upper Riemann sums (1.3),

\begin{equation*} U(f, P, [a,b]) - L(f, P, [a,b]) = \sum_{j=1}^{n} (x_j - x_{j-1}) \Big( \sup_{[x_{j-1}, x_j]} f - \inf_{[x_{j-1}, x_j]} f \Big). \end{equation*}

For each \(j\) the number \(x_j - x_{j-1}\) is strictly positive, and the infimum of a nonempty set of real numbers is less than or equal to its supremum, so every summand above is nonnegative. The hypothesis says the whole sum equals \(0\). A finite sum of nonnegative numbers equals \(0\) only if each summand equals \(0\), and since \(x_j - x_{j-1} > 0\) we conclude

\begin{equation*} \sup_{[x_{j-1}, x_j]} f = \inf_{[x_{j-1}, x_j]} f \quad \text{for each } j = 1, \ldots, n. \end{equation*}

Fix \(j\). For every \(x \in [x_{j-1}, x_j]\) we have \(\inf_{[x_{j-1},x_j]} f \le f(x) \le \sup_{[x_{j-1},x_j]} f\), and the two outer quantities are equal; hence \(f\) is constant on \([x_{j-1}, x_j]\). Write \(c_j\) for that constant value.

It remains to see that \(c_1 = c_2 = \cdots = c_n\). For \(1 \le j \le n-1\) the point \(x_j\) belongs to both \([x_{j-1}, x_j]\) and \([x_j, x_{j+1}]\), so

\begin{equation*} c_j = f(x_j) = c_{j+1}. \end{equation*}

Thus all the \(c_j\) are equal to a common value \(c\). Since \([a,b] = [x_0, x_1] \cup [x_1, x_2] \cup \cdots \cup [x_{n-1}, x_n]\), every \(x \in [a,b]\) lies in some \([x_{j-1}, x_j]\) and therefore satisfies \(f(x) = c_j = c\). Hence \(f\) is constant on \([a,b]\).

The function \(f\) takes only the values \(0\) and \(1\), so it is bounded and the lower and upper Riemann sums are defined.

Fix \(\varepsilon\) with \(0 < \varepsilon < \frac{t-s}{2}\), so that

\begin{equation*} s < s + \varepsilon < t - \varepsilon < t . \end{equation*}

Let \(P_\varepsilon\) be the partition of \([a,b]\) whose points are the distinct elements of

\begin{equation*} \{a,\; s,\; s+\varepsilon,\; t-\varepsilon,\; t,\; b\}, \end{equation*}

listed in increasing order. (All six numbers lie in \([a,b]\) and are listed in nondecreasing order; the only possible coincidences are \(s = a\) and \(t = b\), and deleting repetitions leaves a genuine partition of \([a,b]\).) Its subintervals are

\begin{equation*} [a,s] \ (\text{present only if } a < s), \quad [s, s+\varepsilon], \quad [s+\varepsilon, t-\varepsilon], \quad [t-\varepsilon, t], \quad [t,b] \ (\text{present only if } t < b). \end{equation*}

Lower sum. Every \(x \in [s+\varepsilon, t-\varepsilon]\) satisfies \(s < x < t\), so \(f \equiv 1\) there and \(\inf_{[s+\varepsilon,\,t-\varepsilon]} f = 1\). On every other subinterval the infimum of \(f\) is at least \(0\), since \(f \ge 0\). Hence

\begin{equation*} L(f, P_\varepsilon, [a,b]) \ge (t - \varepsilon) - (s + \varepsilon) = (t-s) - 2\varepsilon . \end{equation*}

Therefore \(L(f, [a,b]) \ge (t-s) - 2\varepsilon\) for every \(\varepsilon \in \big(0, \tfrac{t-s}{2}\big)\), and letting \(\varepsilon \to 0\) gives

\begin{equation*} L(f, [a,b]) \ge t - s . \end{equation*}

Upper sum. If \(x \in [a,s]\) then \(x \le s\), so \(f(x) = 0\); thus \(\sup_{[a,s]} f = 0\). If \(x \in [t,b]\) then \(x \ge t\), so \(f(x) = 0\); thus \(\sup_{[t,b]} f = 0\). On the remaining three subintervals we use only \(\sup f \le 1\). Hence

\begin{equation*} U(f, P_\varepsilon, [a,b]) \le 0 \cdot (s-a) + 1 \cdot \varepsilon + 1 \cdot \big((t-\varepsilon)-(s+\varepsilon)\big) + 1 \cdot \varepsilon + 0 \cdot (b-t) = t - s . \end{equation*}

Therefore \(U(f, [a,b]) \le t-s\).

Combining the two displays with 1.8 gives

\begin{equation*} t - s \le L(f, [a,b]) \le U(f, [a,b]) \le t - s , \end{equation*}

so all these quantities are equal. Thus \(f\) is Riemann integrable on \([a,b]\) and \(\int_a^b f = t - s\).

Throughout, recall from 1.7 that \(L(f,[a,b]) = \sup_P L(f,P,[a,b])\) and \(U(f,[a,b]) = \inf_P U(f,P,[a,b])\), so that

\begin{equation*} L(f, P, [a,b]) \le L(f, [a,b]) \quad\text{and}\quad U(f,[a,b]) \le U(f, P, [a,b]) \end{equation*}

for every partition \(P\) of \([a,b]\).

Suppose first that the \(\varepsilon\)-condition holds. Let \(\varepsilon > 0\) and choose a partition \(P\) of \([a,b]\) with \(U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon\). Then, using the two inequalities above and 1.8,

\begin{equation*} 0 \le U(f,[a,b]) - L(f,[a,b]) \le U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon . \end{equation*}

Since this holds for every \(\varepsilon > 0\), we get \(U(f,[a,b]) - L(f,[a,b]) = 0\). Thus \(L(f,[a,b]) = U(f,[a,b])\), which by 1.9 says exactly that \(f\) is Riemann integrable.

Conversely, suppose \(f\) is Riemann integrable, so \(L(f,[a,b]) = U(f,[a,b])\), and let \(\varepsilon > 0\). Because \(L(f,[a,b])\) is the supremum of the lower Riemann sums, the number \(L(f,[a,b]) - \frac{\varepsilon}{2}\) is not an upper bound for them, so there is a partition \(P_1\) of \([a,b]\) with

\begin{equation*} L(f, P_1, [a,b]) > L(f, [a,b]) - \tfrac{\varepsilon}{2} . \end{equation*}

Because \(U(f,[a,b])\) is the infimum of the upper Riemann sums, there is likewise a partition \(P_2\) of \([a,b]\) with

\begin{equation*} U(f, P_2, [a,b]) < U(f, [a,b]) + \tfrac{\varepsilon}{2} . \end{equation*}

Let \(P\) be the partition of \([a,b]\) obtained by merging the lists defining \(P_1\) and \(P_2\); then the lists defining \(P_1\) and \(P_2\) are both sublists of the list defining \(P\), so 1.5 gives

\begin{equation*} L(f, P, [a,b]) \ge L(f, P_1, [a,b]) \quad\text{and}\quad U(f, P, [a,b]) \le U(f, P_2, [a,b]). \end{equation*}

Combining these with the previous two displays and with \(L(f,[a,b]) = U(f,[a,b])\),

\begin{equation*} \begin{aligned} U(f, P, [a,b]) - L(f, P, [a,b]) &\le U(f, P_2, [a,b]) - L(f, P_1, [a,b]) \\ &< \Big( U(f,[a,b]) + \tfrac{\varepsilon}{2} \Big) - \Big( L(f,[a,b]) - \tfrac{\varepsilon}{2} \Big) \\ &= \varepsilon . \end{aligned} \end{equation*}

Thus \(P\) is a partition of the required kind, completing the proof.

Both \(f\) and \(g\) are bounded, hence so is \(f+g\); thus the Riemann sums of \(f+g\) are defined.

Step 1: a pointwise comparison of Riemann sums. Let \(A\) be a nonempty subset of \([a,b]\). For every \(x \in A\) we have \(f(x) + g(x) \ge \inf_A f + \inf_A g\), so \(\inf_A f + \inf_A g\) is a lower bound for \(\{(f+g)(x) : x \in A\}\) and therefore

\begin{equation*} \inf_A f + \inf_A g \le \inf_A (f+g). \end{equation*}

Symmetrically, \(\sup_A f + \sup_A g\) is an upper bound for that set, so

\begin{equation*} \sup_A (f+g) \le \sup_A f + \sup_A g . \end{equation*}

Applying these with \(A = [x_{j-1}, x_j]\) for each subinterval of an arbitrary partition \(P\) of \([a,b]\), multiplying by \(x_j - x_{j-1} > 0\), and summing over \(j\) gives

\begin{equation*} L(f, P, [a,b]) + L(g, P, [a,b]) \le L(f+g, P, [a,b]) \le U(f+g, P, [a,b]) \le U(f, P, [a,b]) + U(g, P, [a,b]), \end{equation*}

where the middle inequality holds because an infimum never exceeds the corresponding supremum.

Step 2: integrability. Let \(\varepsilon > 0\). Since \(f\) and \(g\) are Riemann integrable, Exercise 3 in this section gives partitions \(P_1\) and \(P_2\) of \([a,b]\) with

\begin{equation*} U(f, P_1, [a,b]) - L(f, P_1, [a,b]) < \tfrac{\varepsilon}{2}, \qquad U(g, P_2, [a,b]) - L(g, P_2, [a,b]) < \tfrac{\varepsilon}{2}. \end{equation*}

Let \(P\) be the partition obtained by merging the lists defining \(P_1\) and \(P_2\). By 1.5, passing from \(P_1\) to \(P\) can only increase the lower sum of \(f\) and decrease its upper sum, so \(U(f,P,[a,b]) - L(f,P,[a,b]) < \frac{\varepsilon}{2}\); for the same reason \(U(g,P,[a,b]) - L(g,P,[a,b]) < \frac{\varepsilon}{2}\). Now Step 1 gives

\begin{equation*} \begin{aligned} U(f+g, P, [a,b]) - L(f+g, P, [a,b]) &\le \big( U(f,P,[a,b]) + U(g,P,[a,b]) \big) - \big( L(f,P,[a,b]) + L(g,P,[a,b]) \big) \\ &< \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{aligned} \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, Exercise 3 shows that \(f + g\) is Riemann integrable on \([a,b]\).

Step 3: the value of the integral. Keep \(\varepsilon > 0\) and the partition \(P\) from Step 2. Using \(\int_a^b (f+g) = U(f+g,[a,b]) \le U(f+g,P,[a,b])\) together with Step 1,

\begin{equation*} \int_a^b (f+g) \le U(f,P,[a,b]) + U(g,P,[a,b]) < \Big( L(f,P,[a,b]) + \tfrac{\varepsilon}{2} \Big) + \Big( L(g,P,[a,b]) + \tfrac{\varepsilon}{2} \Big) \le \int_a^b f + \int_a^b g + \varepsilon, \end{equation*}

where the last step uses \(L(f,P,[a,b]) \le L(f,[a,b]) = \int_a^b f\) and similarly for \(g\). In the same way, using \(\int_a^b (f+g) = L(f+g,[a,b]) \ge L(f+g,P,[a,b])\) and Step 1,

\begin{equation*} \int_a^b (f+g) \ge L(f,P,[a,b]) + L(g,P,[a,b]) > \Big( U(f,P,[a,b]) - \tfrac{\varepsilon}{2} \Big) + \Big( U(g,P,[a,b]) - \tfrac{\varepsilon}{2} \Big) \ge \int_a^b f + \int_a^b g - \varepsilon , \end{equation*}

where the last step uses \(U(f,P,[a,b]) \ge U(f,[a,b]) = \int_a^b f\) and similarly for \(g\). Hence

\begin{equation*} \left| \int_a^b (f+g) - \Big( \int_a^b f + \int_a^b g \Big) \right| < \varepsilon \end{equation*}

for every \(\varepsilon > 0\), which forces \(\int_a^b (f+g) = \int_a^b f + \int_a^b g\).

Since \(f\) is bounded, so is \(-f\).

Step 1: negation swaps infima and suprema. Let \(A\) be a nonempty subset of \([a,b]\) and put \(\alpha = \sup_A f\). For every \(x \in A\) we have \(f(x) \le \alpha\), hence \(-f(x) \ge -\alpha\), so \(-\alpha\) is a lower bound for \(\{-f(x) : x \in A\}\). If \(\beta\) is any lower bound for that set, then \(-\beta\) is an upper bound for \(\{f(x) : x \in A\}\), so \(\alpha \le -\beta\), i.e. \(\beta \le -\alpha\). Thus \(-\alpha\) is the greatest lower bound:

\begin{equation*} \inf_A (-f) = - \sup_A f . \end{equation*}

Replacing \(f\) by \(-f\) in this identity (and using \(-(-f) = f\)) gives \(\inf_A f = -\sup_A(-f)\), that is,

\begin{equation*} \sup_A (-f) = - \inf_A f . \end{equation*}

Step 2: negation swaps lower and upper Riemann sums. Let \(P\) be the partition \(a = x_0 < \cdots < x_n = b\). Applying Step 1 with \(A = [x_{j-1}, x_j]\),

\begin{equation*} L(-f, P, [a,b]) = \sum_{j=1}^n (x_j - x_{j-1}) \inf_{[x_{j-1},x_j]} (-f) = - \sum_{j=1}^n (x_j - x_{j-1}) \sup_{[x_{j-1},x_j]} f = -\,U(f, P, [a,b]), \end{equation*}

and symmetrically \(U(-f, P, [a,b]) = -\,L(f, P, [a,b])\).

Step 3: negation swaps lower and upper Riemann integrals. For any nonempty set \(S \subseteq \mathbb{R}\) we have \(\sup(-S) = -\inf S\) and \(\inf(-S) = -\sup S\), by the argument of Step 1 applied to \(S\). Taking \(S\) to be the set of upper Riemann sums of \(f\) over all partitions of \([a,b]\), Step 2 shows that the set of lower Riemann sums of \(-f\) is exactly \(-S\), so

\begin{equation*} L(-f, [a,b]) = \sup_P L(-f, P, [a,b]) = \sup(-S) = -\inf S = -\,U(f, [a,b]) . \end{equation*}

Likewise, taking \(T\) to be the set of lower Riemann sums of \(f\),

\begin{equation*} U(-f, [a,b]) = \inf_P U(-f, P, [a,b]) = \inf(-T) = -\sup T = -\,L(f, [a,b]) . \end{equation*}

Step 4: conclusion. Because \(f\) is Riemann integrable, \(L(f,[a,b]) = U(f,[a,b]) = \int_a^b f\). Hence Step 3 gives

\begin{equation*} L(-f, [a,b]) = -\int_a^b f = U(-f, [a,b]) . \end{equation*}

So the lower and upper Riemann integrals of \(-f\) agree; by 1.9, \(-f\) is Riemann integrable on \([a,b]\) and \(\int_a^b (-f) = -\int_a^b f\).

Define \(h : [a,b] \to \mathbb{R}\) by \(h = g - f\). Then \(h(x) = 0\) for all except finitely many \(x \in [a,b]\), and \(g = f + h\). By Exercise 4 in this section it suffices to prove that \(h\) is Riemann integrable on \([a,b]\) with \(\int_a^b h = 0\): granting that, \(g = f + h\) is Riemann integrable and \(\int_a^b g = \int_a^b f + \int_a^b h = \int_a^b f\).

So let \(c_1 < c_2 < \cdots < c_m\) be the points of \([a,b]\) at which \(h\) is nonzero (if there are none then \(h \equiv 0\), whose lower and upper Riemann sums are all \(0\), so \(h\) is Riemann integrable with integral \(0\) and we are done). Set

\begin{equation*} M = \max\{ |h(c_1)|, \ldots, |h(c_m)| \} > 0 . \end{equation*}

Every value of \(h\) lies in \([-M, M]\), so \(h\) is bounded.

Construction of good partitions. Let \(\delta > 0\). Let \(P_\delta\) be the partition of \([a,b]\) whose points are the distinct elements of

\begin{equation*} \{a, b\} \cup \big( \{ c_i - \delta,\; c_i + \delta \;:\; i = 1, \ldots, m \} \cap (a,b) \big), \end{equation*}

listed in increasing order. This is a genuine partition of \([a,b]\): it is a finite increasing list of points of \([a,b]\) beginning at \(a\) and ending at \(b\).

Claim: if \([u,w]\) is a subinterval of \(P_\delta\) and \(c_i \in [u,w]\), then \([u,w] \subseteq [c_i - \delta, c_i + \delta]\). Indeed, suppose \(u < c_i - \delta\). Since also \(c_i - \delta < c_i \le w\), the number \(c_i - \delta\) lies in \((u,w)\), and since \(a \le u < c_i - \delta < w \le b\) it lies in \((a,b)\). Hence \(c_i - \delta\) is one of the points of \(P_\delta\) strictly between the consecutive partition points \(u\) and \(w\), which is impossible. So \(u \ge c_i - \delta\), and the symmetric argument gives \(w \le c_i + \delta\), proving the claim.

Estimating the Riemann sums. Call a subinterval of \(P_\delta\) bad if it contains at least one of \(c_1, \ldots, c_m\), and good otherwise. On a good subinterval \(I\) we have \(h \equiv 0\), so \(\sup_I h = \inf_I h = 0\) and that subinterval contributes \(0\) to both \(L(h, P_\delta, [a,b])\) and \(U(h, P_\delta, [a,b])\). On a bad subinterval \(I\) the values of \(h\) lie in \([-M,M]\), so \(|\sup_I h| \le M\) and \(|\inf_I h| \le M\).

For a fixed \(i\), the bad subintervals containing \(c_i\) (there are at most two of them, since \(c_i\) can be an endpoint shared by two subintervals) are all contained in \([c_i - \delta, c_i + \delta]\) by the claim, and distinct subintervals of a partition have disjoint interiors; hence their lengths sum to at most \(2\delta\). Since every bad subinterval contains some \(c_i\), the total length of the bad subintervals is at most \(2m\delta\). Therefore

\begin{equation*} |U(h, P_\delta, [a,b])| \le 2m\delta M \quad\text{and}\quad |L(h, P_\delta, [a,b])| \le 2m\delta M . \end{equation*}

Conclusion. Let \(\varepsilon > 0\) and take \(\delta = \dfrac{\varepsilon}{2mM}\) above. Then

\begin{equation*} U(h, [a,b]) \le U(h, P_\delta, [a,b]) \le \varepsilon \quad\text{and}\quad L(h, [a,b]) \ge L(h, P_\delta, [a,b]) \ge -\varepsilon . \end{equation*}

As \(\varepsilon > 0\) was arbitrary, \(U(h,[a,b]) \le 0\) and \(L(h,[a,b]) \ge 0\). Combined with \(L(h,[a,b]) \le U(h,[a,b])\) from 1.8, this gives

\begin{equation*} L(h, [a,b]) = U(h, [a,b]) = 0 . \end{equation*}

Thus \(h\) is Riemann integrable on \([a,b]\) with \(\int_a^b h = 0\), and by the first paragraph \(g\) is Riemann integrable on \([a,b]\) with \(\int_a^b g = \int_a^b f\).

Write \(P_n\) for the partition with points \(x_j^{(n)} = a + \dfrac{j(b-a)}{2^n}\) for \(j = 0, 1, \ldots, 2^n\); each of its \(2^n\) subintervals has length \(\dfrac{b-a}{2^n}\). Since \(f\) is bounded we may set \(M = \sup_{[a,b]} |f| < \infty\), so every value of \(f\) lies in \([-M, M]\).

A refinement estimate. Suppose \(Q\) is a partition of \([a,b]\) all of whose subintervals have length at most \(\ell\), and \(Q^{\prime}\) is a partition whose defining list is obtained from that of \(Q\) by adjoining \(p\) additional points. Then

\begin{equation*} L(f, Q^{\prime}, [a,b]) \le L(f, Q, [a,b]) + 2pM\ell . \end{equation*}

To see this, first take \(p = 1\), say the adjoined point is \(v\). If \(v\) is already a point of \(Q\) the two sums are equal and there is nothing to prove. Otherwise \(v\) lies in the interior of exactly one subinterval \([u,w]\) of \(Q\), and the two lower sums have identical terms except for those coming from \([u,w]\), so

\begin{equation*} \begin{aligned} L(f, Q^{\prime}, [a,b]) - L(f, Q, [a,b]) &= (v-u) \inf_{[u,v]} f + (w-v) \inf_{[v,w]} f - (w-u) \inf_{[u,w]} f \\ &\le (v-u)M + (w-v)M + (w-u)M \\ &= 2M(w-u) \le 2M\ell , \end{aligned} \end{equation*}

using \(|\inf_A f| \le M\) for every nonempty \(A \subseteq [a,b]\). For general \(p\), adjoin the \(p\) points one at a time; every intermediate partition is a refinement of \(Q\), so all of its subintervals still have length at most \(\ell\), and applying the case \(p = 1\) at each of the \(p\) steps and summing gives the estimate.

The lower integral. Since \(x_j^{(n)} = a + \dfrac{2j(b-a)}{2^{n+1}} = x_{2j}^{(n+1)}\), the list defining \(P_n\) is a sublist of the list defining \(P_{n+1}\). Hence 1.5 gives

\begin{equation*} L(f, P_n, [a,b]) \le L(f, P_{n+1}, [a,b]) \quad \text{for all } n \in \mathbb{Z}^{+}, \end{equation*}

and by the definition 1.7 of the lower Riemann integral,

\begin{equation*} L(f, P_n, [a,b]) \le L(f, [a,b]) \quad \text{for all } n \in \mathbb{Z}^{+}. \end{equation*}

So \(\big( L(f, P_n, [a,b]) \big)_{n \in \mathbb{Z}^{+}}\) is an increasing sequence bounded above by \(L(f,[a,b])\); therefore it converges and

\begin{equation*} \lim_{n \to \infty} L(f, P_n, [a,b]) \le L(f, [a,b]) . \end{equation*}

For the reverse inequality, let \(\varepsilon > 0\). Because \(L(f,[a,b])\) is the supremum of the lower Riemann sums, there is a partition \(P\) of \([a,b]\) with

\begin{equation*} L(f, P, [a,b]) > L(f, [a,b]) - \varepsilon . \end{equation*}

Say \(P\) is the list \(a = y_0 < y_1 < \cdots < y_k = b\), so \(P\) has \(k - 1\) points other than \(a\) and \(b\). For each \(n\) let \(Q_n\) be the partition obtained by merging the lists defining \(P_n\) and \(P\). Then the list defining \(Q_n\) is obtained from that of \(P_n\) by adjoining at most \(k-1\) points, and every subinterval of \(P_n\) has length \(\dfrac{b-a}{2^n}\); so the refinement estimate gives

\begin{equation*} L(f, Q_n, [a,b]) \le L(f, P_n, [a,b]) + \frac{2(k-1)M(b-a)}{2^n} . \end{equation*}

On the other hand, the list defining \(P\) is a sublist of the list defining \(Q_n\), so \(L(f, P, [a,b]) \le L(f, Q_n, [a,b])\) by 1.5. Combining the last three displays,

\begin{equation*} L(f, P_n, [a,b]) \ge L(f, P, [a,b]) - \frac{2(k-1)M(b-a)}{2^n} > L(f, [a,b]) - \varepsilon - \frac{2(k-1)M(b-a)}{2^n} . \end{equation*}

Choose \(N \in \mathbb{Z}^{+}\) with \(\dfrac{2(k-1)M(b-a)}{2^N} < \varepsilon\) (possible because \(2^n \to \infty\)). Then for all \(n \ge N\),

\begin{equation*} L(f, [a,b]) - 2\varepsilon < L(f, P_n, [a,b]) \le L(f, [a,b]) . \end{equation*}

Hence \(\big| \lim_{n \to \infty} L(f, P_n, [a,b]) - L(f,[a,b]) \big| \le 2\varepsilon\) for every \(\varepsilon > 0\), and therefore

\begin{equation*} \lim_{n \to \infty} L(f, P_n, [a,b]) = L(f, [a,b]) . \end{equation*}

The upper integral. Apply the result just proved to the bounded function \(-f\), and use the identities established in Steps 2 and 3 of Exercise 5 in this section, whose proofs used only that the function is bounded:

\begin{equation*} L(-f, P, [a,b]) = -\,U(f, P, [a,b]) \ \text{ for every partition } P, \qquad L(-f, [a,b]) = -\,U(f, [a,b]) . \end{equation*}

Therefore

\begin{equation*} U(f, [a,b]) = -\,L(-f, [a,b]) = -\lim_{n \to \infty} L(-f, P_n, [a,b]) = \lim_{n \to \infty} \big( -L(-f, P_n, [a,b]) \big) = \lim_{n \to \infty} U(f, P_n, [a,b]), \end{equation*}

as desired.

For \(n \in \mathbb{Z}^+\) let \(P_n\) denote the equally spaced partition \(x_0, x_1, \dots, x_n\) of \([a,b]\), where

\begin{equation*} x_j = a + \frac{j(b-a)}{n} \qquad \text{for } j = 0, 1, \dots, n, \end{equation*}

so that \(x_j - x_{j-1} = \frac{b-a}{n}\) for each \(j\). Write

\begin{equation*} R_n = \frac{b-a}{n} \sum_{j=1}^{n} f(x_j), \end{equation*}

which is the quantity whose limit we must compute.

Step 1: \(R_n\) is trapped between the lower and upper Riemann sums for \(P_n\).

For each \(j \in \{1,\dots,n\}\) we have \(x_j \in [x_{j-1},x_j]\), and therefore

\begin{equation*} \inf_{[x_{j-1},x_j]} f \;\le\; f(x_j) \;\le\; \sup_{[x_{j-1},x_j]} f . \end{equation*}

Multiplying by \(x_j - x_{j-1} = \frac{b-a}{n} > 0\) and summing over \(j\) gives

\begin{equation*} L(f, P_n, [a,b]) \;\le\; R_n \;\le\; U(f, P_n, [a,b]). \end{equation*}

So it suffices to prove that both \(L(f,P_n,[a,b])\) and \(U(f,P_n,[a,b])\) converge to \(\int_a^b f\) as \(n \to \infty\); the squeeze then finishes the proof.

Step 2: \(U(f,P_n,[a,b]) \to U(f,[a,b])\).

Because \(f\) is Riemann integrable it is by definition bounded, so we may set

\begin{equation*} M = \sup_{[a,b]} |f| < \infty . \end{equation*}

Let \(\varepsilon > 0\). By the definition 1.7 of the upper Riemann integral as an infimum over partitions, there is a partition \(P\) of \([a,b]\) with

\begin{equation*} U(f, P, [a,b]) < U(f,[a,b]) + \varepsilon . \end{equation*}

Say \(P\) is the list \(y_0, y_1, \dots, y_m\); thus \(P\) has at most \(m-1\) points lying in the open interval \((a,b)\).

Fix \(n \in \mathbb{Z}^+\) and let \(P_n^{\prime}\) be the partition obtained by merging the lists defining \(P_n\) and \(P\). Since the list defining \(P_n\) is a sublist of the list defining \(P_n^{\prime}\), and likewise for \(P\), two applications of 1.5 give

\begin{equation*} U(f, P_n^{\prime}, [a,b]) \le U(f, P, [a,b]) \qquad \text{and} \qquad L(f,P,[a,b]) \le L(f,P_n^{\prime},[a,b]). \end{equation*}

We now compare \(U(f,P_n,[a,b])\) with \(U(f,P_n^{\prime},[a,b])\). The partition \(P_n^{\prime}\) is obtained from \(P_n\) by adjoining those points of \(P\) that are not already in \(P_n\); there are at most \(m-1\) such points (only the points \(y_1,\dots,y_{m-1}\) of \((a,b)\) can be new, since \(y_0 = a\) and \(y_m = b\) already belong to \(P_n\)). Call a subinterval \([x_{j-1},x_j]\) of \(P_n\) affected if at least one new point lies in its interior; at most \(m-1\) subintervals are affected.

If \([x_{j-1},x_j]\) is not affected, then \([x_{j-1},x_j]\) is still a subinterval of \(P_n^{\prime}\) and it contributes the identical term \((x_j - x_{j-1}) \sup_{[x_{j-1},x_j]} f\) to both upper sums. If \([x_{j-1},x_j]\) is affected, it is subdivided in \(P_n^{\prime}\) into consecutive subintervals \([t_{i-1},t_i]\), \(i = 1,\dots,r\), with \(t_0 = x_{j-1}\) and \(t_r = x_j\). Its contribution to \(U(f,P_n,[a,b])\) is at most \((x_j - x_{j-1}) M\), while its total contribution to \(U(f,P_n^{\prime},[a,b])\) is

\begin{equation*} \sum_{i=1}^{r} (t_i - t_{i-1}) \sup_{[t_{i-1},t_i]} f \;\ge\; \sum_{i=1}^{r} (t_i - t_{i-1})(-M) \;=\; -(x_j - x_{j-1}) M . \end{equation*}

Hence the affected subinterval \([x_{j-1},x_j]\) contributes at most \(2M(x_j - x_{j-1}) = \frac{2M(b-a)}{n}\) more to \(U(f,P_n,[a,b])\) than to \(U(f,P_n^{\prime},[a,b])\). Summing over the at most \(m-1\) affected subintervals,

\begin{equation*} U(f,P_n,[a,b]) \;\le\; U(f,P_n^{\prime},[a,b]) + \frac{2M(m-1)(b-a)}{n} \;\le\; U(f,P,[a,b]) + \frac{2M(m-1)(b-a)}{n}. \end{equation*}

Combining with the choice of \(P\),

\begin{equation*} U(f,P_n,[a,b]) \;<\; U(f,[a,b]) + \varepsilon + \frac{2M(m-1)(b-a)}{n}. \end{equation*}

Note that \(m\) and \(M\) do not depend on \(n\). Thus there exists \(N \in \mathbb{Z}^+\) such that \(\frac{2M(m-1)(b-a)}{n} < \varepsilon\) for all \(n \ge N\), and then

\begin{equation*} U(f,[a,b]) \;\le\; U(f,P_n,[a,b]) \;<\; U(f,[a,b]) + 2\varepsilon \qquad \text{for all } n \ge N, \end{equation*}

where the first inequality holds because \(U(f,[a,b])\) is the infimum of the upper sums (1.7). As \(\varepsilon > 0\) was arbitrary, this proves

\begin{equation*} \lim_{n \to \infty} U(f,P_n,[a,b]) = U(f,[a,b]). \end{equation*}

Step 3: \(L(f,P_n,[a,b]) \to L(f,[a,b])\).

The argument is symmetric. Choosing \(P\) with \(L(f,P,[a,b]) > L(f,[a,b]) - \varepsilon\) and estimating in the same way (each affected subinterval contributes at most \(\frac{2M(b-a)}{n}\) less to \(L(f,P_n,[a,b])\) than to \(L(f,P_n^{\prime},[a,b])\), because its contribution to \(L(f,P_n,[a,b])\) is at least \(-(x_j - x_{j-1})M\) and its contribution to \(L(f,P_n^{\prime},[a,b])\) is at most \((x_j-x_{j-1})M\)) yields

\begin{equation*} L(f,P_n,[a,b]) \;\ge\; L(f,P_n^{\prime},[a,b]) - \frac{2M(m-1)(b-a)}{n} \;\ge\; L(f,P,[a,b]) - \frac{2M(m-1)(b-a)}{n}, \end{equation*}

and hence \(L(f,[a,b]) - 2\varepsilon < L(f,P_n,[a,b]) \le L(f,[a,b])\) for all large \(n\). Thus \(L(f,P_n,[a,b]) \to L(f,[a,b])\).

Step 4: Conclusion.

Because \(f\) is Riemann integrable, 1.9 gives

\begin{equation*} L(f,[a,b]) = U(f,[a,b]) = \int_a^b f . \end{equation*}

By Steps 2 and 3 both \(L(f,P_n,[a,b])\) and \(U(f,P_n,[a,b])\) converge to \(\int_a^b f\). Since Step 1 gives

\begin{equation*} L(f,P_n,[a,b]) \le R_n \le U(f,P_n,[a,b]), \end{equation*}

the squeeze theorem yields \(R_n \to \int_a^b f\), which is exactly the assertion

\begin{equation*} \int_a^b f = \lim_{n \to \infty} \frac{b-a}{n} \sum_{j=1}^{n} f\!\left(a + \tfrac{j(b-a)}{n}\right). \qquad \blacksquare \end{equation*}

Throughout, we use the following criterion, which is Exercise 3 in Section 1A. Because it is used repeatedly below and in later exercises, we prove it here.

Riemann criterion. A bounded function \(g \colon [\alpha,\beta] \to \mathbb{R}\) is Riemann integrable if and only if for every \(\varepsilon > 0\) there is a partition \(P\) of \([\alpha,\beta]\) with

\begin{equation*} U(g,P,[\alpha,\beta]) - L(g,P,[\alpha,\beta]) < \varepsilon . \end{equation*}

Proof of the criterion. Suppose first that such partitions exist. Fix \(\varepsilon > 0\) and choose \(P\) as above. By 1.7 we have \(U(g,[\alpha,\beta]) \le U(g,P,[\alpha,\beta])\) and \(L(g,[\alpha,\beta]) \ge L(g,P,[\alpha,\beta])\), so

\begin{equation*} 0 \le U(g,[\alpha,\beta]) - L(g,[\alpha,\beta]) \le U(g,P,[\alpha,\beta]) - L(g,P,[\alpha,\beta]) < \varepsilon, \end{equation*}

the first inequality coming from 1.8. Since \(\varepsilon > 0\) is arbitrary, \(L(g,[\alpha,\beta]) = U(g,[\alpha,\beta])\), so \(g\) is Riemann integrable by 1.9.

Conversely, suppose \(g\) is Riemann integrable and let \(\varepsilon > 0\). By 1.7 there are partitions \(P_1, P_2\) of \([\alpha,\beta]\) with

\begin{equation*} U(g,P_1,[\alpha,\beta]) < U(g,[\alpha,\beta]) + \tfrac{\varepsilon}{2}, \qquad L(g,P_2,[\alpha,\beta]) > L(g,[\alpha,\beta]) - \tfrac{\varepsilon}{2}. \end{equation*}

Let \(P\) be the partition obtained by merging the lists defining \(P_1\) and \(P_2\). By 1.5,

\begin{equation*} U(g,P,[\alpha,\beta]) \le U(g,P_1,[\alpha,\beta]) \quad \text{and} \quad L(g,P,[\alpha,\beta]) \ge L(g,P_2,[\alpha,\beta]). \end{equation*}

Since \(U(g,[\alpha,\beta]) = L(g,[\alpha,\beta])\), subtracting gives

\begin{equation*} U(g,P,[\alpha,\beta]) - L(g,P,[\alpha,\beta]) < \big(U(g,[\alpha,\beta]) + \tfrac{\varepsilon}{2}\big) - \big(L(g,[\alpha,\beta]) - \tfrac{\varepsilon}{2}\big) = \varepsilon, \end{equation*}

which completes the proof of the criterion.

Now we prove the exercise. Since \(f\) is bounded on \([a,b]\), its restriction to \([c,d] \subseteq [a,b]\) is bounded, so the lower and upper Riemann sums and integrals of \(f\) on \([c,d]\) are defined.

Let \(\varepsilon > 0\). Since \(f\) is Riemann integrable on \([a,b]\), the criterion supplies a partition \(P\) of \([a,b]\) with

\begin{equation*} U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon . \end{equation*}

Let \(P^{\prime}\) be the partition of \([a,b]\) obtained by adjoining the points \(c\) and \(d\) to the list defining \(P\) (they may already be present). The list defining \(P\) is a sublist of the list defining \(P^{\prime}\), so 1.5 gives

\begin{equation*} U(f,P^{\prime},[a,b]) \le U(f,P,[a,b]) \quad \text{and} \quad L(f,P^{\prime},[a,b]) \ge L(f,P,[a,b]), \end{equation*}

and therefore

\begin{equation*} U(f,P^{\prime},[a,b]) - L(f,P^{\prime},[a,b]) \le U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon . \end{equation*}

Write \(P^{\prime}\) as the list \(x_0, x_1, \dots, x_n\) with \(a = x_0 < x_1 < \cdots < x_n = b\). Since \(c\) and \(d\) appear in this list and \(c < d\), there are indices \(p < q\) with \(x_p = c\) and \(x_q = d\). Then

\begin{equation*} Q : \quad x_p, x_{p+1}, \dots, x_q \end{equation*}

is a partition of \([c,d]\).

For each \(j \in \{p+1, \dots, q\}\) the subinterval \([x_{j-1},x_j]\) is contained in \([c,d]\), so the supremum and infimum of \(f\) over \([x_{j-1},x_j]\) are unchanged when \(f\) is restricted to \([c,d]\). Hence, writing \(f\) also for its restriction to \([c,d]\),

\begin{equation*} \begin{aligned} U(f,Q,[c,d]) - L(f,Q,[c,d]) &= \sum_{j=p+1}^{q} (x_j - x_{j-1})\Big( \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \Big) \\ &\le \sum_{j=1}^{n} (x_j - x_{j-1})\Big( \sup_{[x_{j-1},x_j]} f - \inf_{[x_{j-1},x_j]} f \Big) \\ &= U(f,P^{\prime},[a,b]) - L(f,P^{\prime},[a,b]) \\ &< \varepsilon , \end{aligned} \end{equation*}

where the inequality in the second line holds because every omitted term is nonnegative (each factor \(x_j - x_{j-1}\) is positive and each supremum is at least the corresponding infimum).

Thus for every \(\varepsilon > 0\) there is a partition \(Q\) of \([c,d]\) with \(U(f,Q,[c,d]) - L(f,Q,[c,d]) < \varepsilon\). By the criterion, \(f\) is Riemann integrable on \([c,d]\). \(\blacksquare\)

Since \(f\) is bounded on \([a,b]\), its restrictions to \([a,c]\) and to \([c,b]\) are bounded, so all four lower and upper Riemann integrals appearing below are defined. We write \(f\) for each of these restrictions as well.

Splitting a partition. Suppose \(P\) is a partition \(x_0, x_1, \dots, x_n\) of \([a,b]\) whose list contains \(c\), say \(c = x_p\) with \(0 < p < n\). Then

\begin{equation*} P_1 : \; x_0, \dots, x_p \quad \text{and} \quad P_2 : \; x_p, \dots, x_n \end{equation*}

are partitions of \([a,c]\) and \([c,b]\) respectively. Each subinterval \([x_{j-1},x_j]\) with \(j \le p\) is contained in \([a,c]\) and each with \(j > p\) is contained in \([c,b]\), so the relevant infima and suprema are unaffected by restricting the domain of \(f\). Splitting the defining sums in 1.3 at \(j = p\) therefore gives

\begin{equation*} L(f,P,[a,b]) = L(f,P_1,[a,c]) + L(f,P_2,[c,b]), \qquad U(f,P,[a,b]) = U(f,P_1,[a,c]) + U(f,P_2,[c,b]). \qquad (\ast) \end{equation*}

Conversely, if \(P_1\) is any partition of \([a,c]\) and \(P_2\) is any partition of \([c,b]\), then concatenating their lists (identifying the shared endpoint \(c\)) produces a partition \(P\) of \([a,b]\) containing \(c\), and \((\ast)\) again holds.

Step 1: additivity of the lower Riemann integral, namely

\begin{equation*} L(f,[a,b]) = L(f,[a,c]) + L(f,[c,b]). \end{equation*}

For the inequality \(\ge\): let \(P_1\) be a partition of \([a,c]\) and \(P_2\) a partition of \([c,b]\), and let \(P\) be their concatenation. By \((\ast)\) and 1.7,

\begin{equation*} L(f,P_1,[a,c]) + L(f,P_2,[c,b]) = L(f,P,[a,b]) \le L(f,[a,b]). \end{equation*}

Taking the supremum over \(P_1\) and then over \(P_2\) (the two are chosen independently) gives \(L(f,[a,c]) + L(f,[c,b]) \le L(f,[a,b])\).

For the inequality \(\le\): let \(P\) be any partition of \([a,b]\) and let \(P^{\prime}\) be obtained by adjoining \(c\) to the list defining \(P\). By 1.5 and then \((\ast)\) and 1.7,

\begin{equation*} L(f,P,[a,b]) \le L(f,P^{\prime},[a,b]) = L(f,P^{\prime}_1,[a,c]) + L(f,P^{\prime}_2,[c,b]) \le L(f,[a,c]) + L(f,[c,b]). \end{equation*}

Taking the supremum over all partitions \(P\) of \([a,b]\) gives \(L(f,[a,b]) \le L(f,[a,c]) + L(f,[c,b])\). This proves Step 1.

Step 2: additivity of the upper Riemann integral, namely

\begin{equation*} U(f,[a,b]) = U(f,[a,c]) + U(f,[c,b]). \end{equation*}

The proof is the mirror image of Step 1. For \(\le\): given partitions \(P_1\) of \([a,c]\) and \(P_2\) of \([c,b]\) with concatenation \(P\), we have \(U(f,[a,b]) \le U(f,P,[a,b]) = U(f,P_1,[a,c]) + U(f,P_2,[c,b])\); take the infimum over \(P_1\) and over \(P_2\). For \(\ge\): given a partition \(P\) of \([a,b]\), adjoin \(c\) to get \(P^{\prime}\) and use 1.5 to obtain

\begin{equation*} U(f,P,[a,b]) \ge U(f,P^{\prime},[a,b]) = U(f,P^{\prime}_1,[a,c]) + U(f,P^{\prime}_2,[c,b]) \ge U(f,[a,c]) + U(f,[c,b]); \end{equation*}

take the infimum over \(P\).

Step 3: the equivalence. Subtracting the identity of Step 1 from that of Step 2 gives

\begin{equation*} U(f,[a,b]) - L(f,[a,b]) = \Big( U(f,[a,c]) - L(f,[a,c]) \Big) + \Big( U(f,[c,b]) - L(f,[c,b]) \Big). \end{equation*}

By 1.8 each of the three differences above is nonnegative. Hence the left side equals \(0\) if and only if both terms on the right equal \(0\). By 1.9, the left side equals \(0\) exactly when \(f\) is Riemann integrable on \([a,b]\), the first term on the right equals \(0\) exactly when \(f\) is Riemann integrable on \([a,c]\), and the second exactly when \(f\) is Riemann integrable on \([c,b]\). This is precisely the asserted equivalence.

Step 4: the formula. Suppose the conditions hold. Then by 1.9 and Step 1,

\begin{equation*} \int_a^b f = L(f,[a,b]) = L(f,[a,c]) + L(f,[c,b]) = \int_a^c f + \int_c^b f . \qquad \blacksquare \end{equation*}

First note that \(F\) is well defined: if \(t \in (a,b]\) then \(a \le a < t \le b\), so by Exercise 9 in Section 1A the function \(f\) is Riemann integrable on \([a,t]\) and hence \(\int_a^t f\) makes sense.

Because \(f\) is Riemann integrable it is bounded, so we may set

\begin{equation*} M = \sup_{[a,b]} |f| < \infty . \end{equation*}

We prove the following Lipschitz estimate, which immediately gives continuity.

Claim. If \(s, t \in [a,b]\) with \(s < t\), then \(|F(t) - F(s)| \le M(t-s)\).

Proof of the claim. We first show that

\begin{equation*} F(t) - F(s) = \int_s^t f . \qquad (\dagger) \end{equation*}

If \(s = a\), then \(F(s) = 0\) and \(F(t) = \int_a^t f = \int_s^t f\), so \((\dagger)\) holds. If \(s > a\), then \(a < s < t\), so \(s \in (a,t)\). By Exercise 9 in Section 1A the function \(f\) is Riemann integrable on \([a,t]\) (and in particular bounded there), and by Exercise 10 in Section 1A applied on the interval \([a,t]\) with interior point \(s\),

\begin{equation*} \int_a^t f = \int_a^s f + \int_s^t f , \end{equation*}

which is exactly \(F(t) = F(s) + \int_s^t f\), proving \((\dagger)\).

Now \(f\) is Riemann integrable on \([s,t]\) by Exercise 9 in Section 1A, so 1.13 applies on \([s,t]\) and gives

\begin{equation*} (t-s) \inf_{[s,t]} f \;\le\; \int_s^t f \;\le\; (t-s) \sup_{[s,t]} f . \end{equation*}

Since \([s,t] \subseteq [a,b]\) we have \(-M \le \inf_{[s,t]} f\) and \(\sup_{[s,t]} f \le M\), hence

\begin{equation*} -M(t-s) \le \int_s^t f \le M(t-s), \qquad \text{that is,} \qquad \left| \int_s^t f \right| \le M(t-s). \end{equation*}

Combining with \((\dagger)\) proves the claim.

Note that the claim is symmetric in \(s\) and \(t\): it says that \(|F(t) - F(s)| \le M|t-s|\) for all \(s, t \in [a,b]\), since for \(s = t\) both sides are \(0\) and for \(s > t\) we may interchange the names of \(s\) and \(t\).

Continuity now follows. Let \(\varepsilon > 0\) and put \(\delta = \dfrac{\varepsilon}{M+1} > 0\). If \(s, t \in [a,b]\) satisfy \(|s-t| < \delta\), then

\begin{equation*} |F(t) - F(s)| \le M|t-s| \le M\delta = \frac{M}{M+1}\,\varepsilon < \varepsilon, \end{equation*}

where the final inequality holds because \(0 \le M < M+1\). (The nonstrict inequalities matter here: if \(M = 0\) the first two steps are equalities.) Thus \(F\) is uniformly continuous on \([a,b]\), and in particular continuous at every point of \([a,b]\). \(\blacksquare\)

Step 1: an oscillation inequality. Let \(A\) be a nonempty subset of \([a,b]\). We claim

\begin{equation*} \sup_A |f| - \inf_A |f| \;\le\; \sup_A f - \inf_A f . \end{equation*}

Indeed, for any real-valued bounded function \(g\) on \(A\) we have

\begin{equation*} \sup_A g - \inf_A g = \sup\{ g(x) - g(y) : x, y \in A \}, \end{equation*}

because \(\sup_A g - \inf_A g = \sup_{x \in A} g(x) + \sup_{y \in A}(-g(y))\) and the supremum of a sum over independent variables is the sum of the suprema. Applying this with \(g = |f|\) and with \(g = f\), the claim reduces to showing that

\begin{equation*} |f|(x) - |f|(y) \le \sup\{ f(u) - f(v) : u,v \in A \} \qquad \text{for all } x,y \in A, \end{equation*}

which holds because the reverse triangle inequality gives

\begin{equation*} |f(x)| - |f(y)| \le \big| |f(x)| - |f(y)| \big| \le |f(x) - f(y)| = \max\{ f(x)-f(y), \, f(y)-f(x) \}, \end{equation*}

and both \(f(x)-f(y)\) and \(f(y)-f(x)\) are of the form \(f(u)-f(v)\) with \(u,v \in A\).

Step 2: \(|f|\) is Riemann integrable. Since \(f\) is bounded, so is \(|f|\), so the lower and upper Riemann sums of \(|f|\) are defined. Let \(P\) be an arbitrary partition \(x_0, \dots, x_n\) of \([a,b]\). Applying Step 1 with \(A = [x_{j-1},x_j]\) for each \(j\) (each such \(A\) is nonempty) and multiplying by \(x_j - x_{j-1} > 0\), then summing over \(j\), gives by 1.3

\begin{equation*} U(|f|,P,[a,b]) - L(|f|,P,[a,b]) \;\le\; U(f,P,[a,b]) - L(f,P,[a,b]). \end{equation*}

Let \(\varepsilon > 0\). By the Riemann criterion (Exercise 3 in Section 1A, proved in the solution to Exercise 9), there is a partition \(P\) of \([a,b]\) with \(U(f,P,[a,b]) - L(f,P,[a,b]) < \varepsilon\); the displayed inequality then gives \(U(|f|,P,[a,b]) - L(|f|,P,[a,b]) < \varepsilon\). By the same criterion, \(|f|\) is Riemann integrable.

Step 3: monotonicity of the Riemann integral. Suppose \(g, h \colon [a,b] \to \mathbb{R}\) are Riemann integrable with \(g(x) \le h(x)\) for all \(x \in [a,b]\). Then for every subinterval \([x_{j-1},x_j]\) we have \(\inf_{[x_{j-1},x_j]} g \le \inf_{[x_{j-1},x_j]} h\), so \(L(g,P,[a,b]) \le L(h,P,[a,b])\) for every partition \(P\) of \([a,b]\). Taking the supremum over \(P\) and using 1.7 and 1.9,

\begin{equation*} \int_a^b g = L(g,[a,b]) \le L(h,[a,b]) = \int_a^b h . \end{equation*}

Step 4: the inequality. By Step 2 the function \(|f|\) is Riemann integrable, and by Exercise 5 in Section 1A the function \(-|f|\) is Riemann integrable with

\begin{equation*} \int_a^b (-|f|) = - \int_a^b |f| . \end{equation*}

For every \(x \in [a,b]\) we have \(-|f(x)| \le f(x) \le |f(x)|\). Applying Step 3 to the pair \((f, |f|)\) and to the pair \((-|f|, f)\) gives

\begin{equation*}

• \int_a^b |f| = \int_a^b (-|f|) \;\le\; \int_a^b f \;\le\; \int_a^b |f| . \end{equation*}

A real number lying between \(-r\) and \(r\) has absolute value at most \(r\), so

\begin{equation*} \left| \int_a^b f \right| \le \int_a^b |f| . \qquad \blacksquare \end{equation*}

Step 1: \(f\) is bounded. For every \(x \in [a,b]\) we have \(a \le x \le b\), so monotonicity gives \(f(a) \le f(x) \le f(b)\). Hence \(f\) is bounded, and the lower and upper Riemann sums and integrals of \(f\) are defined.

Step 2: computing the sums for an equally spaced partition. Fix \(n \in \mathbb{Z}^+\) and let \(P_n\) be the partition \(x_0, x_1, \dots, x_n\) of \([a,b]\) given by

\begin{equation*} x_j = a + \frac{j(b-a)}{n}, \qquad j = 0, 1, \dots, n, \end{equation*}

so \(x_j - x_{j-1} = \frac{b-a}{n}\) for each \(j\).

Fix \(j \in \{1,\dots,n\}\). If \(x \in [x_{j-1},x_j]\), then monotonicity gives \(f(x_{j-1}) \le f(x) \le f(x_j)\) (this holds also when \(x\) equals one of the endpoints). Since \(x_{j-1}\) and \(x_j\) themselves belong to \([x_{j-1},x_j]\), these bounds are attained, so

\begin{equation*} \inf_{[x_{j-1},x_j]} f = f(x_{j-1}) \qquad \text{and} \qquad \sup_{[x_{j-1},x_j]} f = f(x_j). \end{equation*}

Therefore, by 1.3,

\begin{equation*} U(f,P_n,[a,b]) - L(f,P_n,[a,b]) = \frac{b-a}{n} \sum_{j=1}^{n} \big( f(x_j) - f(x_{j-1}) \big). \end{equation*}

The sum telescopes: \(\sum_{j=1}^{n}\big(f(x_j) - f(x_{j-1})\big) = f(x_n) - f(x_0) = f(b) - f(a)\). Hence

\begin{equation*} U(f,P_n,[a,b]) - L(f,P_n,[a,b]) = \frac{(b-a)\big(f(b) - f(a)\big)}{n}. \end{equation*}

Step 3: conclusion. Let \(\varepsilon > 0\). The number \((b-a)(f(b)-f(a))\) is a fixed nonnegative real number (nonnegative since \(a < b\) implies \(f(a) \le f(b)\)), so by the Archimedean property of \(\mathbb{R}\) there exists \(n \in \mathbb{Z}^+\) with

\begin{equation*} \frac{(b-a)\big(f(b)-f(a)\big)}{n} < \varepsilon . \end{equation*}

For this \(n\), Step 2 gives \(U(f,P_n,[a,b]) - L(f,P_n,[a,b]) < \varepsilon\).

Thus for every \(\varepsilon > 0\) there is a partition of \([a,b]\) whose upper and lower Riemann sums differ by less than \(\varepsilon\). By the Riemann criterion (Exercise 3 in Section 1A, proved in the solution to Exercise 9), \(f\) is Riemann integrable on \([a,b]\). \(\blacksquare\)

For \(n \in \mathbb{Z}^+\) set

\begin{equation*} \varepsilon_n = \sup_{x \in [a,b]} |f_n(x) - f(x)| \in [0,\infty]. \end{equation*}

Uniform convergence says exactly that for each \(\eta > 0\) there is \(N\) with \(|f_n(x) - f(x)| < \eta\) for all \(n \ge N\) and all \(x \in [a,b]\); hence \(\varepsilon_n \le \eta\) for all \(n \ge N\). In particular \(\varepsilon_n\) is finite for all large \(n\) and \(\varepsilon_n \to 0\).

Step 1: \(f\) is bounded. Choose \(N\) with \(\varepsilon_N \le 1\). Since \(f_N\) is Riemann integrable it is bounded, say \(|f_N| \le M\) on \([a,b]\). Then for every \(x \in [a,b]\),

\begin{equation*} |f(x)| \le |f(x) - f_N(x)| + |f_N(x)| \le 1 + M . \end{equation*}

So \(f\) is bounded and its lower and upper Riemann sums and integrals are defined. Note also that each \(f_n\) is bounded (being Riemann integrable), so from now on \(\varepsilon_n < \infty\) for every \(n \in \mathbb{Z}^+\).

Step 2: an oscillation comparison. Let \(A \subseteq [a,b]\) be nonempty and let \(n\) be such that \(\varepsilon_n < \infty\). For every \(x \in A\) we have \(f(x) \le f_n(x) + \varepsilon_n\) and \(f(x) \ge f_n(x) - \varepsilon_n\), hence

\begin{equation*} \sup_A f \le \sup_A f_n + \varepsilon_n \qquad \text{and} \qquad \inf_A f \ge \inf_A f_n - \varepsilon_n . \end{equation*}

Subtracting,

\begin{equation*} \sup_A f - \inf_A f \;\le\; \Big( \sup_A f_n - \inf_A f_n \Big) + 2\varepsilon_n . \end{equation*}

Applying this with \(A = [x_{j-1},x_j]\) for each subinterval of a partition \(P\) of \([a,b]\), multiplying by \(x_j - x_{j-1} > 0\) and summing (the lengths sum to \(b-a\)), we obtain by 1.3

\begin{equation*} U(f,P,[a,b]) - L(f,P,[a,b]) \;\le\; \Big( U(f_n,P,[a,b]) - L(f_n,P,[a,b]) \Big) + 2(b-a)\varepsilon_n . \qquad (\star) \end{equation*}

Step 3: \(f\) is Riemann integrable. Let \(\eta > 0\). Since \(\varepsilon_n \to 0\), choose \(n\) with

\begin{equation*} \varepsilon_n < \frac{\eta}{4(b-a)} . \end{equation*}

Since \(f_n\) is Riemann integrable, the Riemann criterion (Exercise 3 in Section 1A, proved in the solution to Exercise 9) provides a partition \(P\) of \([a,b]\) with

\begin{equation*} U(f_n,P,[a,b]) - L(f_n,P,[a,b]) < \frac{\eta}{2} . \end{equation*}

Then \((\star)\) gives

\begin{equation*} U(f,P,[a,b]) - L(f,P,[a,b]) < \frac{\eta}{2} + 2(b-a) \cdot \frac{\eta}{4(b-a)} = \eta . \end{equation*}

As \(\eta > 0\) was arbitrary, the same criterion shows that \(f\) is Riemann integrable on \([a,b]\).

Step 4: convergence of the integrals. Fix \(n\) with \(\varepsilon_n < \infty\). By Exercise 5 in Section 1A the function \(-f_n\) is Riemann integrable with \(\int_a^b (-f_n) = -\int_a^b f_n\), and by Step 3 the function \(f\) is Riemann integrable; so by Exercise 4 in Section 1A the function \(f - f_n\) is Riemann integrable with

\begin{equation*} \int_a^b (f - f_n) = \int_a^b f - \int_a^b f_n . \end{equation*}

By Exercise 12 in Section 1A the function \(|f - f_n|\) is Riemann integrable and

\begin{equation*} \left| \int_a^b f - \int_a^b f_n \right| = \left| \int_a^b (f - f_n) \right| \le \int_a^b |f - f_n| . \end{equation*}

Finally 1.13, applied to the Riemann integrable function \(|f-f_n|\), gives

\begin{equation*} \int_a^b |f - f_n| \le (b-a) \sup_{[a,b]} |f - f_n| = (b-a)\varepsilon_n . \end{equation*}

Combining the last two displays,

\begin{equation*} \left| \int_a^b f - \int_a^b f_n \right| \le (b-a) \varepsilon_n \qquad \text{whenever } \varepsilon_n < \infty . \end{equation*}

Now let \(\eta > 0\). Since \(\varepsilon_n \to 0\), there is \(N \in \mathbb{Z}^+\) such that \(\varepsilon_n < \dfrac{\eta}{b-a}\) for all \(n \ge N\) (in particular \(\varepsilon_n < \infty\) for such \(n\)), and then

\begin{equation*} \left| \int_a^b f - \int_a^b f_n \right| \le (b-a)\varepsilon_n < \eta \qquad \text{for all } n \ge N . \end{equation*}

Hence

\begin{equation*} \int_a^b f = \lim_{n \to \infty} \int_a^b f_n . \qquad \blacksquare \end{equation*}

We show that \(f\) is Riemann integrable with \(\int_0^1 f = 0\).

Step 0: \(f\) is bounded, with \(0 \le f \le 1\).

If \(a\) is irrational then \(f(a) = 0\). If \(a\) is rational, then the smallest positive integer \(n\) with \(a = m/n\) for some integer \(m\) satisfies \(n \ge 1\), so \(f(a) = 1/n \in (0,1]\). Hence \(0 \le f(a) \le 1\) for all \(a \in [0,1]\), so \(f\) is a bounded function on \([0,1]\) and the lower and upper Riemann integrals of 1.7 are defined.

Step 1: \(L(f,[0,1]) = 0\).

Let \(P\) be the partition \(0 = x_0 < x_1 < \cdots < x_n = 1\) of \([0,1]\). Each subinterval \([x_{j-1}, x_j]\) has positive length and therefore contains an irrational number, at which \(f\) takes the value \(0\). Since \(f \ge 0\), this gives

\begin{equation*} \inf_{[x_{j-1},x_j]} f = 0 \qquad \text{for each } j = 1, \dots, n . \end{equation*}

Hence \(L(f, P, [0,1]) = \sum_{j=1}^{n} (x_j - x_{j-1}) \cdot 0 = 0\). As \(P\) was an arbitrary partition, taking the supremum over all \(P\) in 1.7 gives \(L(f,[0,1]) = 0\).

Step 2: the level sets of \(f\) are finite.

For \(N \in \mathbb{Z}^+\) put

\begin{equation*} A_N = \Bigl\{ a \in [0,1] : f(a) \ge \tfrac{1}{N} \Bigr\} . \end{equation*}

Suppose \(a \in A_N\). Then \(f(a) > 0\), so \(a\) is rational; writing \(n\) for the smallest positive integer with \(a = m/n\) for some integer \(m\), we have \(1/n = f(a) \ge 1/N\), hence \(n \le N\). Since \(a \in [0,1]\) and \(a = m/n\) with \(n > 0\), we get \(0 \le m \le n\). Therefore

\begin{equation*} A_N \subseteq \Bigl\{ \tfrac{m}{n} : n \in \{1, \dots, N\},\ m \in \{0, 1, \dots, n\} \Bigr\}, \end{equation*}

which is a finite set. Let \(K_N\) denote the number of elements of \(A_N\); note \(K_N \ge 1\) because \(0 = \frac{0}{1}\) lies in \(A_N\).

Moreover, if \(x \in [0,1] \setminus A_N\) then \(f(x) < 1/N\); since every value of \(f\) lies in \(\{0\} \cup \{1/n : n \in \mathbb{Z}^+\}\), this forces \(f(x) \le \frac{1}{N+1}\).

Step 3: \(U(f,[0,1]) = 0\).

Let \(\varepsilon > 0\). Choose \(N \in \mathbb{Z}^+\) with \(\frac{1}{N} < \frac{\varepsilon}{2}\), and set \(K = K_N\) as in Step 2. Now choose \(n \in \mathbb{Z}^+\) with

\begin{equation*} n > \frac{4K}{\varepsilon}, \end{equation*}

and let \(P_n\) be the equally spaced partition \(0, \frac{1}{n}, \frac{2}{n}, \dots, 1\) of \([0,1]\), whose subintervals \(I_j = \bigl[\frac{j-1}{n}, \frac{j}{n}\bigr]\) each have length \(\frac{1}{n}\).

Split the index set \(\{1, \dots, n\}\) into

\begin{equation*} J = \{ j : I_j \cap A_N \ne \emptyset \}, \qquad J^{\prime} = \{1,\dots,n\} \setminus J . \end{equation*}

Each point of \([0,1]\) lies in at most two of the intervals \(I_1, \dots, I_n\) (in two only when it is one of the interior partition points), so each of the \(K\) points of \(A_N\) contributes at most two indices to \(J\). Hence \(J\) has at most \(2K\) elements.

For \(j \in J\) we use only the crude bound \(\sup_{I_j} f \le 1\) from Step 0. For \(j \in J^{\prime}\) the interval \(I_j\) misses \(A_N\), so by the last sentence of Step 2 every \(x \in I_j\) satisfies \(f(x) \le \frac{1}{N+1} < \frac{1}{N}\), whence \(\sup_{I_j} f \le \frac{1}{N}\). Therefore

\begin{equation*} \begin{aligned} U(f, P_n, [0,1]) &= \sum_{j \in J} \frac{1}{n} \sup_{I_j} f \; + \; \sum_{j \in J^{\prime}} \frac{1}{n} \sup_{I_j} f \\ &\le \frac{2K}{n} \cdot 1 \; + \; \Bigl( \sum_{j \in J^{\prime}} \frac{1}{n} \Bigr) \frac{1}{N} \\ &\le \frac{2K}{n} + \frac{1}{N} \\ &< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon , \end{aligned} \end{equation*}

where we used \(\sum_{j \in J^{\prime}} \frac{1}{n} \le 1\) together with the choices \(n > 4K/\varepsilon\) and \(1/N < \varepsilon/2\).

By the definition 1.7 of the upper Riemann integral as an infimum over partitions, \(U(f,[0,1]) \le U(f,P_n,[0,1]) < \varepsilon\). Since \(\varepsilon > 0\) was arbitrary, \(U(f,[0,1]) \le 0\). On the other hand \(L(f,[0,1]) \le U(f,[0,1])\) by 1.8, and \(L(f,[0,1]) = 0\) by Step 1, so \(U(f,[0,1]) \ge 0\).

Hence \(U(f,[0,1]) = 0 = L(f,[0,1])\), so by 1.9 the function \(f\) is Riemann integrable and

\begin{equation*} \int_0^1 f = 0 . \end{equation*}

It is worth contrasting this with 1.14. The function here is discontinuous at every rational number of \([0,1]\), a dense infinite set, yet it is still Riemann integrable, because at each threshold \(1/N\) the set of points where \(f\) is at least that large is only finite.

The whole exercise rests on one identity, which says that passing to \(-f\) interchanges the roles of the lower and upper Riemann integrals:

\begin{equation*} U(f,[a,b]) = -L(-f,[a,b]) . \tag{\ast} \end{equation*}

First a remark about suprema and infima. If \(S\) is a nonempty set of real numbers that is bounded, then

\begin{equation*} \sup S = -\inf(-S), \qquad \text{where } -S = \{-s : s \in S\} . \end{equation*}

Indeed, \(\alpha\) is an upper bound for \(S\) exactly when \(-\alpha\) is a lower bound for \(-S\); hence the least upper bound of \(S\) is the negative of the greatest lower bound of \(-S\). Applying this with \(S\) the set of values of \(f\) on a nonempty subset \(E \subseteq [a,b]\), and noting that the set of values of \(-f\) on \(E\) is exactly \(-S\), we obtain

\begin{equation*} \sup_{E} f = -\inf_{E} (-f) . \tag{1} \end{equation*}

Note that \(-f\) is bounded because \(f\) is, so all quantities below are real numbers.

Now let \(P\) be any partition \(a = x_0 < x_1 < \cdots < x_n = b\) of \([a,b]\). Using the definitions of the upper and lower Riemann sums together with (1) applied on each \([x_{j-1},x_j]\),

\begin{equation*} \begin{aligned} U(f, P, [a,b]) &= \sum_{j=1}^{n} (x_j - x_{j-1}) \sup_{[x_{j-1},x_j]} f \\ &= \sum_{j=1}^{n} (x_j - x_{j-1}) \Bigl( -\inf_{[x_{j-1},x_j]} (-f) \Bigr) \\ &= -\sum_{j=1}^{n} (x_j - x_{j-1}) \inf_{[x_{j-1},x_j]} (-f) \\ &= -L(-f, P, [a,b]) . \end{aligned} \tag{2} \end{equation*}

Taking the infimum over all partitions \(P\) and using 1.7 twice, together with the remark above (in the form \(\inf T = -\sup(-T)\) applied to the set \(T\) of all numbers \(U(f,P,[a,b])\)),

\begin{equation*} U(f,[a,b]) = \inf_{P} U(f,P,[a,b]) = \inf_{P} \bigl( -L(-f,P,[a,b]) \bigr) = -\sup_{P} L(-f,P,[a,b]) = -L(-f,[a,b]), \end{equation*}

which is exactly \((\ast)\). Here the set \(T\) is nonempty and bounded below (by 1.8 applied to \(f\), or simply because \(U(f,P,[a,b]) \ge (b-a)\inf_{[a,b]} f\) for every \(P\)), so the manipulation is legitimate.

With \((\ast)\) in hand the equivalence is immediate. By 1.9, \(f\) is Riemann integrable if and only if

\begin{equation*} L(f,[a,b]) = U(f,[a,b]) . \end{equation*}

Substituting \((\ast)\) into the right side, this holds if and only if

\begin{equation*} L(f,[a,b]) = -L(-f,[a,b]), \end{equation*}

and multiplying both sides by \(-1\) (a reversible operation) this holds if and only if

\begin{equation*} L(-f,[a,b]) = -L(f,[a,b]) , \end{equation*}

as desired.

Two remarks. First, the exercise is genuinely a statement about \(f\) alone: it says that Riemann integrability can be tested using only the lower Riemann integral, applied to \(f\) and to \(-f\).

Second, \((\ast)\) also recovers Exercise 5 in Section 1A. Suppose \(f\) is Riemann integrable, so \(L(f,[a,b]) = U(f,[a,b]) = \int_a^b f\). By the equivalence just proved, \(L(-f,[a,b]) = -\int_a^b f\). Applying \((\ast)\) with \(-f\) in place of \(f\) (and using \(-(-f) = f\)) gives \(U(-f,[a,b]) = -L(f,[a,b]) = -\int_a^b f\) as well. Hence \(L(-f,[a,b]) = U(-f,[a,b])\), so \(-f\) is Riemann integrable and \(\int_a^b (-f) = -\int_a^b f\).

Since \(f\) and \(g\) are bounded, so is \(f+g\), and all six quantities in the statement are real numbers.

Step 1: pointwise inequalities for infima and suprema.

Let \(E \subseteq [a,b]\) be nonempty. For every \(x \in E\) we have \(\inf_E f \le f(x)\) and \(\inf_E g \le g(x)\), so

\begin{equation*} \inf_E f + \inf_E g \le f(x) + g(x) = (f+g)(x) . \end{equation*}

Thus \(\inf_E f + \inf_E g\) is a lower bound for the set of values of \(f+g\) on \(E\), and since \(\inf_E (f+g)\) is the greatest such lower bound,

\begin{equation*} \inf_E f + \inf_E g \le \inf_E (f+g) . \tag{1} \end{equation*}

Symmetrically, for every \(x \in E\) we have \((f+g)(x) = f(x) + g(x) \le \sup_E f + \sup_E g\), so \(\sup_E f + \sup_E g\) is an upper bound for the values of \(f+g\) on \(E\), and therefore

\begin{equation*} \sup_E (f+g) \le \sup_E f + \sup_E g . \tag{2} \end{equation*}

Step 2: the inequalities for a single partition.

Let \(P\) be the partition \(a = x_0 < x_1 < \cdots < x_n = b\) of \([a,b]\). Applying (1) with \(E = [x_{j-1},x_j]\) and multiplying by the positive number \(x_j - x_{j-1}\), then summing over \(j\),

\begin{equation*} L(f,P,[a,b]) + L(g,P,[a,b]) = \sum_{j=1}^{n} (x_j - x_{j-1}) \Bigl( \inf_{[x_{j-1},x_j]} f + \inf_{[x_{j-1},x_j]} g \Bigr) \le L(f+g, P, [a,b]) . \tag{3} \end{equation*}

Applying (2) in the same way,

\begin{equation*} U(f+g, P, [a,b]) \le U(f,P,[a,b]) + U(g,P,[a,b]) . \tag{4} \end{equation*}

Step 3: passing to the lower and upper Riemann integrals.

One cannot simply take a supremum in (3), because the suprema defining \(L(f,[a,b])\) and \(L(g,[a,b])\) may be approached along different partitions. The fix is to refine, using 1.5.

Let \(\varepsilon > 0\). By 1.7, \(L(f,[a,b])\) is the supremum of the lower Riemann sums of \(f\), so there is a partition \(P_1\) of \([a,b]\) with

\begin{equation*} L(f, P_1, [a,b]) > L(f,[a,b]) - \frac{\varepsilon}{2}, \end{equation*}

and likewise a partition \(P_2\) of \([a,b]\) with

\begin{equation*} L(g, P_2, [a,b]) > L(g,[a,b]) - \frac{\varepsilon}{2} . \end{equation*}

Let \(P\) be the partition of \([a,b]\) obtained by merging the lists that define \(P_1\) and \(P_2\). Then the list defining \(P_1\) is a sublist of the list defining \(P\), and the same holds for \(P_2\), so 1.5 gives

\begin{equation*} L(f, P, [a,b]) \ge L(f, P_1, [a,b]) \qquad \text{and} \qquad L(g, P, [a,b]) \ge L(g, P_2, [a,b]) . \end{equation*}

Combining these with (3) and with \(L(f+g,P,[a,b]) \le L(f+g,[a,b])\) (which holds by 1.7, as \(L(f+g,[a,b])\) is a supremum over all partitions),

\begin{equation*} \begin{aligned} L(f+g,[a,b]) &\ge L(f+g, P, [a,b]) \\ &\ge L(f, P, [a,b]) + L(g, P, [a,b]) \\ &\ge L(f, P_1, [a,b]) + L(g, P_2, [a,b]) \\ &> L(f,[a,b]) + L(g,[a,b]) - \varepsilon . \end{aligned} \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, \(L(f,[a,b]) + L(g,[a,b]) \le L(f+g,[a,b])\), which is the first assertion.

The second assertion is proved by the mirror-image argument. Let \(\varepsilon > 0\). By 1.7 there are partitions \(Q_1, Q_2\) of \([a,b]\) with

\begin{equation*} U(f, Q_1, [a,b]) < U(f,[a,b]) + \frac{\varepsilon}{2} \qquad \text{and} \qquad U(g, Q_2, [a,b]) < U(g,[a,b]) + \frac{\varepsilon}{2} . \end{equation*}

Let \(Q\) be the partition obtained by merging the lists defining \(Q_1\) and \(Q_2\). By 1.5, adjoining points decreases upper Riemann sums, so \(U(f,Q,[a,b]) \le U(f,Q_1,[a,b])\) and \(U(g,Q,[a,b]) \le U(g,Q_2,[a,b])\). Hence, using (4),

\begin{equation*} \begin{aligned} U(f+g,[a,b]) &\le U(f+g, Q, [a,b]) \\ &\le U(f, Q, [a,b]) + U(g, Q, [a,b]) \\ &\le U(f, Q_1, [a,b]) + U(g, Q_2, [a,b]) \\ &< U(f,[a,b]) + U(g,[a,b]) + \varepsilon . \end{aligned} \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, \(U(f+g,[a,b]) \le U(f,[a,b]) + U(g,[a,b])\), completing the proof.

Combining the two inequalities just proved with 1.8 gives a quick proof of Exercise 4 in Section 1A: if \(f\) and \(g\) are Riemann integrable, then

\begin{equation*} \int_a^b f + \int_a^b g = L(f,[a,b]) + L(g,[a,b]) \le L(f+g,[a,b]) \le U(f+g,[a,b]) \le U(f,[a,b]) + U(g,[a,b]) = \int_a^b f + \int_a^b g , \end{equation*}

so all the inequalities are equalities; in particular \(L(f+g,[a,b]) = U(f+g,[a,b])\), so \(f+g\) is Riemann integrable with \(\int_a^b (f+g) = \int_a^b f + \int_a^b g\).

Exercise 1B.4 shows that both inequalities in this exercise can be strict, so this last argument really does require the integrability of \(f\) and \(g\).

Take \(f\) to be the function of 1.14 and \(g = 1 - f\). Explicitly, define \(f, g \colon [0,1] \to \mathbb{R}\) by

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x \text{ is rational,} \\ 0 & \text{if } x \text{ is irrational,} \end{cases} \qquad g(x) = \begin{cases} 0 & \text{if } x \text{ is rational,} \\ 1 & \text{if } x \text{ is irrational.} \end{cases} \end{equation*}

Both take only the values \(0\) and \(1\), so both are bounded.

The computation in 1.14 shows that \(L(f,[0,1]) = 0\) and \(U(f,[0,1]) = 1\). We recall the reason. If \([c,d] \subseteq [0,1]\) with \(c < d\), then \([c,d]\) contains an irrational number and also contains a rational number, so

\begin{equation*} \inf_{[c,d]} f = 0 \qquad \text{and} \qquad \sup_{[c,d]} f = 1 . \end{equation*}

Hence for every partition \(P\) of \([0,1]\), say \(0 = x_0 < x_1 < \cdots < x_n = 1\),

\begin{equation*} L(f,P,[0,1]) = \sum_{j=1}^n (x_j - x_{j-1}) \cdot 0 = 0, \qquad U(f,P,[0,1]) = \sum_{j=1}^n (x_j - x_{j-1}) \cdot 1 = 1 . \end{equation*}

Taking the supremum and the infimum over all partitions as in 1.7 gives \(L(f,[0,1]) = 0\) and \(U(f,[0,1]) = 1\).

The same argument applies verbatim to \(g\), since every nonempty subinterval of positive length also contains both a rational and an irrational number and \(g\) likewise takes the values \(0\) and \(1\) on each such subinterval. Hence

\begin{equation*} L(g,[0,1]) = 0 \qquad \text{and} \qquad U(g,[0,1]) = 1 . \end{equation*}

Now \(f + g\) is the constant function \(1\) on \([0,1]\): if \(x\) is rational then \(f(x) + g(x) = 1 + 0 = 1\), and if \(x\) is irrational then \(f(x) + g(x) = 0 + 1 = 1\). For a constant function every lower and every upper Riemann sum equals \(1\), because \(\inf_{[c,d]}(f+g) = \sup_{[c,d]}(f+g) = 1\) on each subinterval and the subinterval lengths of a partition of \([0,1]\) sum to \(1\). Hence

\begin{equation*} L(f+g,[0,1]) = U(f+g,[0,1]) = 1 . \end{equation*}

(Equivalently: \(f+g\) is continuous, so it is Riemann integrable by 1.11, and \(\int_0^1 (f+g) = 1\) by 1.13, since \(\inf_{[0,1]}(f+g) = \sup_{[0,1]}(f+g) = 1\).)

Putting these together:

\begin{equation*} L(f,[0,1]) + L(g,[0,1]) = 0 + 0 = 0 < 1 = L(f+g,[0,1]) , \end{equation*}

and

\begin{equation*} U(f+g,[0,1]) = 1 < 2 = 1 + 1 = U(f,[0,1]) + U(g,[0,1]) . \end{equation*}

Both inequalities of Exercise 1B.3 are therefore strict for this pair, as required.

The mechanism is worth naming. The lower Riemann integral fails to be additive because \(f\) and \(g\) take their small values at different points: on every subinterval, \(f\) dips to \(0\) at the irrationals while \(g\) dips to \(0\) at the rationals, and no single point makes both dip at once. Dually for the suprema. Cancellation of this kind is invisible to the one-function-at-a-time bookkeeping of Riemann sums, and it is precisely the phenomenon that forces the hypothesis of Riemann integrability into Exercise 4 in Section 1A.

Take \(f\) to be the constant function \(0\), and let \(f_k\) be a tall thin triangular spike of area \(1\) sitting on the interval \([0, \frac1k]\). Explicitly, for \(k \in \mathbb{Z}^+\) define \(f_k \colon [0,1] \to \mathbb{R}\) by

\begin{equation*} f_k(x) = \begin{cases} 4k^2 x & \text{if } 0 \le x \le \frac{1}{2k}, \\[2pt] 4k^2 \bigl( \frac{1}{k} - x \bigr) & \text{if } \frac{1}{2k} \le x \le \frac{1}{k}, \\[2pt] 0 & \text{if } \frac{1}{k} \le x \le 1, \end{cases} \end{equation*}

and define \(f \colon [0,1] \to \mathbb{R}\) by \(f(x) = 0\).

Each \(f_k\) is well defined and continuous. The three formulas agree at the two points where the cases overlap: at \(x = \frac{1}{2k}\) the first gives \(4k^2 \cdot \frac{1}{2k} = 2k\) and the second gives \(4k^2 \bigl( \frac1k - \frac{1}{2k} \bigr) = 4k^2 \cdot \frac{1}{2k} = 2k\); at \(x = \frac1k\) the second gives \(0\) and the third gives \(0\). Each piece is a polynomial and hence continuous on its closed subinterval, so \(f_k\) is continuous on \([0,1]\) (this uses that the three subintervals are closed and cover \([0,1]\), with agreement at the overlaps). Also note \(\frac1k \le 1\), so the description makes sense for every \(k \in \mathbb{Z}^+\). The function \(f\) is constant, hence continuous. By 1.11, \(f\) and every \(f_k\) are Riemann integrable on \([0,1]\).

Pointwise convergence. Let \(x \in [0,1]\). If \(x = 0\), then \(f_k(0) = 4k^2 \cdot 0 = 0\) for every \(k\), so \(\lim_{k \to \infty} f_k(0) = 0 = f(0)\). If \(x > 0\), choose \(K \in \mathbb{Z}^+\) with \(K > \frac1x\), which is possible by the Archimedean property. For every \(k \ge K\) we have \(\frac1k \le \frac1K < x \le 1\), so \(x\) lies in the third case and \(f_k(x) = 0\). Hence \(\lim_{k \to \infty} f_k(x) = 0 = f(x)\). Thus \(f(x) = \lim_{k \to \infty} f_k(x)\) for each \(x \in [0,1]\), as required.

The integrals. Clearly \(\int_0^1 f = 0\), since \(\inf_{[0,1]} f = \sup_{[0,1]} f = 0\) and hence \(\int_0^1 f = 0\) by 1.13.

For \(f_k\), first suppose \(k \ge 2\), so that \(\frac1k \in (0,1)\) and \(\frac{1}{2k} \in (0, \frac1k)\). Apply Exercise 10 in Section 1A once to split \([0,1]\) at \(\frac1k\), and again to split \([0,\frac1k]\) at \(\frac{1}{2k}\) (each application is legitimate because \(f_k\) is Riemann integrable on \([0,1]\), hence on every closed subinterval by Exercise 9 in Section 1A):

\begin{equation*} \int_0^1 f_k = \int_0^{1/(2k)} f_k + \int_{1/(2k)}^{1/k} f_k + \int_{1/k}^{1} f_k . \end{equation*}

When \(k = 1\) we have \(\frac1k = 1\), so only the split at \(\frac{1}{2k} = \frac12\) is used and the identity above holds with its last term read as \(0\). On \([0, \frac{1}{2k}]\) the function \(f_k\) agrees with \(x \mapsto 4k^2 x\), whose antiderivative is \(2k^2 x^2\), so by the fundamental theorem of calculus

\begin{equation*} \int_0^{1/(2k)} f_k = 2k^2 \Bigl( \frac{1}{2k} \Bigr)^{\!2} = 2k^2 \cdot \frac{1}{4k^2} = \frac12 . \end{equation*}

On \([\frac{1}{2k}, \frac1k]\) the function \(f_k\) agrees with \(x \mapsto 4k^2 (\frac1k - x)\), whose antiderivative is \(-2k^2 (\frac1k - x)^2\), so

\begin{equation*} \int_{1/(2k)}^{1/k} f_k = \Bigl[ -2k^2 \Bigl( \tfrac1k - x \Bigr)^{\!2} \Bigr]_{x = 1/(2k)}^{x = 1/k} = 0 - \Bigl( -2k^2 \cdot \frac{1}{4k^2} \Bigr) = \frac12 . \end{equation*}

On \([\frac1k, 1]\) the function \(f_k\) is identically \(0\), so that integral is \(0\) by 1.13. Adding,

\begin{equation*} \int_0^1 f_k = \frac12 + \frac12 + 0 = 1 \qquad \text{for every } k \in \mathbb{Z}^+ . \end{equation*}

Geometrically this is just the area of a triangle with base \(\frac1k\) and height \(2k\), namely \(\frac12 \cdot \frac1k \cdot 2k = 1\).

Conclusion. The limit \(\lim_{k \to \infty} \int_0^1 f_k = \lim_{k \to \infty} 1 = 1\) exists, but

\begin{equation*} \int_0^1 f = 0 \ne 1 = \lim_{k \to \infty} \int_0^1 f_k , \end{equation*}

so this sequence has the required property.

Note that this example does not contradict 1.18. There the sequence is assumed uniformly bounded, that is, \(|f_k(x)| \le M\) for all \(k\) and all \(x\) with a single constant \(M\). Here \(\sup_{[0,1]} f_k = f_k \bigl( \frac{1}{2k} \bigr) = 2k \to \infty\), so no such \(M\) exists. The example shows that the uniform boundedness hypothesis in 1.18 cannot be dropped, and it complements 1.17: there the failure was that the pointwise limit need not be Riemann integrable, whereas here the limit function is as well behaved as could be wished and the integrals still fail to converge to the right value. The mass of \(f_k\) escapes by concentrating on shorter and shorter intervals rather than by growing in the limit function.

We prove the two inequalities separately.

First, \(A \subseteq A \cup B\). Because outer measure preserves order (2.5), this gives

\begin{equation*} |A| \le |A \cup B|. \end{equation*}

Second, apply countable subadditivity of outer measure (2.8) to the sequence \(A, B, \emptyset, \emptyset, \dots\), whose union is \(A \cup B\). Since \(|\emptyset| = 0\) (the empty set is finite, so 2.3 applies, or take every \(I_k = \emptyset\) in 2.2), we obtain

\begin{equation*} |A \cup B| \le |A| + |B| = |A| + 0 = |A|. \end{equation*}

Combining the two displayed inequalities yields \(|A \cup B| = |A|\). Note that the argument is valid also when \(|A| = \infty\), since both inequalities are statements in \([0, \infty]\) and no subtraction was performed.

Case 1: \(t = 0\). If \(A = \emptyset\) then \(tA = \emptyset\), and if \(A \ne \emptyset\) then \(tA = \{0\}\). In either case \(tA\) is a finite set, so \(|tA| = 0\) by 2.3. On the other side, \(|t|\,|A| = 0 \cdot |A| = 0\), using the stated convention \(0 \cdot \infty = 0\) in case \(|A| = \infty\). Hence \(|tA| = |t|\,|A|\).

Case 2: \(t \ne 0\). We first record how dilation acts on open intervals: for every open interval \(I\),

\begin{equation*} \ell(tI) = |t|\,\ell(I), \end{equation*}

where \(tI = \{tx : x \in I\}\) is again an open interval. Indeed, if \(I = \emptyset\) then \(tI = \emptyset\) and both sides are \(0\). If \(I = (\alpha, \beta)\) with \(\alpha, \beta \in \mathbb{R}\) and \(\alpha < \beta\), then \(tI = (t\alpha, t\beta)\) when \(t > 0\) and \(tI = (t\beta, t\alpha)\) when \(t < 0\); in both cases \(\ell(tI) = |t|(\beta - \alpha) = |t|\,\ell(I)\). If \(I\) is one of \((-\infty, \alpha)\), \((\alpha, \infty)\), \((-\infty, \infty)\), then \(tI\) is again an unbounded open interval (multiplication by \(t \ne 0\) is a homeomorphism of \(\mathbb{R}\) carrying rays to rays), so \(\ell(tI) = \infty = |t| \cdot \infty = |t|\,\ell(I)\), the middle equality because \(|t| > 0\).

Now we prove \(|tA| \le |t|\,|A|\). If \(|A| = \infty\) this is immediate, so assume \(|A| < \infty\) and let \(I_1, I_2, \dots\) be any sequence of open intervals with \(A \subseteq \bigcup_{k=1}^{\infty} I_k\). Applying the map \(x \mapsto tx\), which is a bijection of \(\mathbb{R}\), we get \(tA \subseteq \bigcup_{k=1}^{\infty} t I_k\), and each \(tI_k\) is an open interval. Hence by the definition 2.2 of outer measure,

\begin{equation*} |tA| \le \sum_{k=1}^{\infty} \ell(t I_k) = |t| \sum_{k=1}^{\infty} \ell(I_k). \end{equation*}

Taking the infimum of the right-hand side over all such sequences \(I_1, I_2, \dots\) gives \(|tA| \le |t|\,|A|\).

Finally we get the reverse inequality by applying the inequality just proved with \(t\) replaced by \(1/t\) and \(A\) replaced by \(tA\). Since \((1/t)(tA) = A\), this yields

\begin{equation*} |A| \le \tfrac{1}{|t|}\,|tA|, \end{equation*}

that is, \(|t|\,|A| \le |tA|\). Combining the two inequalities gives \(|tA| = |t|\,|A|\).

Every element of \(B\) either lies in \(A\) or does not, so

\begin{equation*} B \subseteq (B \setminus A) \cup A. \end{equation*}

Because outer measure preserves order (2.5) and is countably subadditive (2.8, applied to the sequence \(B \setminus A,\ A,\ \emptyset,\ \emptyset, \dots\), which gives finite subadditivity), we obtain

\begin{equation*} |B| \le \bigl| (B \setminus A) \cup A \bigr| \le |B \setminus A| + |A|. \end{equation*}

If \(|B \setminus A| = \infty\), the desired inequality \(|B \setminus A| \ge |B| - |A|\) holds trivially. Otherwise \(|B \setminus A| < \infty\), and since also \(|A| < \infty\) by hypothesis, both terms on the right side of the display above are real numbers; hence \(|B| < \infty\) as well and we may subtract \(|A|\) from both sides of the inequality \(|B| \le |B \setminus A| + |A|\), obtaining

\begin{equation*} |B| - |A| \le |B \setminus A|, \end{equation*}

as desired. Note that the hypothesis \(|A| < \infty\) is what makes the subtraction legitimate; without it the expression \(|B| - |A|\) could be the undefined \(\infty - \infty\).

Together with the Heine–Borel Theorem (2.12), this exercise shows that a subset of \(\mathbb{R}\) has the finite-subcover property if and only if it is closed and bounded.

\(F\) is bounded. Consider the collection

\begin{equation*} \mathcal{C} = \{ (-k, k) : k \in \mathbb{Z}^{+} \} \end{equation*}

of open subsets of \(\mathbb{R}\). Every real number \(x\) satisfies \(|x| < k\) for some \(k \in \mathbb{Z}^{+}\), so \(\bigcup_{k=1}^{\infty} (-k,k) = \mathbb{R} \supseteq F\); thus \(\mathcal{C}\) is an open cover of \(F\) in the sense of 2.10. By hypothesis there exist \(k_1, \dots, k_n \in \mathbb{Z}^{+}\) with

\begin{equation*} F \subseteq (-k_1, k_1) \cup \cdots \cup (-k_n, k_n). \end{equation*}

(If the finite subcover is the empty list, then \(F = \emptyset\) and \(F\) is certainly bounded.) The intervals \((-k,k)\) increase with \(k\), so setting \(K = \max\{k_1, \dots, k_n\}\) the union above equals \((-K, K)\). Hence \(|x| < K\) for every \(x \in F\), so \(F\) is bounded.

\(F\) is closed. We show that the complement \(\mathbb{R} \setminus F\) is open. Let \(b \in \mathbb{R} \setminus F\). For each \(k \in \mathbb{Z}^{+}\) define

\begin{equation*} G_k = \Bigl(-\infty,\ b - \tfrac{1}{k}\Bigr) \cup \Bigl(b + \tfrac{1}{k},\ \infty\Bigr) = \Bigl\{ x \in \mathbb{R} : |x - b| > \tfrac{1}{k} \Bigr\}, \end{equation*}

which is an open subset of \(\mathbb{R}\), being a union of two open intervals. If \(|x - b| > 1/k\) then \(|x - b| > 1/(k+1)\), so \(G_1 \subseteq G_2 \subseteq G_3 \subseteq \cdots\). Moreover

\begin{equation*} \bigcup_{k=1}^{\infty} G_k = \{ x \in \mathbb{R} : x \ne b \} = \mathbb{R} \setminus \{b\}, \end{equation*}

because any \(x \ne b\) satisfies \(|x - b| > 1/k\) for all sufficiently large \(k\). Since \(b \notin F\), we have \(F \subseteq \mathbb{R} \setminus \{b\}\), so \(\{G_k : k \in \mathbb{Z}^{+}\}\) is an open cover of \(F\).

By hypothesis this cover has a finite subcover: there exist \(k_1, \dots, k_n \in \mathbb{Z}^{+}\) with \(F \subseteq G_{k_1} \cup \cdots \cup G_{k_n}\). (Again, if the list is empty then \(F = \emptyset\), which is closed, and we are done.) Because the \(G_k\) are nested increasing, with \(K = \max\{k_1, \dots, k_n\}\) we get \(F \subseteq G_K\). Thus no point of \(F\) lies within distance \(1/K\) of \(b\), that is,

\begin{equation*} \Bigl( b - \tfrac{1}{K},\ b + \tfrac{1}{K} \Bigr) \subseteq \mathbb{R} \setminus F . \end{equation*}

So every point of \(\mathbb{R} \setminus F\) is contained in an open interval contained in \(\mathbb{R} \setminus F\), which means \(\mathbb{R} \setminus F\) is open. Hence \(F\) is closed.

Let \(F_0 \in \mathcal{A}\) be a bounded set. Since every element of \(\mathcal{A}\) is closed, \(F_0\) is a closed bounded subset of \(\mathbb{R}\).

If \(F_0 = \emptyset\), take \(n = 1\) and \(F_1 = F_0\); then \(F_1 = \emptyset\) and we are done. So assume \(F_0 \ne \emptyset\).

Consider the collection

\begin{equation*} \mathcal{C} = \{\, \mathbb{R} \setminus F : F \in \mathcal{A} \,\}. \end{equation*}

Each element of \(\mathcal{C}\) is an open subset of \(\mathbb{R}\), because each \(F \in \mathcal{A}\) is closed. We claim \(\mathcal{C}\) is an open cover of \(F_0\). Indeed, let \(x \in F_0\). Since \(\bigcap_{F \in \mathcal{A}} F = \emptyset\), the point \(x\) does not belong to every element of \(\mathcal{A}\); hence there is some \(F \in \mathcal{A}\) with \(x \notin F\), i.e. \(x \in \mathbb{R} \setminus F \in \mathcal{C}\). Thus \(F_0\) is contained in the union of the sets in \(\mathcal{C}\), proving the claim.

By the Heine–Borel Theorem (2.12), the open cover \(\mathcal{C}\) of the closed bounded set \(F_0\) has a finite subcover: there exist \(m \in \mathbb{Z}^{+}\) and \(F_1, \dots, F_m \in \mathcal{A}\) such that

\begin{equation*} F_0 \subseteq (\mathbb{R} \setminus F_1) \cup \cdots \cup (\mathbb{R} \setminus F_m). \end{equation*}

(The subcover cannot be an empty list, since \(F_0 \ne \emptyset\).) By De Morgan’s laws the right-hand side equals \(\mathbb{R} \setminus (F_1 \cap \cdots \cap F_m)\), so

\begin{equation*} F_0 \subseteq \mathbb{R} \setminus (F_1 \cap \cdots \cap F_m), \end{equation*}

which says precisely that

\begin{equation*} F_0 \cap F_1 \cap \cdots \cap F_m = \emptyset. \end{equation*}

Since \(F_0, F_1, \dots, F_m\) are all elements of \(\mathcal{A}\), relabelling this list as \(F_1, \dots, F_n\) with \(n = m + 1\) gives \(n \in \mathbb{Z}^{+}\) and \(F_1, \dots, F_n \in \mathcal{A}\) with \(F_1 \cap \cdots \cap F_n = \emptyset\), as required.

Remark: the hypothesis that some element of \(\mathcal{A}\) be bounded cannot be dropped. For example, \(\mathcal{A} = \{ [k, \infty) : k \in \mathbb{Z}^{+} \}\) consists of closed sets with \(\bigcap_{F \in \mathcal{A}} F = \emptyset\), yet every finite subfamily has nonempty intersection, since \([k_1, \infty) \cap \cdots \cap [k_n, \infty) = [\max\{k_1,\dots,k_n\}, \infty) \ne \emptyset\).

By 2.14 we know that \(|[a,b]| = b - a\). Each of the three sets in question differs from \([a,b]\) by at most the two-point set \(\{a,b\}\), and finite sets have outer measure \(0\) by 2.3. We can therefore obtain all three equalities from Exercise 1 in this section, which says that adjoining a set of outer measure \(0\) does not change outer measure.

The open interval. Since \([a,b] = (a,b) \cup \{a,b\}\) and \(|\{a,b\}| = 0\), Exercise 1 (with \(A = (a,b)\) and \(B = \{a,b\}\)) gives

\begin{equation*} |(a,b)| = \bigl| (a,b) \cup \{a,b\} \bigr| = |[a,b]| = b - a. \end{equation*}

The half-open interval \([a,b)\). Since \([a,b] = [a,b) \cup \{b\}\) and \(|\{b\}| = 0\), Exercise 1 gives

\begin{equation*} |[a,b)| = \bigl| [a,b) \cup \{b\} \bigr| = |[a,b]| = b - a. \end{equation*}

The half-open interval \((a,b]\). Since \([a,b] = (a,b] \cup \{a\}\) and \(|\{a\}| = 0\), Exercise 1 gives

\begin{equation*} |(a,b]| = \bigl| (a,b] \cup \{a\} \bigr| = |[a,b]| = b - a. \end{equation*}

Hence \(|(a,b)| = |[a,b)| = |(a,b]| = b - a\).

For the reader who prefers an argument not relying on Exercise 1, here is a direct proof for \((a,b)\); the two half-open cases then follow since \((a,b) \subseteq [a,b) \subseteq [a,b]\) and \((a,b) \subseteq (a,b] \subseteq [a,b]\), so that 2.5 squeezes their outer measures between \(|(a,b)|\) and \(|[a,b]|\). For the upper bound, \((a,b) \subseteq [a,b]\), so \(|(a,b)| \le |[a,b]| = b - a\) by 2.5 and 2.14. For the lower bound, let \(\varepsilon\) satisfy \(0 < \varepsilon < (b-a)/2\). Then \([a + \varepsilon, b - \varepsilon] \subseteq (a,b)\), so by 2.5 and 2.14,

\begin{equation*} |(a,b)| \ge \bigl| [a + \varepsilon,\ b - \varepsilon] \bigr| = (b - \varepsilon) - (a + \varepsilon) = b - a - 2\varepsilon . \end{equation*}

Because this holds for every such \(\varepsilon\), letting \(\varepsilon\) tend to \(0\) gives \(|(a,b)| \ge b - a\).

Throughout write \(E = (a,b) \cup (c,d)\). Note first that by Exercise 6 of this section, \(|(a,b)| = b - a\) and \(|(c,d)| = d - c\), and hence by finite subadditivity (a special case of 2.8, applied to the sequence \((a,b),\ (c,d),\ \emptyset,\ \emptyset, \dots\)) we always have

\begin{equation*} |E| \le (b-a) + (d-c) . \end{equation*}

So the content of the exercise is that equality holds exactly when the two intervals are disjoint.

We begin with an elementary observation about when two nonempty open intervals meet.

Claim. \((a,b) \cap (c,d) \ne \emptyset\) if and only if \(c < b\) and \(a < d\); and in that case

\begin{equation*} (a,b) \cap (c,d) = (u,v), \qquad u = \max\{a,c\}, \quad v = \min\{b,d\}, \end{equation*}

with \(u < v\).

Proof of claim. A real number \(x\) lies in \((a,b) \cap (c,d)\) exactly when \(a < x < b\) and \(c < x < d\), i.e. exactly when \(\max\{a,c\} < x < \min\{b,d\}\); so the intersection is the interval \((u,v)\), which is nonempty precisely when \(u < v\). Now \(u < v\) means \(\max\{a,c\} < \min\{b,d\}\), i.e. all four of \(a < b\), \(a < d\), \(c < b\), \(c < d\) hold. Two of these, \(a<b\) and \(c<d\), hold by hypothesis, so \(u < v\) is equivalent to the conjunction of \(c < b\) and \(a < d\). This proves the claim.

Suppose \((a,b) \cap (c,d) = \emptyset\). By the claim, \(b \le c\) or \(d \le a\). The two cases are symmetric under interchanging the pair \((a,b)\) with the pair \((c,d)\), which changes neither \(E\) nor the quantity \((b-a)+(d-c)\); so we may assume \(b \le c\).

We must prove \(|E| \ge (b-a) + (d-c)\); combined with the subadditivity bound displayed above this gives equality. Since \(a < b \le c < d\), we claim that

\begin{equation*} (a,d) \subseteq (a,b) \cup [b,c] \cup (c,d) = E \cup [b,c]. \end{equation*}

Indeed, let \(x \in (a,d)\). If \(x < b\) then \(a < x < b\), so \(x \in (a,b)\). If \(b \le x \le c\) then \(x \in [b,c]\). If \(x > c\) then \(c < x < d\), so \(x \in (c,d)\). These three cases are exhaustive, which proves the inclusion. (Note that the closed interval \([a,d]\) could not be used in place of \((a,d)\) here: neither \(a\) nor \(d\) belongs to \(E \cup [b,c]\).)

Now \(|[b,c]| = c - b\): this is 2.14 if \(b < c\), and if \(b = c\) then \([b,c] = \{b\}\) is a one-point set, so \(|[b,c]| = 0 = c - b\) by 2.3. Also \(|(a,d)| = d - a\) by Exercise 6 of this section, which applies because \(a < d\). Applying 2.5 and then finite subadditivity (2.8) to the displayed inclusion,

\begin{equation*} d - a = |(a,d)| \le |E| + |[b,c]| = |E| + (c - b). \end{equation*}

All quantities here are finite, so we may rearrange:

\begin{equation*} |E| \ge (d - a) - (c - b) = (b - a) + (d - c). \end{equation*}

Together with \(|E| \le (b-a)+(d-c)\) this gives \(|E| = (b-a) + (d-c)\).

Suppose \((a,b) \cap (c,d) \ne \emptyset\). We show \(|E| < (b-a) + (d-c)\), which in particular gives \(|E| \ne (b-a)+(d-c)\) and hence, by contraposition, completes the proof of the equivalence.

Put \(p = \min\{a,c\}\), \(q = \max\{b,d\}\), and let \(u = \max\{a,c\}\), \(v = \min\{b,d\}\) as in the claim, so \(u < v\). Since \(\{p,u\}\) and \(\{a,c\}\) agree as multisets, and likewise \(\{v,q\}\) and \(\{b,d\}\), at least one of the two intervals has left endpoint \(p\) and at least one has right endpoint \(q\). We first check that \(E = (p,q)\). The inclusion \(E \subseteq (p,q)\) is clear, since \((a,b) \subseteq (p,q)\) and \((c,d) \subseteq (p,q)\). Conversely, let \(x \in (p,q)\), so \(p < x < q\). If \(x \le u\) then, since \(p < x\) and \(\{p, u\} = \{a, c\}\), the point \(x\) lies in whichever of \((a,b)\), \((c,d)\) has left endpoint \(p\); indeed that interval is \((p, w)\) for some \(w \in \{b,d\}\) with \(w \ge v > u \ge x\), so \(p < x < w\) and \(x\) belongs to it. If \(x > u\) then, since \(x < q\) and \(\{v,q\} = \{b,d\}\), the point \(x\) lies in whichever of \((a,b)\), \((c,d)\) has right endpoint \(q\); that interval is \((z, q)\) for some \(z \in \{a,c\}\) with \(z \le u < x\), so \(z < x < q\) and \(x\) belongs to it. Hence \((p,q) \subseteq E\), and therefore \(E = (p,q)\).

Since \(p \le u < v \le q\) we have \(p < q\), so Exercise 6 gives

\begin{equation*} |E| = |(p,q)| = q - p . \end{equation*}

Finally, because \(\{p, u\} = \{a, c\}\) and \(\{v, q\} = \{b, d\}\) as multisets, we have \(p + u = a + c\) and \(v + q = b + d\). Therefore

\begin{equation*} (b - a) + (d - c) = (b + d) - (a + c) = (v + q) - (p + u) = (q - p) + (v - u) = |E| + (v - u). \end{equation*}

Since \(v - u > 0\) and \(|E| = q - p\) is finite, this gives

\begin{equation*} |E| < (b-a) + (d-c), \end{equation*}

as claimed. This completes both directions of the proof.

Write \(B = A \cap (-t,t)\) and \(C = A \cap \bigl(\mathbb{R}\setminus(-t,t)\bigr) = A \setminus (-t,t)\), so that \(A = B \cup C\).

The inequality \(|A| \le |B| + |C|\). This is immediate from the countable subadditivity of outer measure (2.8), applied to the sequence \(B, C, \emptyset, \emptyset, \dots\) whose union is \(A\).

The inequality \(|A| \ge |B| + |C|\). If \(|A| = \infty\) there is nothing to prove, so assume \(|A| < \infty\). Let \(\varepsilon > 0\). By the definition of outer measure (2.2) there is a sequence \(I_1, I_2, \dots\) of open intervals with

\begin{equation*} A \subseteq \bigcup_{k=1}^{\infty} I_k \qquad\text{and}\qquad \sum_{k=1}^{\infty} \ell(I_k) \le |A| + \varepsilon . \end{equation*}

Cut each \(I_k\) at the two points \(-t\) and \(t\): put

\begin{equation*} J_k = I_k \cap (-t,t), \qquad K_k = I_k \cap (-\infty,-t), \qquad L_k = I_k \cap (t,\infty). \end{equation*}

Each of \(J_k, K_k, L_k\) is an open interval (possibly empty), being the intersection of two open intervals. I claim that

\begin{equation*} \ell(J_k) + \ell(K_k) + \ell(L_k) = \ell(I_k) \qquad \text{for each } k . \end{equation*}

To see this, first note the one-point version: if \(I\) is an open interval and \(c \in \mathbb{R}\), then

\begin{equation*} \ell\bigl(I \cap (-\infty,c)\bigr) + \ell\bigl(I \cap (c,\infty)\bigr) = \ell(I). \end{equation*}

Indeed, if \(I = \emptyset\) both sides are \(0\). If \(I = (a,b)\) with \(a,b \in \mathbb{R}\), then for \(c \le a\) the two pieces are \(\emptyset\) and \((a,b)\), for \(c \ge b\) they are \((a,b)\) and \(\emptyset\), and for \(a < c < b\) they are \((a,c)\) and \((c,b)\), whose lengths sum to \((c-a)+(b-c) = b-a\); in every case the sum is \(b - a = \ell(I)\). If \(I\) is unbounded, then at least one of the two pieces is an unbounded open interval, so both sides equal \(\infty\).

Now write \(M_k = I_k \cap (-t,\infty)\), which is again an open interval. Applying the one-point version to \(I_k\) with \(c = -t\) gives \(\ell(K_k) + \ell(M_k) = \ell(I_k)\). Applying it to \(M_k\) with \(c = t\) gives \(\ell\bigl(M_k \cap (-\infty,t)\bigr) + \ell\bigl(M_k \cap (t,\infty)\bigr) = \ell(M_k)\); here \(M_k \cap (-\infty,t) = I_k \cap (-t,t) = J_k\), and, because \(t > -t\), \(M_k \cap (t,\infty) = I_k \cap (t,\infty) = L_k\). Hence \(\ell(J_k) + \ell(L_k) = \ell(M_k)\), and adding the two identities gives the claim.

Now \(B = A \cap (-t,t) \subseteq \bigcup_{k} J_k\), so by 2.2,

\begin{equation*} |B| \le \sum_{k=1}^{\infty} \ell(J_k). \end{equation*}

Also

\begin{equation*} C = A \setminus (-t,t) \subseteq \Bigl(\bigcup_{k=1}^{\infty} K_k\Bigr) \cup \Bigl(\bigcup_{k=1}^{\infty} L_k\Bigr) \cup \{-t,t\}, \end{equation*}

because a point of \(A\) outside \((-t,t)\) lies in some \(I_k\), and if it is neither \(-t\) nor \(t\) it then lies in \(K_k\) or in \(L_k\). The sequence \(K_1, L_1, K_2, L_2, \dots\) is a sequence of open intervals, so by 2.2 the union of its terms has outer measure at most \(\sum_{k}\bigl(\ell(K_k) + \ell(L_k)\bigr)\); since \(|\{-t,t\}| = 0\) by 2.3, subadditivity (2.8) gives

\begin{equation*} |C| \le \sum_{k=1}^{\infty} \bigl(\ell(K_k) + \ell(L_k)\bigr). \end{equation*}

Adding the two displayed bounds and using the claim,

\begin{equation*} |B| + |C| \le \sum_{k=1}^{\infty} \bigl(\ell(J_k) + \ell(K_k) + \ell(L_k)\bigr) = \sum_{k=1}^{\infty} \ell(I_k) \le |A| + \varepsilon . \end{equation*}

Because \(\varepsilon > 0\) is arbitrary, \(|B| + |C| \le |A|\).

Combining the two inequalities gives \(|A| = |A \cap (-t,t)| + \bigl|A \cap (\mathbb{R}\setminus(-t,t))\bigr|\).

First note that the limit exists in \([0,\infty]\). If \(0 < s < t\), then \(A \cap (-s,s) \subseteq A \cap (-t,t)\), so \(|A \cap (-s,s)| \le |A \cap (-t,t)|\) because outer measure preserves order (2.5). Thus \(t \mapsto |A \cap (-t,t)|\) is a nondecreasing function of \(t\) on \((0,\infty)\), and hence

\begin{equation*} \lim_{t \to \infty} |A \cap (-t,t)| = \sup_{t > 0} |A \cap (-t,t)| = \lim_{n \to \infty} |A \cap (-n,n)|, \end{equation*}

the last equality because the supremum over \(t > 0\) of a nondecreasing function is already attained as a limit along the positive integers. Call this common value \(L \in [0,\infty]\).

The inequality \(L \le |A|\). For every \(t > 0\) we have \(A \cap (-t,t) \subseteq A\), so \(|A \cap (-t,t)| \le |A|\) by 2.5. Taking the supremum over \(t > 0\) gives \(L \le |A|\).

The inequality \(|A| \le L\). Decompose \(A\) into the annuli cut out by the integers: set

\begin{equation*} C_1 = A \cap (-1,1) \qquad\text{and}\qquad C_n = A \cap \{x \in \mathbb{R} : n - 1 \le |x| < n\} \ \text{ for } n \ge 2 . \end{equation*}

Fix \(n \ge 2\) and apply Exercise 8 of Section 2A (proved above) to the set \(A \cap (-n,n)\) with \(t = n-1\), which is legitimate because \(n - 1 > 0\). Since \((-(n-1), n-1) \subseteq (-n,n)\) we have

\begin{equation*} \bigl(A \cap (-n,n)\bigr) \cap (-(n-1), n-1) = A \cap (-(n-1), n-1), \end{equation*}

and

\begin{equation*} \bigl(A \cap (-n,n)\bigr) \setminus (-(n-1), n-1) = A \cap \{x : n-1 \le |x| < n\} = C_n . \end{equation*}

Hence Exercise 8, applied to the set \(A \cap (-n,n)\), gives

\begin{equation*} |A \cap (-n,n)| = |A \cap (-(n-1), n-1)| + |C_n| \qquad \text{for all } n \ge 2 . \end{equation*}

Since \(|A \cap (-1,1)| = |C_1|\), induction on \(n\) now yields

\begin{equation*} |A \cap (-n,n)| = \sum_{k=1}^{n} |C_k| \qquad \text{for all } n \in \mathbb{Z}^{+}, \end{equation*}

and therefore

\begin{equation*} L = \lim_{n \to \infty} |A \cap (-n,n)| = \sum_{k=1}^{\infty} |C_k| . \end{equation*}

Every \(x \in A\) satisfies \(|x| < n\) for some \(n \in \mathbb{Z}^{+}\), and hence lies in \(C_m\) where \(m\) is the least such \(n\); thus \(A = \bigcup_{k=1}^{\infty} C_k\). By the countable subadditivity of outer measure (2.8),

\begin{equation*} |A| \le \sum_{k=1}^{\infty} |C_k| = L . \end{equation*}

Combining the two inequalities gives \(|A| = L = \lim_{t \to \infty} |A \cap (-t,t)|\).

Write \(E = [0,1] \setminus \mathbb{Q}\), the set of irrational numbers in \([0,1]\).

Upper bound. Since \(E \subseteq [0,1]\), the order-preserving property of outer measure (2.5) together with 2.14 gives

\begin{equation*} |E| \le |[0,1]| = 1 - 0 = 1 . \end{equation*}

Lower bound. The set \(\mathbb{Q}\) is countable, so \([0,1] \cap \mathbb{Q}\) is countable and therefore has outer measure \(0\) by 2.4. Since

\begin{equation*} [0,1] = E \cup \bigl([0,1] \cap \mathbb{Q}\bigr), \end{equation*}

the countable subadditivity of outer measure (2.8), applied to the sequence \(E,\ [0,1] \cap \mathbb{Q},\ \emptyset,\ \emptyset,\dots\), gives

\begin{equation*} 1 = |[0,1]| \le |E| + |[0,1] \cap \mathbb{Q}| = |E| + 0 = |E| , \end{equation*}

where again \(|[0,1]| = 1\) by 2.14.

The two inequalities give \(|[0,1] \setminus \mathbb{Q}| = 1\).

Let \(A = \bigcup_{k=1}^{\infty} I_k\).

Step 0: every open interval \(I\) satisfies \(|I| = \ell(I)\). If \(I = \emptyset\) both sides are \(0\). If \(I = (a,b)\) with \(a,b \in \mathbb{R}\) and \(a < b\), then on the one hand \((a,b) \subseteq [a,b]\), so \(|(a,b)| \le |[a,b]| = b - a\) by 2.5 and 2.14; on the other hand, for every \(\delta\) with \(0 < \delta < (b-a)/2\) we have \([a+\delta, b-\delta] \subseteq (a,b)\), so \(b - a - 2\delta = |[a+\delta,b-\delta]| \le |(a,b)|\) by 2.14 and 2.5, and letting \(\delta \downarrow 0\) gives \(b - a \le |(a,b)|\). Hence \(|(a,b)| = b - a = \ell\bigl((a,b)\bigr)\). If \(I\) is unbounded, then for every \(M > 0\) the interval \(I\) contains a closed interval \([a,b]\) with \(b - a = M\), so \(|I| \ge |[a,b]| = M\) by 2.5 and 2.14; hence \(|I| = \infty = \ell(I)\).

Step 1: the inequality \(|A| \le \sum_{k=1}^{\infty} \ell(I_k)\). The sequence \(I_1, I_2, \dots\) is itself a sequence of open intervals whose union contains \(A\), so the definition of outer measure (2.2) gives \(|A| \le \sum_{k=1}^{\infty} \ell(I_k)\). Disjointness is not needed here.

Step 2: a one-point splitting lemma. I claim that for every \(S \subseteq \mathbb{R}\) and every \(c \in \mathbb{R}\),

\begin{equation*} |S| = \bigl|S \cap (-\infty,c]\bigr| + \bigl|S \cap (c,\infty)\bigr| . \end{equation*}

The inequality \(\le\) follows from countable subadditivity (2.8), since \(S\) is the union of the two sets on the right. For the inequality \(\ge\) we may assume \(|S| < \infty\). Let \(\varepsilon > 0\) and choose open intervals \(J_1, J_2, \dots\) with \(S \subseteq \bigcup_k J_k\) and \(\sum_k \ell(J_k) \le |S| + \varepsilon\). Put \(J_k^{\prime} = J_k \cap (-\infty,c)\) and \(J_k^{\prime\prime} = J_k \cap (c,\infty)\); these are open intervals, and

\begin{equation*} \ell(J_k^{\prime}) + \ell(J_k^{\prime\prime}) = \ell(J_k) \end{equation*}

for each \(k\): this is trivial if \(J_k = \emptyset\); if \(J_k = (a,b)\) is bounded then for \(c \le a\) the pieces are \(\emptyset\) and \((a,b)\), for \(c \ge b\) they are \((a,b)\) and \(\emptyset\), and for \(a < c < b\) they are \((a,c)\) and \((c,b)\) with \((c-a)+(b-c) = b-a\); and if \(J_k\) is unbounded then at least one piece is unbounded, so both sides are \(\infty\). Now

\begin{equation*} S \cap (-\infty,c] \subseteq \Bigl( \bigcup_{k=1}^{\infty} J_k^{\prime} \Bigr) \cup \{c\} \qquad\text{and}\qquad S \cap (c,\infty) \subseteq \bigcup_{k=1}^{\infty} J_k^{\prime\prime} , \end{equation*}

so using 2.8, then 2.3 (a one-point set has outer measure \(0\)) and 2.2,

\begin{equation*} \bigl|S \cap (-\infty,c]\bigr| \le \sum_{k=1}^{\infty} \ell(J_k^{\prime}) \qquad\text{and}\qquad \bigl|S \cap (c,\infty)\bigr| \le \sum_{k=1}^{\infty} \ell(J_k^{\prime\prime}) . \end{equation*}

Adding gives \(|S \cap (-\infty,c]| + |S \cap (c,\infty)| \le \sum_k \ell(J_k) \le |S| + \varepsilon\); since \(\varepsilon > 0\) is arbitrary, the claim follows.

Step 3: splitting at finitely many points. If \(m \in \mathbb{Z}^{+}\), \(c_1 < c_2 < \cdots < c_m\) are real numbers, and \(S \subseteq \mathbb{R}\), then

\begin{equation*} |S| = \bigl|S \cap (-\infty,c_1]\bigr| + \sum_{j=1}^{m-1} \bigl|S \cap (c_j, c_{j+1}]\bigr| + \bigl|S \cap (c_m,\infty)\bigr| . \end{equation*}

Induct on \(m\). The case \(m = 1\) is Step 2. For \(m > 1\), Step 2 at \(c_1\) gives \(|S| = |S \cap (-\infty,c_1]| + |S^{\prime}|\) where \(S^{\prime} = S \cap (c_1,\infty)\). Applying the induction hypothesis to \(S^{\prime}\) with the cut points \(c_2 < \cdots < c_m\) and noting that \(S^{\prime} \cap (-\infty,c_2] = S \cap (c_1,c_2]\), that \(S^{\prime} \cap (c_j,c_{j+1}] = S \cap (c_j,c_{j+1}]\) for \(j \ge 2\), and that \(S^{\prime} \cap (c_m,\infty) = S \cap (c_m,\infty)\), completes the induction.

Step 4: the inequality \(|A| \ge \sum_{k=1}^{n} \ell(I_k)\) for each \(n\). Fix \(n \in \mathbb{Z}^{+}\). If some \(I_k\) with \(k \le n\) is unbounded, then \(|A| \ge |I_k| = \infty\) by 2.5 and Step 0, and the inequality is trivial. So assume \(I_1,\dots,I_n\) are all bounded. Empty intervals contribute \(0\) to \(\sum_{k=1}^{n} \ell(I_k)\), so we may discard them; let \((a_1,b_1),\dots,(a_p,b_p)\) be the nonempty intervals among \(I_1,\dots,I_n\), where \(a_j < b_j\). If \(p = 0\) then \(\sum_{k=1}^{n} \ell(I_k) = 0\) and the inequality is trivial, so assume \(p \ge 1\). If \((a,b)\) and \((a^{\prime},b^{\prime})\) are disjoint nonempty open intervals, then \(b \le a^{\prime}\) or \(b^{\prime} \le a\) (otherwise \(a^{\prime} < b\) and \(a < b^{\prime}\), so \(\max\{a,a^{\prime}\} < \min\{b,b^{\prime}\}\) and the intersection is nonempty). Hence these \(p\) intervals are linearly ordered, and after relabelling

\begin{equation*} a_1 < b_1 \le a_2 < b_2 \le \cdots \le a_p < b_p . \end{equation*}

Let \(c_1 < c_2 < \cdots < c_m\) be the distinct elements of \(\{a_1,b_1,\dots,a_p,b_p\}\), listed in increasing order. For each \(j\), no element of that set lies strictly between \(a_j\) and \(b_j\) (every \(a_i\) and \(b_i\) with \(i < j\) is at most \(a_j\), and every \(a_i\) and \(b_i\) with \(i > j\) is at least \(b_j\)), so \(a_j\) and \(b_j\) are consecutive entries of the list \(c_1,\dots,c_m\). Therefore each interval \((a_j,b_j]\) occurs among the intervals \((c_l, c_{l+1}]\) appearing in Step 3.

Apply Step 3 with \(S = A\) and these cut points, and discard all terms except the \(p\) terms corresponding to the intervals \((a_1,b_1],\dots,(a_p,b_p]\) (all terms are nonnegative, and the discarded terms are distinct from the kept ones because the intervals \((a_j,b_j]\) are distinct members of the partition). This gives

\begin{equation*} |A| \ge \sum_{j=1}^{p} \bigl|A \cap (a_j,b_j]\bigr| . \end{equation*}

Since \((a_j,b_j) \subseteq A \cap (a_j,b_j]\), we get \(|A \cap (a_j,b_j]| \ge |(a_j,b_j)| = b_j - a_j\) by 2.5 and Step 0. Hence

\begin{equation*} |A| \ge \sum_{j=1}^{p} (b_j - a_j) = \sum_{k=1}^{n} \ell(I_k) . \end{equation*}

Step 5: conclusion. Letting \(n \to \infty\) in Step 4 gives \(|A| \ge \sum_{k=1}^{\infty} \ell(I_k)\). Together with Step 1,

\begin{equation*} \Bigl| \bigcup_{k=1}^{\infty} I_k \Bigr| = \sum_{k=1}^{\infty} \ell(I_k), \end{equation*}

as desired.

Throughout, write

\begin{equation*} G = \bigcup_{k=1}^{\infty} \Bigl( r_k - \frac{1}{2^k},\ r_k + \frac{1}{2^k} \Bigr), \qquad\text{so that}\qquad F = \mathbb{R} \setminus G . \end{equation*}

(a) Each set \(\bigl( r_k - 2^{-k}, r_k + 2^{-k} \bigr)\) is an open subset of \(\mathbb{R}\), and an arbitrary union of open sets is open; hence \(G\) is open. Therefore \(F = \mathbb{R} \setminus G\) is the complement of an open set, which is precisely what it means for \(F\) to be closed.

(b) Suppose, for contradiction, that \(I \subseteq F\) is an interval containing two distinct elements \(x\) and \(y\), say \(x < y\). Because \(I\) is an interval containing \(x\) and \(y\), we have \([x,y] \subseteq I\). By the density of the rationals in \(\mathbb{R}\) there is a rational number \(q\) with \(x < q < y\), so \(q \in I \subseteq F\).

On the other hand, the sequence \(r_1, r_2, \dots\) contains every rational number, so \(q = r_j\) for some \(j \in \mathbb{Z}^{+}\). Then

\begin{equation*} q = r_j \in \Bigl( r_j - \frac{1}{2^{\,j}},\ r_j + \frac{1}{2^{\,j}} \Bigr) \subseteq G , \end{equation*}

so \(q \notin F\). This contradiction shows that \(I\) contains at most one element.

(c) First bound the outer measure of \(G\). The sequence of open intervals \(\bigl( r_k - 2^{-k}, r_k + 2^{-k} \bigr)\), \(k \in \mathbb{Z}^{+}\), has union \(G\), so by the definition of outer measure (2.2),

\begin{equation*} |G| \le \sum_{k=1}^{\infty} \ell\Bigl( \Bigl( r_k - \frac{1}{2^k},\ r_k + \frac{1}{2^k} \Bigr) \Bigr) = \sum_{k=1}^{\infty} \frac{2}{2^k} = 2 . \end{equation*}

Now fix \(n \in \mathbb{Z}^{+}\). Since \(F \cup G = \mathbb{R}\), we have \([-n,n] \subseteq F \cup G\), and hence by 2.5 and the subadditivity of outer measure (2.8),

\begin{equation*} |[-n,n]| \le |F \cup G| \le |F| + |G| \le |F| + 2 , \end{equation*}

the first inequality by 2.5, the second by 2.8 applied to the sequence \(F, G, \emptyset, \emptyset, \dots\). By 2.14 we have \(|[-n,n]| = 2n\), and since \(|G| \le 2 < \infty\) we may subtract, obtaining

\begin{equation*} |F| \ge 2n - 2 . \end{equation*}

This holds for every \(n \in \mathbb{Z}^{+}\), so \(|F| = \infty\).

It is worth pausing over what (b) and (c) say together: \(F\) is a closed set that contains no interval of positive length whatsoever (so \(F\) has empty interior and is nowhere dense), and yet \(F\) has infinite outer measure. Outer measure and topological size are genuinely different notions.

The set \([0,1] \cap \mathbb{Q}\) is countably infinite, so we may list it as a sequence \(q_1, q_2, \dots\) containing every rational number in \([0,1]\). Define

\begin{equation*} G = \bigcup_{k=1}^{\infty} \Bigl( q_k - \frac{\varepsilon}{2^{k+2}},\ q_k + \frac{\varepsilon}{2^{k+2}} \Bigr) \qquad\text{and}\qquad F = [0,1] \setminus G . \end{equation*}

\(F\) is closed. Each interval in the union defining \(G\) is open, so \(G\) is open, so \(\mathbb{R} \setminus G\) is closed. Since \([0,1]\) is closed and \(F = [0,1] \cap (\mathbb{R} \setminus G)\) is the intersection of two closed sets, \(F\) is closed. Also \(F \subseteq [0,1]\) by construction.

Every element of \(F\) is irrational. Suppose \(x \in F\) were rational. Then \(x \in [0,1] \cap \mathbb{Q}\), so \(x = q_k\) for some \(k \in \mathbb{Z}^{+}\), and hence

\begin{equation*} x = q_k \in \Bigl( q_k - \frac{\varepsilon}{2^{k+2}},\ q_k + \frac{\varepsilon}{2^{k+2}} \Bigr) \subseteq G , \end{equation*}

contradicting \(x \notin G\). Thus every element of \(F\) is irrational.

The estimate \(|F| > 1 - \varepsilon\). The intervals used to define \(G\) form a sequence of open intervals whose union is \(G\), so by the definition of outer measure (2.2),

\begin{equation*} |G| \le \sum_{k=1}^{\infty} \frac{2\varepsilon}{2^{k+2}} = \frac{\varepsilon}{2} \sum_{k=1}^{\infty} \frac{1}{2^{k}} = \frac{\varepsilon}{2} . \end{equation*}

Since \([0,1] \subseteq F \cup G\) (a point of \([0,1]\) either lies in \(G\) or, failing that, lies in \(F\)), the order-preserving property (2.5) and the subadditivity of outer measure (2.8), together with \(|[0,1]| = 1\) from 2.14, give

\begin{equation*} 1 = |[0,1]| \le |F \cup G| \le |F| + |G| \le |F| + \frac{\varepsilon}{2} . \end{equation*}

Because \(|G| \le \varepsilon/2 < \infty\), we may subtract it. Hence

\begin{equation*} |F| \ge 1 - \frac{\varepsilon}{2} > 1 - \varepsilon , \end{equation*}

the strict inequality because \(\varepsilon > 0\). This \(F\) has all three required properties.

Note that \(F\) contains no interval with two distinct elements: any such interval would contain a rational number, by the density of \(\mathbb{Q}\), and \(F\) contains no rational number. So \(F\) is a closed, nowhere dense subset of \([0,1]\) whose outer measure is nevertheless as close to \(1\) as we please. Contrast this with the fact that a countable set, however dense, has outer measure \(0\) by 2.4.

(a) Call this triangle \(T\). The side from \((0,0)\) to \((20,0)\) is horizontal and the side from \((20,0)\) to \((20,9)\) is vertical, so \(T\) has its right angle at \((20,0)\), with legs of length \(20\) and \(9\). Hence

\begin{equation*} \operatorname{area}(T) = \tfrac{1}{2} \cdot 20 \cdot 9 = 90 . \end{equation*}

(b) The yellow triangle has vertices \((0,0)\), \((11,0)\), \((11,5)\), with its right angle at \((11,0)\) and legs of length \(11\) and \(5\). Hence its area is

\begin{equation*} \tfrac{1}{2} \cdot 11 \cdot 5 = \tfrac{55}{2} = 27.5 . \end{equation*}

(c) The red rectangle has vertices \((11,0)\), \((20,0)\), \((20,5)\), \((11,5)\); its horizontal side has length \(20 - 11 = 9\) and its vertical side has length \(5 - 0 = 5\). Hence its area is

\begin{equation*} 9 \cdot 5 = 45 . \end{equation*}

(d) The blue triangle has vertices \((11,5)\), \((20,5)\), \((20,9)\), with its right angle at \((20,5)\) and legs of length \(20 - 11 = 9\) and \(9 - 5 = 4\). Hence its area is

\begin{equation*} \tfrac{1}{2} \cdot 9 \cdot 4 = 18 . \end{equation*}

(e) The three colored pieces overlap only along parts of their boundaries: the yellow triangle meets the red rectangle in the segment \(\{11\} \times [0,5]\), the red rectangle meets the blue triangle in the segment \([11,20] \times \{5\}\), and the yellow and blue triangles meet only at the single point \((11,5)\). Segments and points contribute no area, so the area of the colored region is the sum of the three areas:

\begin{equation*} 27.5 + 45 + 18 = 90.5 . \end{equation*}

(f) The colored region is simply not the triangle of part (a). There is no paradox, and no failure of additivity; the mistake is the visual assumption that the three pieces fit together to fill \(T\).

The union \(Q\) of the three colored pieces is the quadrilateral with vertices \((0,0)\), \((20,0)\), \((20,9)\), \((11,5)\), taken in that (cyclic) order. The point \((11,5)\) is a genuine corner of \(Q\), not a point on a straight edge. Indeed, the line through \((0,0)\) and \((20,9)\) is

\begin{equation*} y = \tfrac{9}{20} x , \end{equation*}

and at \(x = 11\) this line has height \(\tfrac{99}{20} = 4.95\), whereas the colored region reaches up to height \(5\) there. Equivalently, compare slopes: the hypotenuse of the yellow triangle has slope \(\tfrac{5}{11} = 0.4545\ldots\), the hypotenuse of the blue triangle has slope \(\tfrac{4}{9} = 0.4444\ldots\), and the hypotenuse of \(T\) has slope \(\tfrac{9}{20} = 0.45\). These three numbers are distinct, so the upper boundary of the colored region is a bent path, not a line segment. The bend is tiny (the two slopes differ by about \(0.01\)), which is exactly why the figure is deceptive even though it is drawn accurately to scale.

Since \(5 > 4.95\), the bend goes upward, so \(T \subseteq Q\) and \(Q\) is strictly larger than \(T\). The difference \(Q \setminus T\) is the very thin triangle \(S\) with vertices \((0,0)\), \((11,5)\), \((20,9)\). Its area is computed as follows: take as base the segment from \((0,0)\) to \((20,9)\), of length \(\sqrt{20^2 + 9^2} = \sqrt{481}\); the corresponding height is the distance from \((11,5)\) to the line \(9x - 20y = 0\), namely

\begin{equation*} \frac{|9 \cdot 11 - 20 \cdot 5|}{\sqrt{9^2 + 20^2}} = \frac{1}{\sqrt{481}} . \end{equation*}

Hence

\begin{equation*} \operatorname{area}(S) = \tfrac{1}{2} \cdot \sqrt{481} \cdot \frac{1}{\sqrt{481}} = \tfrac{1}{2} , \end{equation*}

and \(\operatorname{area}(Q) = \operatorname{area}(T) + \operatorname{area}(S) = 90 + 0.5 = 90.5\), in agreement with part (e). (The same value follows from the shoelace formula applied to the vertices \((0,0)\), \((20,0)\), \((20,9)\), \((11,5)\).)

Finally, to address the bracketed remark: this example has nothing to do with the nonadditivity of outer measure (2.18). Here the three pieces are nonoverlapping and their areas do add up to the area of their union, which is \(90.5\). What fails is only the false identification of that union with the triangle \(T\). The genuinely nonadditive sets of 2.18 are produced by the Axiom of Choice and are far too pathological to be drawn; no picture assembled from polygons could ever exhibit nonadditivity.

For \(K \subseteq \mathbb{Z}\) write

\begin{equation*} E_K = \bigcup_{n \in K} (n, n+1], \end{equation*}

so that \(\mathcal{S} = \{E_K : K \subseteq \mathbb{Z}\}\). Everything follows from the observation that the intervals \((n, n+1]\), for \(n \in \mathbb{Z}\), form a partition of \(\mathbb{R}\).

Step 1: each \(x \in \mathbb{R}\) lies in exactly one of the intervals \((n, n+1]\).

Existence: let \(n_x = \lceil x \rceil - 1\), where \(\lceil x \rceil\) denotes the smallest integer that is greater than or equal to \(x\). If \(x \in \mathbb{Z}\) then \(\lceil x \rceil = x\) and \(n_x = x - 1\), so \(n_x < x \le n_x + 1\). If \(x \notin \mathbb{Z}\) then \(\lceil x \rceil - 1 < x < \lceil x \rceil\), so again \(n_x < x \le n_x + 1\). Either way \(x \in (n_x, n_x + 1]\).

Uniqueness: suppose \(m, n \in \mathbb{Z}\), \(m < n\), and \(x \in (m, m+1] \cap (n, n+1]\). Since \(m < n\) and both are integers, \(n \ge m + 1\). Thus \(x \le m + 1 \le n\), contradicting \(x > n\). Hence no \(x\) lies in two of these intervals, and \(n_x\) is the unique integer with \(x \in (n_x, n_x+1]\).

Step 2: for every \(K \subseteq \mathbb{Z}\) and every \(x \in \mathbb{R}\),

\begin{equation*} x \in E_K \iff n_x \in K . \end{equation*}

Indeed, if \(n_x \in K\) then \(x \in (n_x, n_x+1] \subseteq E_K\). Conversely, if \(x \in E_K\) then \(x \in (n, n+1]\) for some \(n \in K\), and Step 1 forces \(n = n_x\), so \(n_x \in K\). In particular \(K \mapsto E_K\) is injective, though we shall not need that.

Now verify the three conditions in the definition of a \(\sigma\)-algebra (2.23).

The empty set: taking \(K = \emptyset\) gives \(E_\emptyset = \emptyset\) (an empty union), so \(\emptyset \in \mathcal{S}\).

Closure under complementation: we claim that

\begin{equation*} \mathbb{R} \setminus E_K = E_{\mathbb{Z} \setminus K} \end{equation*}

for every \(K \subseteq \mathbb{Z}\). Let \(x \in \mathbb{R}\). By Step 2, \(x \notin E_K\) if and only if \(n_x \notin K\), which (as \(n_x \in \mathbb{Z}\)) holds if and only if \(n_x \in \mathbb{Z} \setminus K\), which by Step 2 again holds if and only if \(x \in E_{\mathbb{Z} \setminus K}\). This proves the claim, and since \(\mathbb{Z} \setminus K \subseteq \mathbb{Z}\) we conclude \(\mathbb{R} \setminus E_K \in \mathcal{S}\).

Closure under countable unions: suppose \(E_{K_1}, E_{K_2}, \ldots\) is a sequence of elements of \(\mathcal{S}\), with each \(K_j \subseteq \mathbb{Z}\). Put \(K = \bigcup_{j=1}^\infty K_j \subseteq \mathbb{Z}\). Then for \(x \in \mathbb{R}\), using Step 2 three times,

\begin{equation*} x \in \bigcup_{j=1}^\infty E_{K_j} \iff n_x \in K_j \text{ for some } j \iff n_x \in K \iff x \in E_K . \end{equation*}

Hence \(\bigcup_{j=1}^\infty E_{K_j} = E_K \in \mathcal{S}\).

Thus \(\mathcal{S}\) contains \(\emptyset\), is closed under complementation in \(\mathbb{R}\), and is closed under countable unions; that is, \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\).

(Note in passing that \(\mathbb{R} = E_{\mathbb{Z}} \in \mathcal{S}\), in agreement with 2.25(a), and that every element of \(\mathcal{S}\) is a Borel set, since each \((n, n+1]\) is a Borel set and the unions involved are countable. So \(\mathcal{S}\) is a \(\sigma\)-algebra contained in the Borel \(\sigma\)-algebra.)

Throughout we use the standard strategy for identifying a smallest \(\sigma\)-algebra containing a collection \(\mathcal{A}\): exhibit a candidate collection \(\mathcal{T}\), show (i) \(\mathcal{T}\) is a \(\sigma\)-algebra, (ii) \(\mathcal{A} \subseteq \mathcal{T}\), and (iii) every \(\sigma\)-algebra containing \(\mathcal{A}\) contains \(\mathcal{T}\). By 2.27 the smallest \(\sigma\)-algebra containing \(\mathcal{A}\) is the intersection of all \(\sigma\)-algebras containing \(\mathcal{A}\); (i) and (ii) say \(\mathcal{T}\) is one of the collections being intersected, so the intersection is contained in \(\mathcal{T}\), while (iii) says \(\mathcal{T}\) is contained in each of them, hence in the intersection. Therefore \(\mathcal{T}\) is exactly the smallest such \(\sigma\)-algebra.

Part (a). Let

\begin{equation*} \mathcal{T} = \{E \subseteq X : E \text{ is countable or } X \setminus E \text{ is countable}\}. \end{equation*}

(i) \(\mathcal{T}\) is a \(\sigma\)-algebra on \(X\). First, \(\emptyset\) is countable, so \(\emptyset \in \mathcal{T}\). Next, the defining condition on \(E\) is symmetric in \(E\) and \(X \setminus E\) (because \(X \setminus (X \setminus E) = E\)), so \(E \in \mathcal{T}\) implies \(X \setminus E \in \mathcal{T}\). Finally, suppose \(E_1, E_2, \ldots \in \mathcal{T}\) and let \(E = \bigcup_{k=1}^\infty E_k\). If every \(E_k\) is countable, then \(E\) is a countable union of countable sets and hence is countable, so \(E \in \mathcal{T}\). Otherwise there is some \(m\) with \(E_m\) uncountable; since \(E_m \in \mathcal{T}\), the set \(X \setminus E_m\) must be countable. Because \(E \supseteq E_m\) we have \(X \setminus E \subseteq X \setminus E_m\), so \(X \setminus E\) is countable and again \(E \in \mathcal{T}\). Thus \(\mathcal{T}\) is a \(\sigma\)-algebra.

(ii) \(\mathcal{A} \subseteq \mathcal{T}\), since each \(\{x\}\) is finite and therefore countable.

(iii) Suppose \(\mathcal{S}\) is a \(\sigma\)-algebra on \(X\) with \(\mathcal{A} \subseteq \mathcal{S}\), and suppose \(E \in \mathcal{T}\). If \(E\) is countable, then either \(E = \emptyset \in \mathcal{S}\), or we may list \(E = \{x_1, x_2, \ldots\}\) (a finite or infinite list) and write

\begin{equation*} E = \bigcup_{k} \{x_k\}, \end{equation*}

a countable union of elements of \(\mathcal{A} \subseteq \mathcal{S}\); hence \(E \in \mathcal{S}\) (pad a finite list with copies of \(\emptyset\) to get a sequence). If instead \(X \setminus E\) is countable, the case just treated gives \(X \setminus E \in \mathcal{S}\), and then \(E = X \setminus (X \setminus E) \in \mathcal{S}\). So \(\mathcal{T} \subseteq \mathcal{S}\).

By the strategy above, \(\mathcal{T}\) is the smallest \(\sigma\)-algebra on \(X\) containing \(\mathcal{A}\), which is the first bullet point.

Part (b). Set

\begin{equation*} A_1 = (0,1), \qquad A_2 = [1,\infty), \qquad A_3 = (-\infty, 0]. \end{equation*}

These three sets are nonempty, pairwise disjoint, and their union is \(\mathbb{R}\); that is, \(\{A_1, A_2, A_3\}\) is a partition of \(\mathbb{R}\). For \(K \subseteq \{1,2,3\}\) put \(E_K = \bigcup_{i \in K} A_i\). Listing the eight subsets \(K\) gives

\begin{equation*} \begin{aligned} E_\emptyset &= \emptyset, & E_{\{1\}} &= (0,1), & E_{\{2\}} &= [1,\infty), & E_{\{3\}} &= (-\infty,0],\\ E_{\{1,2\}} &= (0,\infty), & E_{\{1,3\}} &= (-\infty,1), & E_{\{2,3\}} &= (-\infty,0] \cup [1,\infty), & E_{\{1,2,3\}} &= \mathbb{R}, \end{aligned} \end{equation*}

so the collection \(\mathcal{T} = \{E_K : K \subseteq \{1,2,3\}\}\) is precisely the eight-element collection displayed in the exercise. (The eight sets are distinct: since the \(A_i\) are nonempty and pairwise disjoint, \(E_K\) determines \(K\) via \(i \in K \iff A_i \subseteq E_K\).)

(i) \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). Since \(\{A_1,A_2,A_3\}\) partitions \(\mathbb{R}\), each \(x \in \mathbb{R}\) lies in exactly one \(A_{i(x)}\), and \(x \in E_K \iff i(x) \in K\). Exactly as in Exercise 1 this yields

\begin{equation*} \mathbb{R} \setminus E_K = E_{\{1,2,3\} \setminus K} \qquad\text{and}\qquad \bigcup_{j=1}^\infty E_{K_j} = E_{K_1 \cup K_2 \cup \cdots}, \end{equation*}

and \(\emptyset = E_\emptyset \in \mathcal{T}\). Hence \(\mathcal{T}\) is closed under complementation and countable unions and contains \(\emptyset\), so it is a \(\sigma\)-algebra.

(ii) \(\mathcal{A} \subseteq \mathcal{T}\), because \((0,1) = E_{\{1\}}\) and \((0,\infty) = E_{\{1,2\}}\).

(iii) Suppose \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) with \((0,1), (0,\infty) \in \mathcal{S}\). Then

\begin{equation*} A_1 = (0,1) \in \mathcal{S}, \qquad A_2 = [1,\infty) = (0,\infty) \setminus (0,1) \in \mathcal{S}, \qquad A_3 = (-\infty,0] = \mathbb{R} \setminus (0,\infty) \in \mathcal{S}, \end{equation*}

using 2.25(a) and 2.25(b). Since \(\mathcal{S}\) is closed under finite unions (2.25(b)), every \(E_K\) is in \(\mathcal{S}\). Thus \(\mathcal{T} \subseteq \mathcal{S}\).

By the strategy above, \(\mathcal{T}\) is the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing \(\{(0,1), (0,\infty)\}\), which is the second bullet point.

Write \(\mathcal{B}\) for the collection of Borel subsets of \(\mathbb{R}\), which by 2.29 is the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing all open subsets of \(\mathbb{R}\). We prove the two inclusions \(\mathcal{S} \subseteq \mathcal{B}\) and \(\mathcal{B} \subseteq \mathcal{S}\).

Step 1: \(\mathcal{S} \subseteq \mathcal{B}\).

It suffices to show that \((r,s] \in \mathcal{B}\) for all \(r, s \in \mathbb{Q}\), for then \(\mathcal{B}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) containing \(\{(r,s] : r, s \in \mathbb{Q}\}\), and \(\mathcal{S}\), being the smallest such \(\sigma\)-algebra, is contained in \(\mathcal{B}\) (see the remark following 2.27).

If \(r \ge s\) then \((r,s] = \emptyset \in \mathcal{B}\). If \(r < s\) then

\begin{equation*} (r,s] = \bigcap_{k=1}^\infty \left(r,\ s + \tfrac{1}{k}\right), \end{equation*}

because a real number \(x\) satisfies \(r < x \le s\) if and only if \(r < x < s + \frac1k\) for every \(k \in \mathbb{Z}^+\) (the forward implication is clear; conversely, \(x < s + \frac1k\) for all \(k\) forces \(x \le s\)). Each set \(\left(r, s + \frac1k\right)\) is open, hence lies in \(\mathcal{B}\), and \(\mathcal{B}\) is closed under countable intersections by 2.25(c). Thus \((r,s] \in \mathcal{B}\).

Step 2: \(\mathcal{B} \subseteq \mathcal{S}\).

Since \(\mathcal{B}\) is the smallest \(\sigma\)-algebra containing the open subsets of \(\mathbb{R}\), it suffices to show that every open subset of \(\mathbb{R}\) belongs to \(\mathcal{S}\).

So let \(G \subseteq \mathbb{R}\) be open, and let

\begin{equation*} P = \{(r,s) \in \mathbb{Q} \times \mathbb{Q} : (r,s] \subseteq G\}. \end{equation*}

We claim

\begin{equation*} G = \bigcup_{(r,s) \in P} (r,s]. \end{equation*}

The inclusion \(\supseteq\) holds because every set in the union is contained in \(G\) by definition of \(P\). For the inclusion \(\subseteq\), suppose \(x \in G\). Because \(G\) is open there exists \(\varepsilon > 0\) with \((x - \varepsilon, x + \varepsilon) \subseteq G\). By the density of \(\mathbb{Q}\) in \(\mathbb{R}\) choose rationals \(r\) and \(s\) with

\begin{equation*} x - \varepsilon < r < x \qquad\text{and}\qquad x < s < x + \varepsilon . \end{equation*}

Then \(x \in (r,s]\) and \((r,s] \subseteq (x-\varepsilon, x+\varepsilon) \subseteq G\), so \((r,s) \in P\) and \(x\) belongs to the right side. This proves the claim.

Now \(P \subseteq \mathbb{Q} \times \mathbb{Q}\) is a countable set, so the union above is a countable union of elements of \(\mathcal{S}\) (if \(P = \emptyset\) then \(G = \emptyset \in \mathcal{S}\); otherwise list \(P\) as a finite or infinite sequence, padding with repetitions if necessary to obtain a sequence). Because \(\mathcal{S}\) is closed under countable unions, \(G \in \mathcal{S}\).

Thus \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) containing every open subset of \(\mathbb{R}\), so \(\mathcal{B} \subseteq \mathcal{S}\).

Combining the two steps, \(\mathcal{S} = \mathcal{B}\); that is, \(\mathcal{S}\) is exactly the collection of Borel subsets of \(\mathbb{R}\).

Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\).

Step 1: \(\mathcal{S} \subseteq \mathcal{B}\).

Every integer is rational, so \(\{(r,n] : r \in \mathbb{Q},\, n \in \mathbb{Z}\} \subseteq \{(r,s] : r,s \in \mathbb{Q}\}\), and Step 1 of Exercise 3 showed each such half-open interval is a Borel set: either it is empty, or

\begin{equation*} (r,n] = \bigcap_{k=1}^\infty \left(r,\ n + \tfrac1k\right), \end{equation*}

a countable intersection of open sets, which lies in \(\mathcal{B}\) by 2.25(c). Hence \(\mathcal{B}\) is a \(\sigma\)-algebra containing the generating collection, and therefore contains the smallest such \(\sigma\)-algebra, namely \(\mathcal{S}\).

Step 2: \(\mathcal{B} \subseteq \mathcal{S}\).

By Exercise 3, the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing \(\{(r,s] : r,s \in \mathbb{Q}\}\) equals \(\mathcal{B}\). So it suffices to prove that \((r,s] \in \mathcal{S}\) for all \(r,s \in \mathbb{Q}\); the smallest \(\sigma\)-algebra containing all such intervals is then contained in \(\mathcal{S}\), giving \(\mathcal{B} \subseteq \mathcal{S}\).

Let \(r, s \in \mathbb{Q}\). If \(r \ge s\) then \((r,s] = \emptyset \in \mathcal{S}\). So assume \(r < s\), and choose an integer \(n\) with \(n \ge s\) (possible by the Archimedean property). We claim

\begin{equation*} (r,s] = (r,n] \setminus (s,n]. \end{equation*}

To see this, let \(x \in \mathbb{R}\). If \(r < x \le s\), then, since \(s \le n\), we have \(r < x \le n\), so \(x \in (r,n]\); and \(x \le s\) shows \(x \notin (s,n]\). Conversely, suppose \(x \in (r,n]\) and \(x \notin (s,n]\). From \(x \in (r,n]\) we get \(r < x \le n\); the failure of \(s < x \le n\) then forces \(x \le s\). Hence \(r < x \le s\). This proves the claim.

Both \((r,n]\) and \((s,n]\) are in \(\mathcal{S}\) (they are among the generators, as \(r, s \in \mathbb{Q}\) and \(n \in \mathbb{Z}\)), and \(\sigma\)-algebras are closed under set differences by 2.25(b). Therefore \((r,s] \in \mathcal{S}\).

Combining the two steps gives \(\mathcal{S} = \mathcal{B}\), so \(\mathcal{S}\) is the collection of Borel subsets of \(\mathbb{R}\).

Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\).

Step 1: \(\mathcal{S} \subseteq \mathcal{B}\).

Each interval \((r, r+1)\) is an open subset of \(\mathbb{R}\) and hence is a Borel set. Thus \(\mathcal{B}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) containing \(\{(r,r+1) : r \in \mathbb{Q}\}\), and therefore contains the smallest such \(\sigma\)-algebra, which is \(\mathcal{S}\).

Step 2: short rational intervals lie in \(\mathcal{S}\).

We claim that if \(p, q \in \mathbb{Q}\) satisfy \(p < q\) and \(q - p \le 1\), then \((p,q) \in \mathcal{S}\). Indeed, both \(p\) and \(q-1\) are rational, so

\begin{equation*} (p, p+1) \in \mathcal{S} \qquad\text{and}\qquad (q-1, q) = \bigl((q-1),\ (q-1)+1\bigr) \in \mathcal{S}, \end{equation*}

and \(\sigma\)-algebras are closed under finite intersections by 2.25(b). Now

\begin{equation*} (p, p+1) \cap (q-1, q) = \bigl(\max\{p,\ q-1\},\ \min\{p+1,\ q\}\bigr) = (p, q), \end{equation*}

where the last equality uses \(q - p \le 1\), which gives \(q - 1 \le p\) and \(q \le p+1\). Hence \((p,q) \in \mathcal{S}\).

Step 3: \(\mathcal{B} \subseteq \mathcal{S}\).

Since \(\mathcal{B}\) is by 2.29 the smallest \(\sigma\)-algebra containing the open subsets of \(\mathbb{R}\), it is enough to show that every open \(G \subseteq \mathbb{R}\) lies in \(\mathcal{S}\).

Let \(G \subseteq \mathbb{R}\) be open and set

\begin{equation*} P = \{(p,q) \in \mathbb{Q} \times \mathbb{Q} : p < q,\ q - p \le 1,\ (p,q) \subseteq G\}. \end{equation*}

We claim \(G = \bigcup_{(p,q) \in P} (p,q)\). The inclusion \(\supseteq\) is immediate from the definition of \(P\). For \(\subseteq\), let \(x \in G\). Because \(G\) is open there exists \(\varepsilon > 0\) with \((x - \varepsilon, x + \varepsilon) \subseteq G\), and we may shrink \(\varepsilon\) so that \(\varepsilon < \frac12\). By density of \(\mathbb{Q}\) in \(\mathbb{R}\) choose rationals \(p, q\) with

\begin{equation*} x - \varepsilon < p < x \qquad\text{and}\qquad x < q < x + \varepsilon . \end{equation*}

Then \(p < x < q\), and \(q - p < 2\varepsilon < 1\), and \((p,q) \subseteq (x-\varepsilon, x+\varepsilon) \subseteq G\). So \((p,q) \in P\) and \(x \in (p,q)\), proving the claim.

The index set \(P\) is a subset of the countable set \(\mathbb{Q} \times \mathbb{Q}\), so the union above is a countable union (if \(P = \emptyset\) then \(G = \emptyset \in \mathcal{S}\); otherwise enumerate \(P\) as a sequence, repeating terms if \(P\) is finite). By Step 2 each set in the union is in \(\mathcal{S}\), and \(\mathcal{S}\) is closed under countable unions. Hence \(G \in \mathcal{S}\).

Therefore \(\mathcal{S}\) is a \(\sigma\)-algebra containing every open subset of \(\mathbb{R}\), so \(\mathcal{B} \subseteq \mathcal{S}\). Combined with Step 1 this gives \(\mathcal{S} = \mathcal{B}\), so \(\mathcal{S}\) is the collection of Borel subsets of \(\mathbb{R}\).

Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\).

Step 1: \(\mathcal{S} \subseteq \mathcal{B}\).

For each \(r \in \mathbb{Q}\) the set \([r,\infty)\) is a closed subset of \(\mathbb{R}\), because its complement \((-\infty, r)\) is open. By the first bullet point of 2.30, every closed subset of \(\mathbb{R}\) is a Borel set. Hence \(\mathcal{B}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) containing \(\{[r,\infty) : r \in \mathbb{Q}\}\), and therefore contains the smallest such \(\sigma\)-algebra, namely \(\mathcal{S}\).

Step 2: bounded open intervals with rational endpoints lie in \(\mathcal{S}\).

First, for each \(p \in \mathbb{Q}\),

\begin{equation*} (p, \infty) = \bigcup_{k=1}^\infty \left[p + \tfrac1k,\ \infty\right). \end{equation*}

Indeed, if \(x \ge p + \frac1k\) for some \(k \in \mathbb{Z}^+\) then \(x > p\); conversely, if \(x > p\) then by the Archimedean property there is \(k \in \mathbb{Z}^+\) with \(\frac1k \le x - p\), so \(x \in \left[p + \frac1k, \infty\right)\). Each \(p + \frac1k\) is rational, so each set in this union is a generator of \(\mathcal{S}\); as \(\mathcal{S}\) is closed under countable unions, \((p,\infty) \in \mathcal{S}\).

Second, for each \(q \in \mathbb{Q}\),

\begin{equation*} (-\infty, q) = \mathbb{R} \setminus [q, \infty) \in \mathcal{S}, \end{equation*}

since \(\mathcal{S}\) is closed under complementation.

Therefore, for all \(p, q \in \mathbb{Q}\),

\begin{equation*} (p,q) = (p,\infty) \cap (-\infty, q) \in \mathcal{S}, \end{equation*}

using closure of \(\mathcal{S}\) under finite intersections (2.25(b)). (When \(p \ge q\) this set is empty, which is also in \(\mathcal{S}\).)

Step 3: \(\mathcal{B} \subseteq \mathcal{S}\).

By 2.29 it suffices to show that every open subset of \(\mathbb{R}\) belongs to \(\mathcal{S}\). Let \(G \subseteq \mathbb{R}\) be open and put

\begin{equation*} P = \{(p,q) \in \mathbb{Q} \times \mathbb{Q} : p < q \text{ and } (p,q) \subseteq G\}. \end{equation*}

We claim \(G = \bigcup_{(p,q) \in P} (p,q)\). The inclusion \(\supseteq\) holds by the definition of \(P\). For \(\subseteq\), let \(x \in G\); openness of \(G\) gives \(\varepsilon > 0\) with \((x-\varepsilon, x+\varepsilon) \subseteq G\), and by density of \(\mathbb{Q}\) we may choose rationals \(p \in (x - \varepsilon, x)\) and \(q \in (x, x + \varepsilon)\). Then \(x \in (p,q) \subseteq (x-\varepsilon, x+\varepsilon) \subseteq G\), so \((p,q) \in P\) and \(x\) lies in the right side. This proves the claim.

Since \(P \subseteq \mathbb{Q} \times \mathbb{Q}\) is countable, the displayed union is a countable union of sets that lie in \(\mathcal{S}\) by Step 2 (if \(P = \emptyset\) then \(G = \emptyset \in \mathcal{S}\)). Hence \(G \in \mathcal{S}\).

Thus \(\mathcal{S}\) contains every open subset of \(\mathbb{R}\), so \(\mathcal{B} \subseteq \mathcal{S}\). With Step 1 this gives \(\mathcal{S} = \mathcal{B}\), so \(\mathcal{S}\) is the collection of Borel subsets of \(\mathbb{R}\).

Here \(t + B = \{t + b : b \in B\}\). Let \(\mathcal{B}\) denote the collection of Borel subsets of \(\mathbb{R}\), and fix \(t \in \mathbb{R}\).

A short proof. Define \(f : \mathbb{R} \to \mathbb{R}\) by \(f(x) = x - t\). Then \(f\) is continuous, so \(f\) is a Borel measurable function by 2.41. For \(B \subseteq \mathbb{R}\) we have

\begin{equation*} f^{-1}(B) = \{x \in \mathbb{R} : x - t \in B\} = \{t + b : b \in B\} = t + B, \end{equation*}

where the middle equality holds because \(x - t \in B\) if and only if \(x = t + b\) for some \(b \in B\). If \(B\) is a Borel set, then Borel measurability of \(f\) (see 2.40) gives that \(f^{-1}(B) = t + B\) is a Borel set.

A self-contained proof, not using 2.41. Define

\begin{equation*} \mathcal{T} = \{A \subseteq \mathbb{R} : t + A \in \mathcal{B}\}. \end{equation*}

We show that \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) that contains every open subset of \(\mathbb{R}\); it then follows from 2.29 that \(\mathcal{B} \subseteq \mathcal{T}\), which is exactly the assertion that \(t + B\) is a Borel set for every Borel set \(B\).

First, \(t + \emptyset = \emptyset \in \mathcal{B}\), so \(\emptyset \in \mathcal{T}\).

Second, for every \(A \subseteq \mathbb{R}\),

\begin{equation*} t + (\mathbb{R} \setminus A) = \mathbb{R} \setminus (t + A). \end{equation*}

Indeed, for \(x \in \mathbb{R}\) we have \(x \in t + A\) if and only if \(x - t \in A\); hence \(x \in t + (\mathbb{R} \setminus A)\) if and only if \(x - t \notin A\), if and only if \(x \notin t + A\). So if \(A \in \mathcal{T}\), then \(t + A \in \mathcal{B}\), and since \(\mathcal{B}\) is closed under complementation we get \(t + (\mathbb{R} \setminus A) = \mathbb{R} \setminus (t+A) \in \mathcal{B}\), that is, \(\mathbb{R} \setminus A \in \mathcal{T}\).

Third, if \(A_1, A_2, \ldots\) is a sequence of subsets of \(\mathbb{R}\), then

\begin{equation*} t + \bigcup_{k=1}^\infty A_k = \bigcup_{k=1}^\infty (t + A_k), \end{equation*}

because \(x - t \in \bigcup_k A_k\) if and only if \(x - t \in A_k\) for some \(k\). So if each \(A_k \in \mathcal{T}\), then each \(t + A_k \in \mathcal{B}\), and \(\mathcal{B}\) is closed under countable unions, giving \(\bigcup_{k=1}^\infty A_k \in \mathcal{T}\).

Thus \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}\).

Finally, suppose \(G \subseteq \mathbb{R}\) is open; we check \(t + G\) is open. Let \(x \in t + G\), so \(x - t \in G\). Since \(G\) is open there is \(\varepsilon > 0\) with \((x - t - \varepsilon,\ x - t + \varepsilon) \subseteq G\). If \(|y - x| < \varepsilon\) then \(|(y - t) - (x - t)| < \varepsilon\), so \(y - t \in G\) and hence \(y \in t + G\). Thus \((x - \varepsilon, x + \varepsilon) \subseteq t + G\), proving \(t + G\) is open. Every open set is a Borel set, so \(t + G \in \mathcal{B}\) and therefore \(G \in \mathcal{T}\).

So \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) containing all open subsets of \(\mathbb{R}\). Because \(\mathcal{B}\) is the smallest such \(\sigma\)-algebra (2.29), we conclude \(\mathcal{B} \subseteq \mathcal{T}\). In other words, if \(B\) is a Borel subset of \(\mathbb{R}\) and \(t \in \mathbb{R}\), then \(t + B\) is a Borel set. Hence the collection of Borel subsets of \(\mathbb{R}\) is translation invariant.

Suppose \(B \subseteq \mathbb{R}\) is a Borel set and \(t \in \mathbb{R}\).

First dispose of the degenerate case \(t = 0\). If \(B = \emptyset\) then \(0B = \emptyset\), and if \(B \neq \emptyset\) then \(0B = \{0\}\). Both \(\emptyset\) and \(\{0\}\) are Borel sets (\(\{0\}\) is closed, hence Borel by 2.30), so \(tB\) is a Borel set when \(t = 0\).

Now suppose \(t \neq 0\). Define \(m : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} m(x) = \frac{x}{t}. \end{equation*}

Then \(m\) is continuous, and \(\mathbb{R}\) is a Borel set, so 2.41 tells us that \(m\) is a Borel measurable function. Hence \(m^{-1}(B)\) is a Borel set, by the definition 2.40 of Borel measurability.

We now check that \(m^{-1}(B) = tB\). For \(x \in \mathbb{R}\) we have, using 2.31,

\begin{equation*} x \in m^{-1}(B) \iff \frac{x}{t} \in B . \end{equation*}

If \(x/t \in B\), then \(x = t\,(x/t) \in tB\). Conversely, if \(x \in tB\), then \(x = tb\) for some \(b \in B\), and since \(t \neq 0\) this gives \(x/t = b \in B\). Thus

\begin{equation*} m^{-1}(B) = tB, \end{equation*}

and therefore \(tB\) is a Borel set.

Combining the two cases: \(tB\) is a Borel set for every Borel set \(B \subseteq \mathbb{R}\) and every \(t \in \mathbb{R}\). In other words, the collection of Borel subsets of \(\mathbb{R}\) is dilation invariant.

It is worth noting the alternative argument, which avoids 2.41 and mirrors the proof of Exercise 7 in this section. Fix \(t \neq 0\) and let

\begin{equation*} \mathcal{S} = \{ B \subseteq \mathbb{R} : tB \text{ is a Borel set} \}. \end{equation*}

Because \(t\emptyset = \emptyset\), we have \(\emptyset \in \mathcal{S}\). Because multiplication by \(t\) is a bijection of \(\mathbb{R}\) onto \(\mathbb{R}\), we have \(t(\mathbb{R} \setminus B) = \mathbb{R} \setminus tB\) and \(t\bigl(\bigcup_{k=1}^{\infty} B_k\bigr) = \bigcup_{k=1}^{\infty} tB_k\); hence \(\mathcal{S}\) is closed under complementation and under countable unions, so \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). If \(G \subseteq \mathbb{R}\) is open, then \(tG\) is open (multiplication by \(t\) is a homeomorphism) and thus Borel, so \(G \in \mathcal{S}\). Hence \(\mathcal{S}\) is a \(\sigma\)-algebra containing every open subset of \(\mathbb{R}\), and so \(\mathcal{S}\) contains the smallest such \(\sigma\)-algebra, which by 2.29 is the collection of Borel sets. Thus \(tB\) is a Borel set for every Borel set \(B\).

Take \(X = \mathbb{R}\) and \(\mathcal{S} = \{\emptyset, \mathbb{R}\}\). This is a \(\sigma\)-algebra on \(\mathbb{R}\) (it contains \(\emptyset\), it is closed under complementation because \(\mathbb{R} \setminus \emptyset = \mathbb{R}\) and \(\mathbb{R} \setminus \mathbb{R} = \emptyset\), and any union of sets from \(\mathcal{S}\) is either \(\emptyset\) or \(\mathbb{R}\)), so \((\mathbb{R}, \mathcal{S})\) is a measurable space.

Define \(f : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f(x) = \begin{cases} 1 & \text{if } x \geq 0,\\ -1 & \text{if } x < 0. \end{cases} \end{equation*}

Then \(|f|\) is the constant function with value \(1\). For every Borel set \(B \subseteq \mathbb{R}\) we have \(|f|^{-1}(B) = \mathbb{R}\) if \(1 \in B\) and \(|f|^{-1}(B) = \emptyset\) if \(1 \notin B\); in either case \(|f|^{-1}(B) \in \mathcal{S}\). Hence \(|f|\) is \(\mathcal{S}\)-measurable. (This is also the content of the first bullet point of 2.36: when \(\mathcal{S} = \{\emptyset, X\}\), the \(\mathcal{S}\)-measurable functions are exactly the constant functions.)

However \(f\) is not \(\mathcal{S}\)-measurable. Indeed, \((0, \infty)\) is an open, hence Borel, subset of \(\mathbb{R}\), and

\begin{equation*} f^{-1}\bigl((0,\infty)\bigr) = \{x \in \mathbb{R} : f(x) > 0\} = [0, \infty), \end{equation*}

which is neither \(\emptyset\) nor \(\mathbb{R}\), so \(f^{-1}\bigl((0,\infty)\bigr) \notin \mathcal{S}\). Thus \(f\) fails the definition 2.35 of \(\mathcal{S}\)-measurability.

The example shows that the implication in 2.45 (if \(f\) is measurable then \(|f|\) is measurable) cannot be reversed: passing to \(|f|\) destroys exactly the sign information that measurability of \(f\) would require.

More generally, the same phenomenon occurs for any measurable space \((X, \mathcal{S})\) and any set \(E \subseteq X\) with \(E \notin \mathcal{S}\): the function \(f = \chi_E - \chi_{X \setminus E}\) satisfies \(|f| \equiv 1\), which is \(\mathcal{S}\)-measurable, while \(f^{-1}\bigl((0,\infty)\bigr) = E \notin \mathcal{S}\), so \(f\) is not \(\mathcal{S}\)-measurable. Such an \(E\) exists whenever \(\mathcal{S}\) is not the collection of all subsets of \(X\).

Let \(A\) denote the set in question. Here a decimal expansion of \(x \geq 0\) means a representation

\begin{equation*} x = N + \sum_{j=1}^{\infty} e_j 10^{-j}, \qquad N \in \mathbb{Z},\ N \ge 0,\ e_j \in \{0, 1, \ldots, 9\}, \end{equation*}

written \(x = N.e_1 e_2 e_3 \ldots\), and a decimal expansion of a negative number is a minus sign followed by a decimal expansion of its absolute value. Since the digits of the integer part \(N\) form a finite string, the digit 5 can appear infinitely often only among the digits \(e_1, e_2, \ldots\) after the decimal point.

Step 1: reduction to \([0, \infty)\). Because the expansions of \(x\) and of \(-x\) carry the same digit string, \(x \in A\) if and only if \(|x| \in A\). Hence, setting \(A_+ = A \cap [0, \infty)\), we have

\begin{equation*} A = A_+ \cup (-A_+), \end{equation*}

where \(-A_+ = \{-y : y \in A_+\}\). By Exercise 8 of this section (dilation invariance, with \(t = -1\)), \(-A_+\) is a Borel set whenever \(A_+\) is. So it suffices to prove that \(A_+\) is a Borel set.

Step 2: the canonical digits. For \(x \geq 0\) and \(k \in \mathbb{Z}^+\) define

\begin{equation*} d_k(x) = \lfloor 10^k x \rfloor - 10 \lfloor 10^{k-1} x \rfloor . \end{equation*}

Write \(q = \lfloor 10^{k-1} x \rfloor\), so \(q \leq 10^{k-1} x < q + 1\) and hence \(10q \leq 10^k x < 10q + 10\), which gives \(10q \leq \lfloor 10^k x \rfloor \leq 10q + 9\). Thus \(d_k(x) \in \{0, 1, \ldots, 9\}\). Moreover the sum telescopes:

\begin{equation*} \lfloor x \rfloor + \sum_{k=1}^{n} d_k(x) 10^{-k} = \frac{\lfloor 10^n x \rfloor}{10^n}, \end{equation*}

and \(\lfloor 10^n x\rfloor / 10^n \to x\) as \(n \to \infty\) because \(x - 10^{-n} < \lfloor 10^n x \rfloor/10^n \le x\). Hence \(\lfloor x \rfloor . d_1(x) d_2(x) d_3(x)\ldots\) is a decimal expansion of \(x\); call it the canonical expansion.

Step 3: 5 appears infinitely often in some expansion if and only if it appears infinitely often in the canonical one. One direction is immediate, since the canonical string is itself an expansion. For the other direction, suppose \(x = N.e_1e_2e_3\ldots\) is a decimal expansion of \(x\) in which \(e_j = 5\) for infinitely many \(j\). Fix \(k \geq 0\). Since \(e_j = 5\) for infinitely many \(j\), there is some \(i > k\) with \(e_i = 5\); in particular the digits \(e_{k+1}, e_{k+2}, \ldots\) are not all equal to \(9\). Hence

\begin{equation*} \sum_{j > k} e_j 10^{-j} \leq \Bigl( \sum_{j > k} 9 \cdot 10^{-j} \Bigr) - 4 \cdot 10^{-i} = 10^{-k} - 4 \cdot 10^{-i} < 10^{-k}, \end{equation*}

and of course \(\sum_{j>k} e_j 10^{-j} \geq 0\). Setting \(M_k = 10^k N + \sum_{j=1}^{k} e_j 10^{k-j} \in \mathbb{Z}\), the displayed inequality gives

\begin{equation*} M_k \leq 10^k x < M_k + 1, \end{equation*}

so \(\lfloor 10^k x \rfloor = M_k\) for every \(k \geq 0\). Consequently \(\lfloor x \rfloor = M_0 = N\) and, for \(k \in \mathbb{Z}^+\),

\begin{equation*} d_k(x) = M_k - 10 M_{k-1} = e_k . \end{equation*}

So the given expansion is the canonical one, and therefore \(d_k(x) = 5\) for infinitely many \(k\). Hence

\begin{equation*} A_+ = \{ x \geq 0 : d_k(x) = 5 \text{ for infinitely many } k \in \mathbb{Z}^+ \}. \end{equation*}

Step 4: each digit set is Borel. Fix \(k \in \mathbb{Z}^+\) and let

\begin{equation*} D_k = \{ x \geq 0 : d_k(x) = 5 \}. \end{equation*}

We claim \(d_k(x) = 5\) if and only if \(\lfloor 10^k x \rfloor \in \{10m + 5 : m \in \mathbb{Z}\}\). Indeed, if \(d_k(x) = 5\) then \(\lfloor 10^k x\rfloor = 10\lfloor 10^{k-1}x\rfloor + 5\), which has the stated form. Conversely, if \(\lfloor 10^k x \rfloor = 10m + 5\) for some \(m \in \mathbb{Z}\), then \(10m + 5 \leq 10^k x < 10m + 6\), so \(m + \tfrac12 \leq 10^{k-1} x < m + \tfrac35\), whence \(\lfloor 10^{k-1} x \rfloor = m\) and \(d_k(x) = (10m + 5) - 10m = 5\). Since \(x \geq 0\) forces \(\lfloor 10^k x\rfloor \ge 0\), only \(m \geq 0\) occurs, and the claim translates into

\begin{equation*} D_k = \bigcup_{m=0}^{\infty} \left[ \frac{10m + 5}{10^{k}},\ \frac{10m + 6}{10^{k}} \right). \end{equation*}

Each half-open interval \([a, b)\) is a Borel set by 2.30, and \(D_k\) is a countable union of such intervals, so \(D_k\) is a Borel set.

Step 5: assembling. The digit 5 occurs among \(d_1(x), d_2(x), \ldots\) infinitely often exactly when for every \(n \in \mathbb{Z}^+\) there is some \(k \geq n\) with \(d_k(x) = 5\). Therefore

\begin{equation*} A_+ = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} D_k . \end{equation*}

Each inner union is a countable union of Borel sets, hence Borel by the definition 2.23 of a \(\sigma\)-algebra, and the outer intersection is a countable intersection of Borel sets, hence Borel by 2.25(c). Thus \(A_+\) is a Borel set, and by Step 1 so is

\begin{equation*} A = A_+ \cup (-A_+). \end{equation*}

(a) First suppose \(E \in \mathcal{S}\). Then \(E \in \mathcal{T}\) and \(E \subseteq X\), so \(E = E \cap X\), which exhibits \(E\) as an element of \(\{F \cap X : F \in \mathcal{T}\}\) (take \(F = E\)). Thus

\begin{equation*} \mathcal{S} \subseteq \{F \cap X : F \in \mathcal{T}\}. \end{equation*}

For the other inclusion, suppose \(F \in \mathcal{T}\). Since \(X \in \mathcal{T}\) by hypothesis, 2.25(b) gives \(F \cap X \in \mathcal{T}\). Also \(F \cap X \subseteq X\). Hence \(F \cap X \in \mathcal{S}\), proving

\begin{equation*} \{F \cap X : F \in \mathcal{T}\} \subseteq \mathcal{S}. \end{equation*}

The two inclusions together give \(\mathcal{S} = \{F \cap X : F \in \mathcal{T}\}\).

(b) We verify the three conditions in the definition 2.23 of a \(\sigma\)-algebra on \(X\). Note first that every element of \(\mathcal{S}\) is by definition a subset of \(X\), so \(\mathcal{S}\) is indeed a collection of subsets of \(X\); and complementation below must be taken within \(X\), not within \(Y\).

The empty set: \(\emptyset \in \mathcal{T}\) because \(\mathcal{T}\) is a \(\sigma\)-algebra, and \(\emptyset \subseteq X\). Hence \(\emptyset \in \mathcal{S}\).

Closure under complementation in \(X\): suppose \(E \in \mathcal{S}\). Then \(E \in \mathcal{T}\) and \(X \in \mathcal{T}\), so \(X \setminus E \in \mathcal{T}\) by 2.25(b). Since also \(X \setminus E \subseteq X\), we conclude \(X \setminus E \in \mathcal{S}\).

Closure under countable unions: suppose \(E_1, E_2, \ldots\) is a sequence of elements of \(\mathcal{S}\). Each \(E_k \in \mathcal{T}\), so \(\bigcup_{k=1}^{\infty} E_k \in \mathcal{T}\) because \(\mathcal{T}\) is a \(\sigma\)-algebra. Each \(E_k \subseteq X\), so \(\bigcup_{k=1}^{\infty} E_k \subseteq X\). Hence \(\bigcup_{k=1}^{\infty} E_k \in \mathcal{S}\).

Thus \(\mathcal{S}\) is a \(\sigma\)-algebra on \(X\).

The point of this exercise is recorded in the text just before 2.40: if \(X\) is a Borel subset of \(\mathbb{R}\) and \(\mathcal{T}\) is the collection of Borel subsets of \(\mathbb{R}\), then \(\mathcal{S}\) is the collection of Borel sets contained in \(X\), and part (a) says that this is the same as the collection of sets of the form \(F \cap X\) with \(F\) Borel. This is the \(\sigma\)-algebra with respect to which Borel measurable functions on \(X\) (definition 2.40) are measurable.

(a) Fix \(k \in \mathbb{Z}^+\) and suppose \(a \in G_k\). Choose \(\delta > 0\) such that

\begin{equation*} |f(b) - f( c)| < \tfrac{1}{k} \quad \text{for all } b, c \in (a - \delta, a + \delta). \end{equation*}

We claim that \((a - \tfrac{\delta}{2}, a + \tfrac{\delta}{2}) \subseteq G_k\), which shows that \(G_k\) contains an open interval about each of its points and hence is open.

To prove the claim, suppose \(a^{\prime} \in (a - \tfrac{\delta}{2}, a + \tfrac{\delta}{2})\) and set

\begin{equation*} \delta^{\prime} = \tfrac{\delta}{2} - |a^{\prime} - a| > 0 . \end{equation*}

If \(y \in (a^{\prime} - \delta^{\prime}, a^{\prime} + \delta^{\prime})\), then

\begin{equation*} |y - a| \leq |y - a^{\prime}| + |a^{\prime} - a| < \delta^{\prime} + |a^{\prime} - a| = \tfrac{\delta}{2} < \delta, \end{equation*}

so \(y \in (a - \delta, a + \delta)\). Thus \((a^{\prime} - \delta^{\prime}, a^{\prime} + \delta^{\prime}) \subseteq (a - \delta, a + \delta)\), and therefore \(|f(b) - f( c)| < \tfrac{1}{k}\) for all \(b, c \in (a^{\prime} - \delta^{\prime}, a^{\prime} + \delta^{\prime})\). Hence \(a^{\prime} \in G_k\), proving the claim and (a).

(b) Let \(C\) denote the set of points at which \(f\) is continuous.

Suppose \(a \in C\) and let \(k \in \mathbb{Z}^+\). By continuity of \(f\) at \(a\) applied with \(\varepsilon = \tfrac{1}{2k}\), there exists \(\delta > 0\) such that \(|f(y) - f(a)| < \tfrac{1}{2k}\) for all \(y \in (a - \delta, a + \delta)\). Hence for all \(b, c \in (a - \delta, a + \delta)\),

\begin{equation*} |f(b) - f( c)| \leq |f(b) - f(a)| + |f(a) - f( c)| < \tfrac{1}{2k} + \tfrac{1}{2k} = \tfrac{1}{k}. \end{equation*}

Thus \(a \in G_k\). Since \(k\) was arbitrary, \(a \in \bigcap_{k=1}^{\infty} G_k\), and so \(C \subseteq \bigcap_{k=1}^{\infty} G_k\).

Conversely, suppose \(a \in \bigcap_{k=1}^{\infty} G_k\) and let \(\varepsilon > 0\). Choose \(k \in \mathbb{Z}^+\) with \(\tfrac{1}{k} < \varepsilon\). Because \(a \in G_k\), there exists \(\delta > 0\) such that \(|f(b) - f( c)| < \tfrac{1}{k}\) for all \(b, c \in (a - \delta, a + \delta)\). Taking \(c = a\) (which lies in \((a - \delta, a + \delta)\)), we get

\begin{equation*} |f(b) - f(a)| < \tfrac{1}{k} < \varepsilon \quad \text{for all } b \in (a - \delta, a + \delta). \end{equation*}

Hence \(f\) is continuous at \(a\), so \(a \in C\). This gives \(\bigcap_{k=1}^{\infty} G_k \subseteq C\).

The two inclusions show that \(C = \bigcap_{k=1}^{\infty} G_k\).

(c) By (a), each \(G_k\) is an open subset of \(\mathbb{R}\) and hence is a Borel set, because by 2.29 the collection of Borel sets is a \(\sigma\)-algebra containing all open subsets of \(\mathbb{R}\). By 2.25(c) a \(\sigma\)-algebra is closed under countable intersections, so

\begin{equation*} C = \bigcap_{k=1}^{\infty} G_k \end{equation*}

is a Borel set. Thus the set of points at which \(f\) is continuous is a Borel set, as asserted in the last bullet point of 2.30.

Write

\begin{equation*} f = c_1 \chi_{E_1} + \cdots + c_n \chi_{E_n} . \end{equation*}

First suppose \(E_1, \ldots, E_n \in \mathcal{S}\). Fix \(j\). By 2.38, for every \(B \subseteq \mathbb{R}\) the inverse image \(\chi_{E_j}^{-1}(B)\) equals one of \(E_j\), \(X \setminus E_j\), \(X\), \(\emptyset\), according to which of \(0, 1\) belong to \(B\). All four of these sets are in \(\mathcal{S}\): we have \(E_j \in \mathcal{S}\) by hypothesis, \(X \in \mathcal{S}\) and \(\emptyset \in \mathcal{S}\) by 2.25(a) and the definition 2.23, and \(X \setminus E_j \in \mathcal{S}\) by the second bullet point of 2.23. Hence \(\chi_{E_j}^{-1}(B) \in \mathcal{S}\) for every Borel set \(B \subseteq \mathbb{R}\), so \(\chi_{E_j}\) is an \(\mathcal{S}\)-measurable function. Each constant function on \(X\) is \(\mathcal{S}\)-measurable (the inverse image of a set is either \(X\) or \(\emptyset\), both of which are in \(\mathcal{S}\) by 2.25(a)), so 2.46(a) implies that each product \(c_j \chi_{E_j}\) is \(\mathcal{S}\)-measurable. Applying 2.46(a) again, this time to sums, and inducting on the number of summands, we conclude that \(f\) is an \(\mathcal{S}\)-measurable function.

Now suppose \(f\) is \(\mathcal{S}\)-measurable. We claim that

\begin{equation*} f^{-1}(\{c_j\}) = E_j \quad \text{for each } j \in \{1, \ldots, n\}. \end{equation*}

To see this, first note the values of \(f\). If \(x \in E_i\) for some \(i\), then \(\chi_{E_i}(x) = 1\) and, because \(E_1, \ldots, E_n\) are disjoint, \(\chi_{E_m}(x) = 0\) for every \(m \neq i\); hence \(f(x) = c_i\). If \(x \notin E_1 \cup \cdots \cup E_n\), then every term vanishes and \(f(x) = 0\).

Fix \(j\). If \(x \in E_j\), then \(f(x) = c_j\) by the previous paragraph, so \(x \in f^{-1}(\{c_j\})\). Conversely suppose \(f(x) = c_j\). If \(x \notin E_1 \cup \cdots \cup E_n\), then \(f(x) = 0 \neq c_j\), because the numbers \(c_1, \ldots, c_n\) are nonzero; this is a contradiction. Hence \(x \in E_i\) for some \(i\), and then \(c_i = f(x) = c_j\), which forces \(i = j\) because the numbers \(c_1, \ldots, c_n\) are distinct. Thus \(x \in E_j\). This proves the claim.

The set \(\{c_j\}\) is a closed subset of \(\mathbb{R}\) and hence is a Borel set by 2.30. Because \(f\) is \(\mathcal{S}\)-measurable, the definition 2.35 gives

\begin{equation*} E_j = f^{-1}(\{c_j\}) \in \mathcal{S}. \end{equation*}

Since \(j \in \{1, \ldots, n\}\) was arbitrary, \(E_1, \ldots, E_n \in \mathcal{S}\).

Both hypotheses on the \(c_j\) are needed for the forward implication. If some \(c_j\) were \(0\), the corresponding term would contribute nothing and \(E_j\) could be an arbitrary (possibly non-measurable) set. If two of the \(c_j\) coincided, say \(c_1 = c_2\), then \(f = c_1 \chi_{E_1 \cup E_2} + \cdots\) would only detect \(E_1 \cup E_2\), which can be measurable while \(E_1\) and \(E_2\) are not. Disjointness is what makes \(f\) take the value \(c_j\) exactly on \(E_j\).

(a) Denote the left side by \(L\) and the right side by \(R\).

We first unwind the right side. By the definition 2.31 of inverse image, for \(x \in X\) we have

\begin{equation*} x \in (f_j - f_k)^{-1}\Bigl( \bigl(-\tfrac{1}{n}, \tfrac{1}{n}\bigr) \Bigr) \iff -\tfrac{1}{n} < f_j(x) - f_k(x) < \tfrac{1}{n} \iff |f_j(x) - f_k(x)| < \tfrac{1}{n}. \end{equation*}

Hence, reading the intersections as “for all” and the union as “there exists”,

\begin{equation*} x \in R \iff \text{for every } n \in \mathbb{Z}^+ \text{ there exists } j \in \mathbb{Z}^+ \text{ such that } |f_j(x) - f_k(x)| < \tfrac{1}{n} \text{ for all } k \geq j. \end{equation*}

Now fix \(x \in X\) and consider the sequence of real numbers \(f_1(x), f_2(x), \ldots\).

Suppose \(x \in L\), so the sequence has a limit in \(\mathbb{R}\); then it is a Cauchy sequence. Let \(n \in \mathbb{Z}^+\). By the Cauchy condition applied with \(\varepsilon = \tfrac{1}{n}\), there exists \(j \in \mathbb{Z}^+\) such that \(|f_p(x) - f_q(x)| < \tfrac{1}{n}\) for all \(p, q \geq j\). In particular, taking \(p = j\), we get \(|f_j(x) - f_k(x)| < \tfrac{1}{n}\) for all \(k \geq j\). Since \(n\) was arbitrary, \(x \in R\). Thus \(L \subseteq R\).

Conversely, suppose \(x \in R\), and let \(\varepsilon > 0\). Choose \(n \in \mathbb{Z}^+\) with \(\tfrac{2}{n} < \varepsilon\). By the description of \(R\) above, there exists \(j \in \mathbb{Z}^+\) such that \(|f_j(x) - f_k(x)| < \tfrac{1}{n}\) for all \(k \geq j\). Hence for all \(p, q \geq j\),

\begin{equation*} |f_p(x) - f_q(x)| \leq |f_p(x) - f_j(x)| + |f_j(x) - f_q(x)| < \tfrac{1}{n} + \tfrac{1}{n} = \tfrac{2}{n} < \varepsilon . \end{equation*}

Thus \(f_1(x), f_2(x), \ldots\) is a Cauchy sequence of real numbers, and so by the completeness of \(\mathbb{R}\) it has a limit in \(\mathbb{R}\). Hence \(x \in L\), giving \(R \subseteq L\).

The two inclusions establish the asserted equality. In words: the identity is exactly the statement that convergence in \(\mathbb{R}\) is equivalent to the Cauchy condition, with the Cauchy condition written using only the single “anchor” index \(j\) rather than two free indices.

(b) Let

\begin{equation*} L = \{x \in X : \text{the sequence } f_1(x), f_2(x), \ldots \text{ has a limit in } \mathbb{R}\}. \end{equation*}

Fix \(j, k \in \mathbb{Z}^+\). Since \(f_j\) and \(f_k\) are \(\mathcal{S}\)-measurable, 2.46(a) implies that \(f_j - f_k\) is an \(\mathcal{S}\)-measurable function from \(X\) to \(\mathbb{R}\). For each \(n \in \mathbb{Z}^+\) the interval \(\bigl(-\tfrac1n, \tfrac1n\bigr)\) is an open subset of \(\mathbb{R}\) and hence a Borel set, by 2.29. Therefore the definition 2.35 of measurability gives

\begin{equation*} (f_j - f_k)^{-1}\Bigl( \bigl(-\tfrac{1}{n}, \tfrac{1}{n}\bigr) \Bigr) \in \mathcal{S} \end{equation*}

for all \(n, j, k \in \mathbb{Z}^+\).

Now build up the set in (a) from these pieces. For fixed \(n\) and \(j\), the set

\begin{equation*} \bigcap_{k=j}^{\infty} (f_j - f_k)^{-1}\Bigl( \bigl(-\tfrac{1}{n}, \tfrac{1}{n}\bigr) \Bigr) \end{equation*}

is a countable intersection of elements of \(\mathcal{S}\), hence is in \(\mathcal{S}\) by 2.25(c). Next, for fixed \(n\), the set

\begin{equation*} \bigcup_{j=1}^{\infty} \bigcap_{k=j}^{\infty} (f_j - f_k)^{-1}\Bigl( \bigl(-\tfrac{1}{n}, \tfrac{1}{n}\bigr) \Bigr) \end{equation*}

is a countable union of elements of \(\mathcal{S}\), hence is in \(\mathcal{S}\) by the third bullet point in the definition 2.23 of a \(\sigma\)-algebra. Finally,

\begin{equation*} L = \bigcap_{n=1}^{\infty} \Biggl( \bigcup_{j=1}^{\infty} \bigcap_{k=j}^{\infty} (f_j - f_k)^{-1}\Bigl( \bigl(-\tfrac{1}{n}, \tfrac{1}{n}\bigr) \Bigr) \Biggr) \end{equation*}

is a countable intersection of elements of \(\mathcal{S}\), hence \(L \in \mathcal{S}\) by 2.25(c) again.

Thus the set of \(x \in X\) at which the sequence \(f_1(x), f_2(x), \ldots\) converges in \(\mathbb{R}\) is an \(\mathcal{S}\)-measurable subset of \(X\).

This complements 2.48: there the sequence was assumed to converge at every point of \(X\), and the limit function was shown to be \(\mathcal{S}\)-measurable. Here we see that even without that assumption the convergence set \(L\) is itself measurable. Consequently one may define \(f : X \to \mathbb{R}\) by \(f(x) = \lim_{k \to \infty} f_k(x)\) for \(x \in L\) and \(f(x) = 0\) for \(x \in X \setminus L\), and this \(f\) is \(\mathcal{S}\)-measurable. Indeed, let \(\mathcal{S}_L = \{E \in \mathcal{S} : E \subseteq L\}\), which by Exercise 11 of this section is a \(\sigma\)-algebra on \(L\). Each restriction \(f_k|_L\) is \(\mathcal{S}_L\)-measurable, because \((f_k|_L)^{-1}(B) = f_k^{-1}(B) \cap L \in \mathcal{S}_L\) for every Borel set \(B\). Since \(f_1(x), f_2(x), \ldots\) converges for every \(x \in L\), 2.48 shows that \(f|_L\) is \(\mathcal{S}_L\)-measurable. Finally, for every Borel set \(B \subseteq \mathbb{R}\),

\begin{equation*} f^{-1}(B) = \begin{cases} (f|_L)^{-1}(B) \cup (X \setminus L) & \text{if } 0 \in B,\\ (f|_L)^{-1}(B) & \text{if } 0 \notin B, \end{cases} \end{equation*}

and both \((f|_L)^{-1}(B) \in \mathcal{S}_L \subseteq \mathcal{S}\) and \(X \setminus L \in \mathcal{S}\), so \(f^{-1}(B) \in \mathcal{S}\).

Throughout, we use the following basic consequence of the hypotheses: because \(\bigcup_{k=1}^{\infty} E_k = X\) and the sets \(E_1, E_2, \ldots\) are pairwise disjoint, every \(x \in X\) belongs to \(E_k\) for exactly one \(k \in \mathbb{Z}^+\). Denote that index by \(k(x)\).

Part (a). Taking \(K = \emptyset\) gives \(\bigcup_{k \in \emptyset} E_k = \emptyset\), so \(\emptyset \in \mathcal{S}\).

Closure under complementation. Suppose \(E \in \mathcal{S}\), say \(E = \bigcup_{k \in K} E_k\) with \(K \subseteq \mathbb{Z}^+\). We claim

\begin{equation*} X \setminus E = \bigcup_{k \in \mathbb{Z}^+ \setminus K} E_k . \end{equation*}

Indeed, let \(x \in X\). Then \(x \in E\) if and only if \(x \in E_j\) for some \(j \in K\), which (by uniqueness of \(k(x)\)) happens if and only if \(k(x) \in K\). Hence \(x \in X \setminus E\) if and only if \(k(x) \in \mathbb{Z}^+ \setminus K\), which happens if and only if \(x \in \bigcup_{k \in \mathbb{Z}^+ \setminus K} E_k\). This proves the displayed equation, and therefore \(X \setminus E \in \mathcal{S}\).

Closure under countable unions. Suppose \(A_1, A_2, \ldots \in \mathcal{S}\), say \(A_n = \bigcup_{k \in K_n} E_k\) with \(K_n \subseteq \mathbb{Z}^+\). Put \(K = \bigcup_{n=1}^{\infty} K_n \subseteq \mathbb{Z}^+\). Then

\begin{equation*} \bigcup_{n=1}^{\infty} A_n = \bigcup_{n=1}^{\infty} \bigcup_{k \in K_n} E_k = \bigcup_{k \in K} E_k \in \mathcal{S} . \end{equation*}

Thus \(\mathcal{S}\) contains \(\emptyset\) and is closed under complementation and countable unions, so \(\mathcal{S}\) is a \(\sigma\)-algebra on \(X\).

Part (b). Suppose first that \(f : X \to \mathbb{R}\) is constant on \(E_k\) for every \(k \in \mathbb{Z}^+\). Let \(B \subseteq \mathbb{R}\) be a Borel set and set

\begin{equation*} K = \{ k \in \mathbb{Z}^+ : E_k \subseteq f^{-1}(B) \} . \end{equation*}

Clearly \(\bigcup_{k \in K} E_k \subseteq f^{-1}(B)\). Conversely, suppose \(x \in f^{-1}(B)\). Because \(f\) is constant on \(E_{k(x)}\) and \(x \in E_{k(x)}\), we have \(f(y) = f(x) \in B\) for every \(y \in E_{k(x)}\); hence \(E_{k(x)} \subseteq f^{-1}(B)\), so \(k(x) \in K\) and thus \(x \in \bigcup_{k \in K} E_k\). Therefore

\begin{equation*} f^{-1}(B) = \bigcup_{k \in K} E_k \in \mathcal{S} . \end{equation*}

Because \(B\) was an arbitrary Borel subset of \(\mathbb{R}\), the function \(f\) is \(\mathcal{S}\)-measurable.

Conversely, suppose \(f : X \to \mathbb{R}\) is \(\mathcal{S}\)-measurable. Fix \(k \in \mathbb{Z}^+\) and suppose \(x, y \in E_k\); we must show \(f(x) = f(y)\). The singleton \(\{f(x)\}\) is a closed subset of \(\mathbb{R}\) and hence a Borel set, so \(f^{-1}(\{f(x)\}) \in \mathcal{S}\); write \(f^{-1}(\{f(x)\}) = \bigcup_{j \in K} E_j\) for some \(K \subseteq \mathbb{Z}^+\). Now \(x \in f^{-1}(\{f(x)\})\), so \(x \in E_j\) for some \(j \in K\); by uniqueness of the index containing \(x\) we get \(j = k(x) = k\), so \(k \in K\). Consequently \(E_k \subseteq f^{-1}(\{f(x)\})\), and since \(y \in E_k\) this gives \(f(y) = f(x)\). Thus \(f\) is constant on \(E_k\) for every \(k \in \mathbb{Z}^+\).

Part (a). We have \(\emptyset \in \mathcal{S}\) and \(A \cap \emptyset = \emptyset\), so \(\emptyset \in \mathcal{S}_A\).

Closure under complementation. Suppose \(E \in \mathcal{S}_A\). Then \(E \in \mathcal{S}\), so \(X \setminus E \in \mathcal{S}\). If \(A \subseteq E\), then \(A \cap (X \setminus E) = \emptyset\). If instead \(A \cap E = \emptyset\), then every element of \(A\) lies in \(X\) but not in \(E\), so \(A \subseteq X \setminus E\). In either case \(X \setminus E \in \mathcal{S}_A\).

Closure under countable unions. Suppose \(E_1, E_2, \ldots \in \mathcal{S}_A\) and let \(E = \bigcup_{k=1}^{\infty} E_k\). Then \(E \in \mathcal{S}\) because \(\mathcal{S}\) is a \(\sigma\)-algebra. If \(A \subseteq E_k\) for some \(k\), then \(A \subseteq E\). Otherwise the definition of \(\mathcal{S}_A\) forces \(A \cap E_k = \emptyset\) for every \(k\), and then

\begin{equation*} A \cap E = A \cap \bigcup_{k=1}^{\infty} E_k = \bigcup_{k=1}^{\infty} (A \cap E_k) = \emptyset . \end{equation*}

In either case \(E \in \mathcal{S}_A\). Hence \(\mathcal{S}_A\) is a \(\sigma\)-algebra on \(X\).

Part (b). Suppose first that \(f\) is \(\mathcal{S}_A\)-measurable. Because \(\mathcal{S}_A \subseteq \mathcal{S}\), every set of the form \(f^{-1}(B)\) with \(B \subseteq \mathbb{R}\) Borel lies in \(\mathcal{S}_A\) and hence in \(\mathcal{S}\); thus \(f\) is \(\mathcal{S}\)-measurable. To see that \(f\) is constant on \(A\), we may assume \(A \neq \emptyset\) (otherwise there is nothing to prove). Pick \(x \in A\). The singleton \(\{f(x)\}\) is a closed, hence Borel, subset of \(\mathbb{R}\), so \(f^{-1}(\{f(x)\}) \in \mathcal{S}_A\). Since \(x \in A\) and \(x \in f^{-1}(\{f(x)\})\), we have \(A \cap f^{-1}(\{f(x)\}) \neq \emptyset\); the definition of \(\mathcal{S}_A\) therefore forces \(A \subseteq f^{-1}(\{f(x)\})\). In other words, \(f(y) = f(x)\) for every \(y \in A\), so \(f\) is constant on \(A\).

Conversely, suppose \(f\) is \(\mathcal{S}\)-measurable and \(f\) is constant on \(A\). If \(A = \emptyset\), then \(A \cap E = \emptyset\) for every \(E \in \mathcal{S}\), so \(\mathcal{S}_A = \mathcal{S}\) and there is nothing to prove. So assume \(A \neq \emptyset\) and let \(c \in \mathbb{R}\) be the value of \(f\) on \(A\). Let \(B \subseteq \mathbb{R}\) be a Borel set. Then \(f^{-1}(B) \in \mathcal{S}\) because \(f\) is \(\mathcal{S}\)-measurable. If \(c \in B\), then \(f(y) = c \in B\) for every \(y \in A\), so \(A \subseteq f^{-1}(B)\). If \(c \notin B\), then no \(y \in A\) satisfies \(f(y) \in B\), so \(A \cap f^{-1}(B) = \emptyset\). In either case \(f^{-1}(B) \in \mathcal{S}_A\). Because \(B\) was an arbitrary Borel subset of \(\mathbb{R}\), we conclude that \(f\) is \(\mathcal{S}_A\)-measurable.

Let

\begin{equation*} D = \{ x \in X : f \text{ is not continuous at } x \}, \end{equation*}

which by hypothesis is countable. Note first that every countable subset of \(\mathbb{R}\) is a Borel set: each singleton \(\{t\}\) is closed and hence Borel, and a countable set is a countable union of singletons, so it is Borel because the collection of Borel sets is a \(\sigma\)-algebra.

As remarked after 2.40, to prove that \(f\) is Borel measurable it suffices to show that \(f^{-1}\big((a, \infty)\big)\) is a Borel set for every \(a \in \mathbb{R}\) (this is 2.39 applied with \(\mathcal{S}\) equal to the \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}\) contained in \(X\)).

So fix \(a \in \mathbb{R}\) and let

\begin{equation*} U = \{ x \in X \setminus D : f(x) > a \}, \qquad V = \{ x \in D : f(x) > a \} , \end{equation*}

so that \(f^{-1}\big((a,\infty)\big) = U \cup V\). Since \(V \subseteq D\), the set \(V\) is countable and hence Borel.

Fix \(x \in U\). Then \(f\) is continuous at \(x\) and \(f(x) > a\). Continuity of \(f\) at \(x\) means continuity relative to the domain \(X\): for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|f(y) - f(x)| < \varepsilon\) for all \(y \in X\) with \(|y - x| < \delta\). Applying this with \(\varepsilon = f(x) - a > 0\) produces \(\delta_x > 0\) such that \(|f(y) - f(x)| < f(x) - a\), and hence

\begin{equation*} f(y) > f(x) - \big( f(x) - a \big) = a \quad \text{for all } y \in (x - \delta_x, x + \delta_x) \cap X . \end{equation*}

Note that this conclusion applies to every \(y\) in that intersection, including points of \(D\) at which \(f\) fails to be continuous; continuity at \(x\) constrains the values of \(f\) at all nearby points of \(X\), whatever the behavior of \(f\) there. Let

\begin{equation*} G = \bigcup_{x \in U} (x - \delta_x, x + \delta_x) , \end{equation*}

which is an open subset of \(\mathbb{R}\) (a union of open intervals), hence a Borel set. Therefore \(G \cap X\) is a Borel set, being the intersection of two Borel sets.

We claim

\begin{equation*} f^{-1}\big((a,\infty)\big) = (G \cap X) \cup V . \end{equation*}

To see the inclusion \(\subseteq\): if \(x \in f^{-1}\big((a,\infty)\big)\), then either \(x \in D\), in which case \(x \in V\), or \(x \notin D\), in which case \(x \in U\) and so \(x \in (x - \delta_x, x + \delta_x) \cap X \subseteq G \cap X\). For the inclusion \(\supseteq\): certainly \(V \subseteq f^{-1}\big((a,\infty)\big)\); and if \(y \in G \cap X\), then \(y \in (x - \delta_x, x + \delta_x) \cap X\) for some \(x \in U\), and the defining property of \(\delta_x\) gives \(f(y) > a\), so \(y \in f^{-1}\big((a,\infty)\big)\). This proves the claim.

Thus \(f^{-1}\big((a,\infty)\big)\) is the union of the two Borel sets \(G \cap X\) and \(V\), and hence is a Borel set. Because \(a \in \mathbb{R}\) was arbitrary, 2.39 implies that \(f\) is a Borel measurable function.

Because \(f\) is differentiable at every point of \(\mathbb{R}\), it is continuous at every point of \(\mathbb{R}\). Hence \(f\) is a Borel measurable function by 2.41.

For each \(k \in \mathbb{Z}^+\) define \(g_k : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} g_k(x) = \frac{f\big(x + \tfrac{1}{k}\big) - f(x)}{1/k} = k \left( f\Big(x + \tfrac{1}{k}\Big) - f(x) \right) . \end{equation*}

The function \(x \mapsto f\big(x + \tfrac{1}{k}\big)\) is continuous on \(\mathbb{R}\) (it is the composition of the continuous translation \(x \mapsto x + \tfrac1k\) with the continuous function \(f\)), and \(f\) is continuous, so \(g_k\) is continuous on \(\mathbb{R}\). Therefore each \(g_k\) is a Borel measurable function by 2.41.

Now fix \(x \in \mathbb{R}\). By definition of the derivative,

\begin{equation*} f^{\prime}(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} , \end{equation*}

and this limit exists in \(\mathbb{R}\). Taking the particular sequence \(h = \tfrac{1}{k}\), which converges to \(0\) with \(h \neq 0\), we get

\begin{equation*} \lim_{k \to \infty} g_k(x) = f^{\prime}(x) . \end{equation*}

Thus \(g_1, g_2, \ldots\) is a sequence of Borel measurable functions from \(\mathbb{R}\) to \(\mathbb{R}\) whose pointwise limit exists at every \(x \in \mathbb{R}\) and equals \(f^{\prime}(x)\).

Apply 2.48 with \(X = \mathbb{R}\) and \(\mathcal{S}\) the \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}\) (for which \(\mathcal{S}\)-measurable means Borel measurable): the pointwise limit \(f^{\prime}\) of the sequence \(g_1, g_2, \ldots\) is \(\mathcal{S}\)-measurable. Hence \(f^{\prime}\) is a Borel measurable function from \(\mathbb{R}\) to \(\mathbb{R}\).

Characterization. A function \(f : X \to \mathbb{R}\) is \(\mathcal{S}\)-measurable if and only if there exists \(c \in \mathbb{R}\) such that

\begin{equation*} \{ x \in X : f(x) \neq c \} \text{ is countable} , \end{equation*}

that is, if and only if \(f\) is constant outside some countable subset of \(X\).

Proof of the easier direction. Suppose \(c \in \mathbb{R}\) and \(C = \{x \in X : f(x) \neq c\}\) is countable. Let \(B \subseteq \mathbb{R}\) be a Borel set. If \(c \in B\), then \(X \setminus C \subseteq f^{-1}(B)\), so \(X \setminus f^{-1}(B) \subseteq C\) is countable and hence \(f^{-1}(B) \in \mathcal{S}\). If \(c \notin B\), then \(f^{-1}(B) \subseteq C\), so \(f^{-1}(B)\) is countable and hence \(f^{-1}(B) \in \mathcal{S}\). In either case \(f^{-1}(B) \in \mathcal{S}\), so \(f\) is \(\mathcal{S}\)-measurable.

Proof of the other direction. Suppose \(f : X \to \mathbb{R}\) is \(\mathcal{S}\)-measurable.

If \(X\) is countable, then \(\{x \in X : f(x) \neq 0\}\) is countable and we may take \(c = 0\). So assume from now on that \(X\) is uncountable. Then no subset of \(X\) is both countable and cocountable (otherwise \(X\) would be the union of two countable sets and hence countable). Also, a countable union of countable subsets of \(X\) is countable.

For \(a \in \mathbb{R}\) write \(E_a = f^{-1}\big((a, \infty)\big) = \{x \in X : f(x) > a\}\). Each \(E_a\) belongs to \(\mathcal{S}\), so each \(E_a\) is either countable or cocountable, and (by the previous paragraph) exactly one of these holds. Let

\begin{equation*} T = \{ a \in \mathbb{R} : E_a \text{ is countable} \} . \end{equation*}

\(T\) is nonempty. Suppose instead that \(T = \emptyset\). Then \(E_n\) is cocountable for every \(n \in \mathbb{Z}^+\), so \(X \setminus \bigcap_{n=1}^{\infty} E_n = \bigcup_{n=1}^{\infty} (X \setminus E_n)\) is countable, whence \(\bigcap_{n=1}^{\infty} E_n\) is nonempty because \(X\) is uncountable. But \(\bigcap_{n=1}^{\infty} E_n = \{x \in X : f(x) > n \text{ for all } n \in \mathbb{Z}^+\} = \emptyset\), since \(f\) takes values in \(\mathbb{R}\). This contradiction shows \(T \neq \emptyset\).

\(T\) is bounded below. Because \(f\) is real-valued, \(X = \bigcup_{n=1}^{\infty} E_{-n}\). As \(X\) is uncountable and a countable union of countable sets is countable, there exists \(m \in \mathbb{Z}^+\) with \(E_{-m}\) uncountable, so \(-m \notin T\). Now observe that \(T\) is closed upward: if \(a \in T\) and \(b > a\), then \(E_b \subseteq E_a\), so \(E_b\) is countable and \(b \in T\). Consequently, if some \(a \le -m\) were in \(T\), then \(-m \in T\) (as \(-m \ge a\); the case \(-m = a\) is immediate and the case \(-m > a\) follows from upward closure), a contradiction. Hence \(a > -m\) for every \(a \in T\), so \(T\) is bounded below.

Let \(c = \inf T \in \mathbb{R}\). We claim \(\{x \in X : f(x) \neq c\}\) is countable.

First, since \(c + \tfrac1n > c = \inf T\), for each \(n \in \mathbb{Z}^+\) there is \(a \in T\) with \(a < c + \tfrac1n\), and upward closure gives \(c + \tfrac1n \in T\); that is, \(E_{c + 1/n}\) is countable. Therefore

\begin{equation*} \{ x \in X : f(x) > c \} = \bigcup_{n=1}^{\infty} E_{c + \frac{1}{n}} \end{equation*}

is a countable union of countable sets, hence countable.

Second, for each \(n \in \mathbb{Z}^+\) we have \(c - \tfrac1n < \inf T\), so \(c - \tfrac1n \notin T\); thus \(E_{c - 1/n}\) is not countable and therefore is cocountable, so \(X \setminus E_{c - 1/n} = \{x \in X : f(x) \le c - \tfrac1n\}\) is countable. Therefore

\begin{equation*} \{ x \in X : f(x) < c \} = \bigcup_{n=1}^{\infty} \Big\{ x \in X : f(x) \le c - \tfrac{1}{n} \Big\} \end{equation*}

is countable as well. (The displayed equality holds because \(f(x) < c\) if and only if \(c - f(x) > 0\), which happens if and only if \(c - f(x) \ge \tfrac1n\) for some \(n \in \mathbb{Z}^+\).)

Combining the two paragraphs, \(\{x \in X : f(x) \neq c\} = \{f > c\} \cup \{f < c\}\) is countable, as claimed. This completes the proof of the characterization.

Recall that for \(t > 0\) and \(s \in \mathbb{R}\) the power \(t^s\) is defined by \(t^s = e^{s \ln t}\). Since \(f(x) > 0\) for every \(x \in X\), this gives

\begin{equation*} f^g(x) = f(x)^{g(x)} = e^{\,g(x) \ln f(x)} \qquad \text{for all } x \in X , \end{equation*}

so \(f^g = \exp \circ \big( g \cdot (\ln \circ f) \big)\). We build this function up in three steps.

Step 1: \(\ln \circ f\) is \(\mathcal{S}\)-measurable. The set \((0, \infty)\) is an open, hence Borel, subset of \(\mathbb{R}\), and the natural logarithm \(\ln : (0, \infty) \to \mathbb{R}\) is continuous. Thus \(\ln\) is a real-valued Borel measurable function by 2.41. Because \(f(x) > 0\) for all \(x \in X\), the range of \(f\) is contained in \((0, \infty)\), which is the domain of \(\ln\). Hence 2.44 applies and shows that \(\ln \circ f : X \to \mathbb{R}\) is an \(\mathcal{S}\)-measurable function.

Step 2: \(g \cdot (\ln \circ f)\) is \(\mathcal{S}\)-measurable. Both \(g\) and \(\ln \circ f\) are \(\mathcal{S}\)-measurable real-valued functions on \(X\), so their product is \(\mathcal{S}\)-measurable by 2.46(a). Denote this product by \(h\), so \(h(x) = g(x) \ln f(x)\) for all \(x \in X\).

Step 3: \(\exp \circ h\) is \(\mathcal{S}\)-measurable. The exponential function \(\exp : \mathbb{R} \to \mathbb{R}\) is continuous on the Borel set \(\mathbb{R}\), hence is Borel measurable by 2.41. Its domain \(\mathbb{R}\) contains the range of \(h\), so 2.44 shows that \(\exp \circ h\) is an \(\mathcal{S}\)-measurable function.

Since \(f^g = \exp \circ h\) by the displayed formula above, we conclude that \(f^g\) is an \(\mathcal{S}\)-measurable function.

Following the pattern of the proof of 2.39, let

\begin{equation*} \mathcal{T} = \{ A \subseteq [-\infty, \infty] : f^{-1}(A) \in \mathcal{S} \} . \end{equation*}

We must show that every Borel subset of \([-\infty, \infty]\) (in the sense of 2.50) belongs to \(\mathcal{T}\).

Step 1: \(\mathcal{T}\) is a \(\sigma\)-algebra on \([-\infty, \infty]\). First, \(f^{-1}(\emptyset) = \emptyset \in \mathcal{S}\), so \(\emptyset \in \mathcal{T}\). Next, if \(A \in \mathcal{T}\), then, because \(f\) maps \(X\) into \([-\infty, \infty]\),

\begin{equation*} f^{-1}\big([-\infty, \infty] \setminus A\big) = X \setminus f^{-1}(A) \in \mathcal{S} , \end{equation*}

so \([-\infty, \infty] \setminus A \in \mathcal{T}\); thus \(\mathcal{T}\) is closed under complementation. Finally, if \(A_1, A_2, \ldots \in \mathcal{T}\), then

\begin{equation*} f^{-1}\left( \bigcup_{k=1}^{\infty} A_k \right) = \bigcup_{k=1}^{\infty} f^{-1}(A_k) \in \mathcal{S} , \end{equation*}

so \(\bigcup_{k=1}^{\infty} A_k \in \mathcal{T}\); thus \(\mathcal{T}\) is closed under countable unions. Hence \(\mathcal{T}\) is a \(\sigma\)-algebra on \([-\infty, \infty]\). In particular, \(\mathcal{T}\) is closed under countable intersections and under set differences (by 2.25).

Step 2: \(\{\infty\}\), \(\{-\infty\}\), and \(\mathbb{R}\) belong to \(\mathcal{T}\). By hypothesis \((a, \infty] \in \mathcal{T}\) for every \(a \in \mathbb{R}\). Now

\begin{equation*} \bigcap_{n=1}^{\infty} (n, \infty] = \{\infty\} , \end{equation*}

because an element \(t \in [-\infty, \infty]\) satisfies \(t > n\) for every \(n \in \mathbb{Z}^+\) if and only if \(t = \infty\). Hence \(\{\infty\} \in \mathcal{T}\). Similarly

\begin{equation*} \bigcup_{n=1}^{\infty} (-n, \infty] = (-\infty, \infty] , \end{equation*}

because every \(t \in [-\infty, \infty]\) other than \(-\infty\) satisfies \(t > -n\) for some \(n \in \mathbb{Z}^+\). Hence \((-\infty, \infty] \in \mathcal{T}\), and taking complements gives \(\{-\infty\} \in \mathcal{T}\). Therefore

\begin{equation*} \mathbb{R} = [-\infty, \infty] \setminus \big( \{\infty\} \cup \{-\infty\} \big) \in \mathcal{T} . \end{equation*}

Step 3: every Borel subset of \(\mathbb{R}\) belongs to \(\mathcal{T}\). Let

\begin{equation*} \mathcal{T}_{\mathbb{R}} = \{ A \in \mathcal{T} : A \subseteq \mathbb{R} \} . \end{equation*}

We check that \(\mathcal{T}_{\mathbb{R}}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). Indeed \(\emptyset \in \mathcal{T}_{\mathbb{R}}\); if \(A \in \mathcal{T}_{\mathbb{R}}\), then

\begin{equation*} \mathbb{R} \setminus A = \mathbb{R} \cap \big( [-\infty, \infty] \setminus A \big) \in \mathcal{T} \end{equation*}

because \(\mathbb{R} \in \mathcal{T}\) by Step 2 and \(\mathcal{T}\) is closed under complementation and finite intersections, and clearly \(\mathbb{R} \setminus A \subseteq \mathbb{R}\), so \(\mathbb{R} \setminus A \in \mathcal{T}_{\mathbb{R}}\); and \(\mathcal{T}_{\mathbb{R}}\) is closed under countable unions because \(\mathcal{T}\) is and because a countable union of subsets of \(\mathbb{R}\) is a subset of \(\mathbb{R}\).

Moreover, for every \(a \in \mathbb{R}\) we have

\begin{equation*} (a, \infty) = (a, \infty] \setminus \{\infty\} \in \mathcal{T} , \end{equation*}

and \((a, \infty) \subseteq \mathbb{R}\), so \((a, \infty) \in \mathcal{T}_{\mathbb{R}}\).

Thus \(\mathcal{T}_{\mathbb{R}}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) containing \(\{ (a, \infty) : a \in \mathbb{R} \}\). The second half of the proof of 2.39 shows that any such \(\sigma\)-algebra contains every Borel subset of \(\mathbb{R}\); for completeness we repeat the argument. Closure under complementation gives \((-\infty, b] \in \mathcal{T}_{\mathbb{R}}\) for all \(b \in \mathbb{R}\), and then closure under finite intersections (2.25) gives \((a, b] = (a, \infty) \cap (-\infty, b] \in \mathcal{T}_{\mathbb{R}}\) for all \(a, b \in \mathbb{R}\). Because

\begin{equation*} (a, b) = \bigcup_{k=1}^{\infty} \Big( a, b - \tfrac{1}{k} \Big] \quad \text{and} \quad (-\infty, b) = \bigcup_{k=1}^{\infty} \Big( -k, b - \tfrac{1}{k} \Big] , \end{equation*}

and because \((a, \infty)\) and \(\mathbb{R}\) are already in \(\mathcal{T}_{\mathbb{R}}\), the \(\sigma\)-algebra \(\mathcal{T}_{\mathbb{R}}\) contains every open interval of \(\mathbb{R}\) (bounded or unbounded). Every open subset of \(\mathbb{R}\) is a countable union of open intervals: if \(G \subseteq \mathbb{R}\) is open, then for each \(x \in G\) there are rationals \(p < x < q\) with \((p, q) \subseteq G\), and \(G\) is the union of the countably many such rational intervals. Hence \(\mathcal{T}_{\mathbb{R}}\) contains every open subset of \(\mathbb{R}\), and therefore, being a \(\sigma\)-algebra, it contains the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing the open sets, which is the collection of Borel subsets of \(\mathbb{R}\).

Step 4: conclusion. Let \(C \subseteq [-\infty, \infty]\) be a Borel set. By 2.50, \(B := C \cap \mathbb{R}\) is a Borel subset of \(\mathbb{R}\), and \(C\) equals one of

\begin{equation*} B, \quad B \cup \{\infty\}, \quad B \cup \{-\infty\}, \quad B \cup \{\infty, -\infty\} . \end{equation*}

By Step 3 we have \(B \in \mathcal{T}_{\mathbb{R}} \subseteq \mathcal{T}\), by Step 2 we have \(\{\infty\}, \{-\infty\} \in \mathcal{T}\), and \(\mathcal{T}\) is closed under finite unions. Hence \(C \in \mathcal{T}\) in all four cases, which means \(f^{-1}( C) \in \mathcal{S}\).

Because \(C\) was an arbitrary Borel subset of \([-\infty, \infty]\), the function \(f\) is \(\mathcal{S}\)-measurable, completing the proof of 2.52.

Recall that \(f\) increasing means \(f(s) \le f(t)\) for all \(s, t \in B\) with \(s < t\), and that continuity of \(f\) at \(x \in B\) means: for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|f(t) - f(x)| < \varepsilon\) for all \(t \in B \cap (x - \delta, x + \delta)\) (continuity is relative to the domain \(B\)).

Call \(x \in B\) a left limit point of \(B\) if \(B \cap (x - \delta, x) \neq \emptyset\) for every \(\delta > 0\), and a right limit point of \(B\) if \(B \cap (x, x + \delta) \neq \emptyset\) for every \(\delta > 0\).

For \(x \in B\) define

\begin{equation*} \alpha(x) = \begin{cases} \sup\{f(t) : t \in B,\ t < x\} & \text{if } B \cap (-\infty, x) \neq \emptyset, \\ f(x) & \text{otherwise,}\end{cases} \qquad \beta(x) = \begin{cases} \inf\{f(t) : t \in B,\ t > x\} & \text{if } B \cap (x, \infty) \neq \emptyset, \\ f(x) & \text{otherwise.}\end{cases} \end{equation*}

Because \(f\) is increasing, \(f(t) \le f(x)\) for every \(t \in B\) with \(t < x\), so the set defining \(\alpha(x)\) is a nonempty set of real numbers bounded above by \(f(x)\); hence \(\alpha(x) \in \mathbb{R}\) and \(\alpha(x) \le f(x)\). Symmetrically \(\beta(x) \in \mathbb{R}\) and \(f(x) \le \beta(x)\).

Step 1: a characterization of the discontinuity points. Let \(D\) denote the set of \(x \in B\) at which \(f\) is not continuous. We claim that \(x \in D\) if and only if at least one of the following holds:

• (i) \(x\) is a left limit point of \(B\) and \(\alpha(x) < f(x)\);
• (ii) \(x\) is a right limit point of \(B\) and \(f(x) < \beta(x)\).

First suppose neither (i) nor (ii) holds; we show \(f\) is continuous at \(x\). Let \(\varepsilon > 0\). We produce \(\delta_1 > 0\) controlling the left side. If \(x\) is not a left limit point of \(B\), choose \(\delta_1 > 0\) with \(B \cap (x - \delta_1, x) = \emptyset\); then the only element of \(B \cap (x - \delta_1, x]\) is \(x\) itself, so \(|f(t) - f(x)| = 0 < \varepsilon\) for every \(t \in B \cap (x - \delta_1, x]\). If \(x\) is a left limit point of \(B\), then \(B \cap (-\infty, x) \neq \emptyset\), so \(\alpha(x)\) is genuinely the supremum written above; the failure of (i), combined with \(\alpha(x) \le f(x)\), forces \(\alpha(x) = f(x)\). By the definition of supremum there exists \(t_0 \in B\) with \(t_0 < x\) and \(f(t_0) > f(x) - \varepsilon\). Put \(\delta_1 = x - t_0 > 0\). For \(t \in B \cap (x - \delta_1, x]\) we have \(t_0 < t \le x\), hence \(f(x) - \varepsilon < f(t_0) \le f(t) \le f(x)\), so \(|f(t) - f(x)| < \varepsilon\).

Symmetrically, using that (ii) fails, there exists \(\delta_2 > 0\) such that \(|f(t) - f(x)| < \varepsilon\) for all \(t \in B \cap [x, x + \delta_2)\). Taking \(\delta = \min\{\delta_1, \delta_2\}\) gives \(|f(t) - f(x)| < \varepsilon\) for all \(t \in B \cap (x - \delta, x + \delta)\). Thus \(f\) is continuous at \(x\), so \(x \notin D\).

Conversely, suppose (i) holds and let \(\varepsilon = f(x) - \alpha(x) > 0\). Given any \(\delta > 0\), since \(x\) is a left limit point there exists \(t \in B \cap (x - \delta, x)\), and then \(f(t) \le \alpha(x) = f(x) - \varepsilon\), so \(|f(t) - f(x)| \ge \varepsilon\). Hence \(f\) is not continuous at \(x\). The case where (ii) holds is symmetric. This proves the claim.

Step 2: assigning disjoint intervals. For \(x \in D\) define a nonempty open interval \(I_x \subseteq \mathbb{R}\) by

\begin{equation*} I_x = \begin{cases} \bigl(\alpha(x),\, f(x)\bigr) & \text{if (i) holds for } x, \\ \bigl(f(x),\, \beta(x)\bigr) & \text{if (i) fails for } x \text{ (so (ii) holds).}\end{cases} \end{equation*}

In either case \(I_x\) is a nonempty bounded open interval.

We claim that \(I_x \cap I_y = \emptyset\) whenever \(x, y \in D\) with \(x < y\). It suffices to show \(\sup I_x \le \inf I_y\).

Case A: \(I_x = (\alpha(x), f(x))\), so \(\sup I_x = f(x)\). If \(I_y = (\alpha(y), f(y))\), then since \(x \in B\) and \(x < y\) we get \(\inf I_y = \alpha(y) = \sup\{f(t) : t \in B,\ t < y\} \ge f(x) = \sup I_x\). If instead \(I_y = (f(y), \beta(y))\), then \(\inf I_y = f(y) \ge f(x) = \sup I_x\) because \(f\) is increasing.

Case B: \(I_x = (f(x), \beta(x))\), so (ii) holds for \(x\) and \(\sup I_x = \beta(x)\). If \(I_y = (f(y), \beta(y))\), then \(\inf I_y = f(y) \ge \beta(x)\), because \(y \in B\) and \(y > x\) imply \(\beta(x) = \inf\{f(t) : t \in B,\ t > x\} \le f(y)\). If instead \(I_y = (\alpha(y), f(y))\), use that \(x\) is a right limit point of \(B\): taking \(\delta = y - x\) gives some \(s \in B\) with \(x < s < y\). Then \(\beta(x) \le f(s)\) (as \(s > x\)) and \(f(s) \le \alpha(y)\) (as \(s < y\)), so \(\sup I_x = \beta(x) \le \alpha(y) = \inf I_y\).

Thus \(\{I_x : x \in D\}\) is a pairwise disjoint collection of nonempty open intervals, and \(x \mapsto I_x\) is injective on \(D\) (distinct \(x\) give disjoint, hence distinct, nonempty intervals).

Step 3: countability. For each \(x \in D\) choose (using the density of \(\mathbb{Q}\) in \(\mathbb{R}\)) a rational number \(q_x \in I_x\). If \(x, y \in D\) with \(x \neq y\), then \(I_x \cap I_y = \emptyset\), so \(q_x \neq q_y\). Hence \(x \mapsto q_x\) is an injective map from \(D\) into \(\mathbb{Q}\), and therefore \(D\) is countable.

So \(f\) is continuous at every element of \(B\) except for the countable set \(D \subseteq B\).

First note that \(f\) is injective: if \(x \neq y\), say \(x < y\), then \(f(x) < f(y)\), so \(f(x) \neq f(y)\). Hence \(f\) is a bijection from \(\mathbb{R}\) onto \(f(\mathbb{R})\) and the inverse function \(g = f^{-1} : f(\mathbb{R}) \to \mathbb{R}\) is well defined; it is characterized by \(g(f(x)) = x\) for all \(x \in \mathbb{R}\).

We also record the following consequence of strict monotonicity: for \(x, u \in \mathbb{R}\),

\begin{equation*} f(x) < f(u) \implies x < u. \end{equation*}

Indeed, if \(x \ge u\) then \(f(x) \ge f(u)\) (with equality only when \(x = u\)), contradicting \(f(x) < f(u)\).

Now fix \(y_0 \in f(\mathbb{R})\) and let \(x_0 = g(y_0)\), so \(y_0 = f(x_0)\). Let \(\varepsilon > 0\). Because \(f\) is strictly increasing,

\begin{equation*} f(x_0 - \varepsilon) < f(x_0) = y_0 < f(x_0 + \varepsilon). \end{equation*}

Hence

\begin{equation*} \delta = \min\bigl\{\, y_0 - f(x_0 - \varepsilon),\ f(x_0 + \varepsilon) - y_0 \,\bigr\} > 0 . \end{equation*}

Suppose \(y \in f(\mathbb{R})\) with \(|y - y_0| < \delta\). Then

\begin{equation*} f(x_0 - \varepsilon) \le y_0 - \delta < y < y_0 + \delta \le f(x_0 + \varepsilon). \end{equation*}

Write \(x = g(y)\), so \(y = f(x)\). From \(f(x_0 - \varepsilon) < f(x)\) we get \(x_0 - \varepsilon < x\) by the displayed implication above, and from \(f(x) < f(x_0 + \varepsilon)\) we get \(x < x_0 + \varepsilon\). Therefore

\begin{equation*} |g(y) - g(y_0)| = |x - x_0| < \varepsilon . \end{equation*}

Thus for every \(y_0 \in f(\mathbb{R})\) and every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|g(y) - g(y_0)| < \varepsilon\) for all \(y \in f(\mathbb{R})\) with \(|y - y_0| < \delta\). In other words, \(g = f^{-1}\) is continuous at every point of its domain \(f(\mathbb{R})\), as desired.

Note that continuity of \(f\) was never used; the only inputs were that \(f\) is strictly increasing and that its domain is all of \(\mathbb{R}\) (the latter guaranteeing that \(x_0 \pm \varepsilon\) lie in the domain).

Let \(g = f^{-1} : f(\mathbb{R}) \to \mathbb{R}\) be the inverse function, which exists because \(f\) is injective (strictly increasing), and which is continuous by Exercise 23 of this section. The key observation is

\begin{equation*} g^{-1}(B) = \{\, y \in f(\mathbb{R}) : g(y) \in B \,\} = \{\, f(x) : x \in \mathbb{R},\ x \in B \,\} = f(B), \end{equation*}

where the middle equality holds because every \(y \in f(\mathbb{R})\) has the form \(y = f(x)\) for a unique \(x \in \mathbb{R}\), and then \(g(y) = x\).

So by 2.41 (every continuous real-valued function defined on a Borel subset of \(\mathbb{R}\) is Borel measurable) it suffices to prove that the domain \(f(\mathbb{R})\) of \(g\) is a Borel set: then \(g\) is Borel measurable, hence \(f(B) = g^{-1}(B)\) is a Borel set for every Borel set \(B \subseteq \mathbb{R}\).

Proof that \(f(\mathbb{R})\) is a Borel set. For \(x \in \mathbb{R}\) put

\begin{equation*} f(x^-) = \sup\{f(t) : t < x\}, \qquad f(x^+) = \inf\{f(t) : t > x\}, \end{equation*}

which are real numbers with \(f(x^-) \le f(x) \le f(x^+)\) (the sets involved are nonempty and bounded by \(f(x)\) on the appropriate side). Put

\begin{equation*} L = \inf\{f(t) : t \in \mathbb{R}\} \in [-\infty, \infty), \qquad M = \sup\{f(t) : t \in \mathbb{R}\} \in (-\infty, \infty], \end{equation*}

with the convention that \((-\infty, L] = \emptyset\) if \(L = -\infty\) and \([M, \infty) = \emptyset\) if \(M = \infty\). Finally let

\begin{equation*} D = \{\, x \in \mathbb{R} : f(x^-) < f(x) \text{ or } f(x) < f(x^+) \,\}, \end{equation*}

which is countable. Indeed, every real number is both a left limit point and a right limit point of \(\mathbb{R}\), so the characterization proved in Step 1 of Exercise 22 of this section (applied with \(B = \mathbb{R}\), where \(\alpha(x) = f(x^-)\) and \(\beta(x) = f(x^+)\)) says that \(D\) is exactly the set of points at which \(f\) is not continuous; that exercise shows this set is countable.

We claim

\begin{equation*} \mathbb{R} \setminus f(\mathbb{R}) = (-\infty, L] \ \cup\ [M, \infty) \ \cup \bigcup_{x \in D} \Bigl( \bigl[f(x^-), f(x)\bigr) \cup \bigl(f(x), f(x^+)\bigr] \Bigr). \end{equation*}

First, none of the sets on the right meets \(f(\mathbb{R})\).

• If \(L \in \mathbb{R}\) and \(f(u) = L\) for some \(u\), then \(f(u - 1) < f(u) = L\), contradicting the definition of \(L\) as an infimum. So the infimum \(L\) is not attained, and since \(f(t) \ge L\) for all \(t\), we get \((-\infty, L] \cap f(\mathbb{R}) = \emptyset\). Symmetrically \([M, \infty) \cap f(\mathbb{R}) = \emptyset\).
• Fix \(x\) and suppose \(f(x) < f(x^+)\). The infimum \(f(x^+)\) is not attained: if \(f(u) = f(x^+)\) with \(u > x\), then for \(t\) with \(x < t < u\) we would have \(f(t) < f(u) = f(x^+) = \inf\{f(s) : s > x\}\), a contradiction; and if \(u \le x\) then \(f(u) \le f(x) < f(x^+)\). Moreover any value \(f(u)\) lying in \((f(x), f(x^+)]\) must have \(u > x\) (since \(u \le x\) gives \(f(u) \le f(x)\)), whence \(f(u) \ge f(x^+)\), so \(f(u) = f(x^+)\), which we just excluded. Hence \((f(x), f(x^+)] \cap f(\mathbb{R}) = \emptyset\). Symmetrically \([f(x^-), f(x)) \cap f(\mathbb{R}) = \emptyset\).

For the reverse inclusion, let \(y \notin f(\mathbb{R})\) and suppose \(L < y < M\) (otherwise \(y \le L\) or \(y \ge M\), so \(y\) lies in one of the first two sets). Let

\begin{equation*} T = \{\, t \in \mathbb{R} : f(t) < y \,\}. \end{equation*}

Since \(y > L\), the set \(T\) is nonempty. Since \(y < M\), there is \(s\) with \(f(s) > y\); for \(t \in T\) we then have \(f(t) < y < f(s)\), which forces \(t < s\) (as \(t \ge s\) would give \(f(t) \ge f(s)\)). So \(T\) is bounded above, and \(x = \sup T \in \mathbb{R}\).

If \(t < x\), then there is \(t^{\prime} \in T\) with \(t < t^{\prime} \le x\), so \(f(t) \le f(t^{\prime}) < y\); taking the supremum over \(t < x\) gives \(f(x^-) \le y\). If \(t > x\), then \(t \notin T\), so \(f(t) \ge y\), and in fact \(f(t) > y\) because \(y \notin f(\mathbb{R})\); taking the infimum over \(t > x\) gives \(f(x^+) \ge y\). Also \(f(x) \neq y\) since \(y \notin f(\mathbb{R})\). There are two cases.

• If \(f(x) > y\), then \(f(x^-) \le y < f(x)\), so \(y \in [f(x^-), f(x))\) and in particular \(f(x^-) < f(x)\), so \(x \in D\).
• If \(f(x) < y\), then \(f(x) < y \le f(x^+)\), so \(y \in (f(x), f(x^+)]\) and in particular \(f(x) < f(x^+)\), so \(x \in D\).

Either way \(y\) belongs to the right side, proving the claimed equality.

The right side is a union of \(2\) intervals together with a countable family of intervals indexed by the countable set \(D\); each interval is a Borel set, and the collection of Borel sets is closed under countable unions (it is a \(\sigma\)-algebra), so \(\mathbb{R} \setminus f(\mathbb{R})\) is a Borel set. Hence \(f(\mathbb{R})\) is a Borel set, being the complement of a Borel set.

This completes the proof: \(g\) is a continuous function on the Borel set \(f(\mathbb{R})\), hence Borel measurable by 2.41, hence \(f(B) = g^{-1}(B)\) is a Borel set.

For each \(k \in \mathbb{Z}^+\) define \(f_k : B \to \mathbb{R}\) by

\begin{equation*} f_k(x) = f(x) + \frac{x}{k}. \end{equation*}

Each \(f_k\) is a well-defined real-valued function on \(B\).

\(f_k\) is strictly increasing: suppose \(x, y \in B\) with \(x < y\). Because \(f\) is increasing, \(f(x) \le f(y)\). Because \(k > 0\) and \(x < y\), we have \(x/k < y/k\). Adding the two inequalities gives

\begin{equation*} f_k(x) = f(x) + \frac{x}{k} < f(y) + \frac{y}{k} = f_k(y). \end{equation*}

Thus \(f_k(x) < f_k(y)\) whenever \(x < y\), i.e. \(f_k\) is strictly increasing.

Pointwise convergence: fix \(x \in B\). Then \(x\) is a fixed real number, so

\begin{equation*} \lim_{k \to \infty} f_k(x) = \lim_{k \to \infty} \Bigl( f(x) + \frac{x}{k} \Bigr) = f(x) + \lim_{k \to \infty} \frac{x}{k} = f(x) + 0 = f(x). \end{equation*}

Hence \(f_1, f_2, \ldots\) is a sequence of strictly increasing functions from \(B\) to \(\mathbb{R}\) converging pointwise on \(B\) to \(f\), as required.

Remark. Any strictly increasing function \(\varphi : \mathbb{R} \to \mathbb{R}\) would serve equally well in place of the identity: setting \(f_k(x) = f(x) + \varphi(x)/k\) gives strictly increasing functions converging pointwise to \(f\). Taking \(\varphi\) bounded, say \(\varphi(x) = \arctan x\), has the extra feature that the convergence is uniform on \(B\) and that each \(f_k\) is bounded whenever \(f\) is.

If \(B = \emptyset\) then \(f\) is the empty function and the constant function \(g = 0\) on \(\mathbb{R}\) works (it is increasing and the condition on \(B\) is vacuous). So assume \(B \neq \emptyset\).

Because \(f\) is bounded, the set \(f(B) = \{f(t) : t \in B\}\) is a nonempty bounded set of real numbers; let

\begin{equation*} m = \inf f(B) \in \mathbb{R}, \qquad M = \sup f(B) \in \mathbb{R}. \end{equation*}

Define \(g : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} g(x) = \begin{cases} m & \text{if } B \cap (-\infty, x] = \emptyset, \\[2pt] \sup\{\, f(t) : t \in B,\ t \le x \,\} & \text{if } B \cap (-\infty, x] \neq \emptyset. \end{cases} \end{equation*}

\(g\) is well defined and real-valued: in the second case the set \(\{f(t) : t \in B,\ t \le x\}\) is a nonempty subset of \(f(B)\), hence bounded above by \(M\), so its supremum is a real number lying in \([m, M]\). In the first case \(g(x) = m \in \mathbb{R}\). In particular \(m \le g(x) \le M\) for all \(x \in \mathbb{R}\).

\(g\) is increasing: suppose \(x < y\). If \(B \cap (-\infty, x] = \emptyset\), then \(g(x) = m \le g(y)\), because \(g(y)\) is either \(m\) itself or the supremum of a nonempty subset of \(f(B)\), and every element of \(f(B)\) is at least \(m\). If \(B \cap (-\infty, x] \neq \emptyset\), then \(B \cap (-\infty, y] \neq \emptyset\) as well and

\begin{equation*} \{\, f(t) : t \in B,\ t \le x \,\} \subseteq \{\, f(t) : t \in B,\ t \le y \,\}, \end{equation*}

so the supremum over the larger set is at least the supremum over the smaller set, i.e. \(g(x) \le g(y)\).

\(g\) extends \(f\): let \(x \in B\). Then \(x \in B \cap (-\infty, x]\), so the second case of the definition applies and

\begin{equation*} g(x) = \sup\{\, f(t) : t \in B,\ t \le x \,\}. \end{equation*}

For \(t \in B\) with \(t \le x\) we have \(f(t) \le f(x)\): this is clear if \(t = x\), and follows from \(f\) being increasing if \(t < x\). So \(f(x)\) is an upper bound of the set, and \(f(x)\) belongs to the set (take \(t = x\)). Hence the supremum equals \(f(x)\), i.e. \(g(x) = f(x)\).

Thus \(g : \mathbb{R} \to \mathbb{R}\) is an increasing function with \(g|_B = f\), as required.

Remark. Boundedness is used only to guarantee that \(g\) takes values in \(\mathbb{R}\) rather than in \([-\infty, \infty]\); without it the same formula still defines an increasing function from \(\mathbb{R}\) to \([-\infty, \infty]\) extending \(f\). Some hypothesis of this kind is genuinely needed: let \(B = (0,1)\) and \(f(x) = 1/(1 - x)\), which is increasing on \(B\) but unbounded above. If \(g : \mathbb{R} \to \mathbb{R}\) were an increasing extension, then \(g(1) \ge g(x) = 1/(1 - x)\) for every \(x \in (0,1)\), and letting \(x\) increase to \(1\) makes the right side arbitrarily large, so no real number \(g(1)\) can work.

The statement is false; here is a counterexample.

Let \(X = \{1, 2\}\) and \(\mathcal{S} = \{\emptyset, X\}\), which is a \(\sigma\)-algebra on \(X\). Define \(f : X \to [-\infty, \infty]\) by

\begin{equation*} f(1) = \infty, \qquad f(2) = -\infty . \end{equation*}

The hypothesis holds. Indeed, for \(a \in \mathbb{R}\),

\begin{equation*} f^{-1}\bigl((a, \infty)\bigr) = \{\, x \in X : a < f(x) < \infty \,\} = \emptyset \in \mathcal{S}, \end{equation*}

because neither value of \(f\) lies in \(\mathbb{R}\): \(f(1) = \infty \notin (a, \infty)\) and \(f(2) = -\infty \notin (a, \infty)\).

The conclusion fails. By 2.50, a subset of \([-\infty, \infty]\) is a Borel set exactly when its intersection with \(\mathbb{R}\) is a Borel subset of \(\mathbb{R}\); since \(\{\infty\} \cap \mathbb{R} = \emptyset\) is a Borel set, \(\{\infty\}\) is a Borel subset of \([-\infty, \infty]\). But

\begin{equation*} f^{-1}\bigl(\{\infty\}\bigr) = \{1\} \notin \mathcal{S} . \end{equation*}

Hence \(f\) is not \(\mathcal{S}\)-measurable in the sense of 2.51.

What goes wrong, and what remains true. Let

\begin{equation*} \mathcal{T} = \{\, A \subseteq [-\infty, \infty] : f^{-1}(A) \in \mathcal{S} \,\}. \end{equation*}

Exactly as in the proof of 2.39 (using 2.33), \(\mathcal{T}\) is a \(\sigma\)-algebra on \([-\infty, \infty]\). The hypothesis says \((a, \infty) \in \mathcal{T}\) for all \(a \in \mathbb{R}\). For \(a < b\) the complement of \((b, \infty)\) in \([-\infty, \infty]\) is \([-\infty, b] \cup \{\infty\}\), and intersecting it with \((a, \infty)\) gives

\begin{equation*} (a, \infty) \cap \bigl( [-\infty, b] \cup \{\infty\} \bigr) = (a, b], \end{equation*}

since \(\infty \notin (a, \infty)\). Hence \((a, b] \in \mathcal{T}\) for all \(a < b\), and therefore \(\mathbb{R} = \bigcup_{k=1}^{\infty} (-k, k] \in \mathcal{T}\). It follows that

\begin{equation*} \mathcal{T}|_{\mathbb{R}} = \{\, A \subseteq \mathbb{R} : A \in \mathcal{T} \,\} \end{equation*}

is a \(\sigma\)-algebra on \(\mathbb{R}\): it contains \(\emptyset\), it is closed under countable unions because \(\mathcal{T}\) is, and if \(A \in \mathcal{T}|_{\mathbb{R}}\) then \(\mathbb{R} \setminus A = \bigl( [-\infty, \infty] \setminus A \bigr) \cap \mathbb{R} \in \mathcal{T}|_{\mathbb{R}}\) using \(\mathbb{R} \in \mathcal{T}\) and 2.25(b). Every open subset of \(\mathbb{R}\) is a countable union of sets of the form \((a, b]\) (write \((c, d) = \bigcup_{k} (c, d_k]\) with \(d_k\) increasing to \(d\), and use that every open subset of \(\mathbb{R}\) is a countable union of open intervals), so \(\mathcal{T}|_{\mathbb{R}}\) contains every open subset of \(\mathbb{R}\) and hence every Borel subset of \(\mathbb{R}\). So the hypothesis does imply

\begin{equation*} f^{-1}(B) \in \mathcal{S} \quad \text{for every Borel set } B \subseteq \mathbb{R}. \end{equation*}

What it does not control is the two points \(\infty\) and \(-\infty\): no countable combination of the sets \((a, \infty)\), \(a \in \mathbb{R}\), isolates \(\{\infty\}\), and the counterexample above exploits precisely this. Replacing the hypothesis by \(f^{-1}\bigl((a, \infty]\bigr) \in \mathcal{S}\) for all \(a \in \mathbb{R}\) repairs the statement: that is 2.52, since then

\begin{equation*} f^{-1}\bigl(\{\infty\}\bigr) = \bigcap_{k=1}^{\infty} f^{-1}\bigl((k, \infty]\bigr) \in \mathcal{S} \end{equation*}

and \(f^{-1}\bigl(\{-\infty\}\bigr) = X \setminus \bigcup_{k=1}^{\infty} f^{-1}\bigl((-k, \infty]\bigr) \in \mathcal{S}\).

Here \(B \subseteq \mathbb{R}\) and, by 2.40, \(f\) being Borel measurable means that \(f^{-1}(A)\) is a Borel set for every Borel set \(A \subseteq \mathbb{R}\).

First note that \(B\) itself is a Borel set: as observed after 2.40,

\begin{equation*} B = f^{-1}(\mathbb{R}), \end{equation*}

and \(\mathbb{R}\) is a Borel subset of \(\mathbb{R}\), so \(B\) is a Borel set. Consequently \(\mathbb{R} \setminus B\) is a Borel set, since the collection of Borel sets is a \(\sigma\)-algebra on \(\mathbb{R}\) and hence closed under complementation.

Now let \(A \subseteq \mathbb{R}\) be a Borel set. We compute \(g^{-1}(A)\) by splitting \(\mathbb{R}\) into \(B\) and \(\mathbb{R} \setminus B\):

\begin{equation*} g^{-1}(A) = \bigl( g^{-1}(A) \cap B \bigr) \cup \bigl( g^{-1}(A) \cap (\mathbb{R} \setminus B) \bigr). \end{equation*}

For \(x \in B\) we have \(g(x) = f(x)\), so \(g^{-1}(A) \cap B = f^{-1}(A)\) (note \(f^{-1}(A) \subseteq B\) automatically, since \(B\) is the domain of \(f\)). For \(x \in \mathbb{R} \setminus B\) we have \(g(x) = 0\), so

\begin{equation*} g^{-1}(A) \cap (\mathbb{R} \setminus B) = \begin{cases} \mathbb{R} \setminus B & \text{if } 0 \in A, \\ \emptyset & \text{if } 0 \notin A. \end{cases} \end{equation*}

Therefore

\begin{equation*} g^{-1}(A) = \begin{cases} f^{-1}(A) \cup (\mathbb{R} \setminus B) & \text{if } 0 \in A, \\ f^{-1}(A) & \text{if } 0 \notin A. \end{cases} \end{equation*}

In both cases \(g^{-1}(A)\) is a Borel set: \(f^{-1}(A)\) is a Borel set because \(f\) is Borel measurable, \(\mathbb{R} \setminus B\) is a Borel set as shown above, and the union of two Borel sets is a Borel set by 2.25(b).

Since \(g^{-1}(A)\) is a Borel set for every Borel set \(A \subseteq \mathbb{R}\), the function \(g : \mathbb{R} \to \mathbb{R}\) is Borel measurable, as desired.

(Alternatively, one could verify only the sets \((a, \infty)\), as permitted by 2.39 and the remark following 2.40: for \(a \ge 0\) we get \(g^{-1}\bigl((a, \infty)\bigr) = f^{-1}\bigl((a, \infty)\bigr)\), and for \(a < 0\) we get \(g^{-1}\bigl((a, \infty)\bigr) = f^{-1}\bigl((a, \infty)\bigr) \cup (\mathbb{R} \setminus B)\), in each case a Borel set.)

Take \(X = \mathbb{R}\) and let

\begin{equation*} \mathcal{S} = \{E \subseteq \mathbb{R} : E \text{ is countable or } \mathbb{R} \setminus E \text{ is countable}\}. \end{equation*}

By the first bullet point of 2.28, \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) (indeed it is the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing all the singletons), so \((X, \mathcal{S})\) is a measurable space.

For \(t \in \mathbb{R}\) define \(f_t : \mathbb{R} \to [0, 1]\) by

\begin{equation*} f_t = \chi_{\{t\}} \ \text{ if } t \in [0,1], \qquad f_t = 0 \ \text{ if } t \notin [0,1], \end{equation*}

where \(\chi_A\) denotes the characteristic function of \(A\) (so \(\chi_A(x) = 1\) if \(x \in A\) and \(\chi_A(x) = 0\) otherwise). Each \(f_t\) maps \(\mathbb{R}\) into \(\{0,1\} \subseteq [0,1]\), as required.

Each \(f_t\) is \(\mathcal{S}\)-measurable. If \(t \notin [0,1]\) then \(f_t\) is the constant function \(0\), and the inverse image under \(f_t\) of any set is either \(\emptyset\) or \(\mathbb{R}\), both of which lie in \(\mathcal{S}\). If \(t \in [0,1]\) then for a Borel set \(B \subseteq \mathbb{R}\) the inverse image \(f_t^{-1}(B)\) is one of the four sets \(\emptyset\), \(\{t\}\), \(\mathbb{R} \setminus \{t\}\), \(\mathbb{R}\), according to which of \(0, 1\) belong to \(B\). The set \(\{t\}\) is countable, and \(\mathbb{R} \setminus \{t\}\) has countable complement, namely \(\{t\}\); also \(\emptyset\) is countable and \(\mathbb{R}\) has empty (hence countable) complement. So all four of these sets are in \(\mathcal{S}\), and \(f_t\) is \(\mathcal{S}\)-measurable.

Now compute the pointwise supremum. Fix \(x \in \mathbb{R}\).

If \(x \in [0,1]\), then \(f_x = \chi_{\{x\}}\), so \(f_x(x) = 1\); since every \(f_t\) takes values in \([0,1]\) we get \(f(x) = \sup\{f_t(x) : t \in \mathbb{R}\} = 1\).

If \(x \notin [0,1]\), then for each \(t \in \mathbb{R}\) we have \(f_t(x) = 0\): either \(t \notin [0,1]\), in which case \(f_t\) is identically \(0\); or \(t \in [0,1]\), in which case \(t \neq x\) (because \(x \notin [0,1]\)) and so \(\chi_{\{t\}}(x) = 0\). Hence \(f(x) = 0\).

Therefore \(f = \chi_{[0,1]}\). This function is not \(\mathcal{S}\)-measurable, because

\begin{equation*} f^{-1}\big((\tfrac{1}{2}, \infty)\big) = [0,1], \end{equation*}

and \([0,1]\) is not in \(\mathcal{S}\): the interval \([0,1]\) is uncountable, and its complement \((-\infty, 0) \cup (1, \infty)\) is uncountable as well, since it contains the interval \((1,2)\). (Every open interval \((a,b)\) with \(a < b\) is uncountable: the map \(s \mapsto a + (b-a)\frac{e^{s}}{1 + e^{s}}\) is a bijection from \(\mathbb{R}\) onto \((a,b)\), and \(\mathbb{R}\) is uncountable by Cantor’s theorem. Hence \([0,1] \supseteq (0,1)\) and \((-\infty,0) \cup (1,\infty) \supseteq (1,2)\) are both uncountable.) Since \((\tfrac{1}{2}, \infty)\) is a Borel subset of \(\mathbb{R}\) whose inverse image under \(f\) is not in \(\mathcal{S}\), the function \(f\) fails to be \(\mathcal{S}\)-measurable.

Thus \(\{f_t\}_{t \in \mathbb{R}}\) is a family of \(\mathcal{S}\)-measurable functions from \(X\) to \([0,1]\) whose pointwise supremum is not \(\mathcal{S}\)-measurable.

Note where the argument in 2.53 breaks down. There one writes, for \(a \in \mathbb{R}\),

\begin{equation*} h^{-1}\big((a, \infty]\big) = \bigcup_{k=1}^{\infty} f_k^{-1}\big((a, \infty]\big), \end{equation*}

and this union is in \(\mathcal{S}\) because \(\sigma\)-algebras are closed under countable unions. With an index set of cardinality \(|\mathbb{R}|\) the corresponding identity

\begin{equation*} f^{-1}\big((a, \infty)\big) = \bigcup_{t \in \mathbb{R}} f_t^{-1}\big((a, \infty)\big) \end{equation*}

is still true, but it is an uncountable union, and \(\sigma\)-algebras need not be closed under such unions. The example above realizes exactly this failure: each \(f_t^{-1}\big((\tfrac{1}{2},\infty)\big)\) is a singleton or empty, hence in \(\mathcal{S}\), but the union of the singletons \(\{t\}\) over \(t \in [0,1]\) is the non-measurable set \([0,1]\).

Fix \(x \in \mathbb{R}\). The claim is proved in two steps: first the inner limit over \(k\) is evaluated for each fixed \(j \in \mathbb{Z}^{+}\), and then the outer limit over \(j\) is evaluated.

Step 1: the inner limit. Fix \(j \in \mathbb{Z}^{+}\) and set \(c = \cos(j!\,\pi x)\), so \(c \in [-1, 1]\) and \(c^{2k} = (c^2)^k\) with \(c^2 \in [0,1]\). Hence

\begin{equation*} \lim_{k \to \infty} c^{2k} = \lim_{k \to \infty} (c^2)^k = \begin{cases} 1 & \text{if } c^2 = 1,\\ 0 & \text{if } c^2 < 1, \end{cases} \end{equation*}

because \(1^k = 1\) for every \(k\), while \(0 \le r < 1\) implies \(r^k \to 0\). (The case \(c = 0\) is included: then \(c^{2k} = 0\) for every \(k \in \mathbb{Z}^{+}\).) Thus the inner limit exists for every \(j\) and every \(x\), and it equals \(1\) exactly when \(|\cos(j!\,\pi x)| = 1\).

Now \(|\cos \theta| = 1\) if and only if \(\theta\) is an integer multiple of \(\pi\). Applying this with \(\theta = j!\,\pi x\) gives

\begin{equation*} |\cos(j!\,\pi x)| = 1 \iff j!\,\pi x \in \{m\pi : m \in \mathbb{Z}\} \iff j!\,x \in \mathbb{Z}. \end{equation*}

Therefore, defining \(g_j : \mathbb{R} \to \mathbb{R}\) by \(g_j(x) = \lim_{k \to \infty} \big(\cos(j!\,\pi x)\big)^{2k}\), we have shown

\begin{equation*} g_j = \chi_{A_j}, \qquad \text{where } A_j = \{x \in \mathbb{R} : j!\,x \in \mathbb{Z}\} = \Big\{ \tfrac{m}{j!} : m \in \mathbb{Z} \Big\}, \end{equation*}

with \(\chi_{A_j}\) the characteristic function of \(A_j\).

Step 2: the outer limit. Suppose first that \(x\) is irrational. If we had \(j!\,x \in \mathbb{Z}\) for some \(j \in \mathbb{Z}^{+}\), then \(x = (j!\,x)/j!\) would be a quotient of integers with nonzero denominator, hence rational, a contradiction. Thus \(j!\,x \notin \mathbb{Z}\) for every \(j \in \mathbb{Z}^{+}\), so \(g_j(x) = 0\) for every \(j\), and consequently

\begin{equation*} \lim_{j \to \infty} g_j(x) = 0. \end{equation*}

Now suppose \(x\) is rational and write \(x = p/q\) with \(p \in \mathbb{Z}\) and \(q \in \mathbb{Z}^{+}\). If \(j \ge q\) then \(q\) is one of the factors in the product \(j! = 1 \cdot 2 \cdots q \cdots j\), so \(j!/q \in \mathbb{Z}\) and hence

\begin{equation*} j!\,x = \frac{j!\,p}{q} = p \cdot \frac{j!}{q} \in \mathbb{Z}. \end{equation*}

Thus \(g_j(x) = 1\) for every \(j \ge q\), and consequently

\begin{equation*} \lim_{j \to \infty} g_j(x) = 1. \end{equation*}

Combining the two cases,

\begin{equation*} \lim_{j \to \infty} \left( \lim_{k \to \infty} \big( \cos(j!\,\pi x) \big)^{2k} \right) = \begin{cases} 1 & \text{if } x \text{ is rational},\\ 0 & \text{if } x \text{ is irrational}, \end{cases} \end{equation*}

for every \(x \in \mathbb{R}\), as desired.

Remark. The double limit above is \(\chi_{\mathbb{Q}}\), whose restriction to \([0,1]\) is the function of Example 1.17 that is not Riemann integrable. Each function \(x \mapsto \big(\cos(j!\,\pi x)\big)^{2k}\) is continuous on \(\mathbb{R}\), hence Borel measurable by 2.41. Applying 2.48 twice, first in \(k\) and then in \(j\), shows again that \(\chi_{\mathbb{Q}}\) is Borel measurable. This illustrates the point made just before 2.48: unlike the Riemann integrable functions, the Borel measurable functions are closed under pointwise limits, so no iteration of such limits can escape measurability.

Suppose, seeking a contradiction, that \((X, \mathcal{S}, \mu)\) is a measure space with \[ \{\mu(E) : E \in \mathcal{S}\} = [0, 1). \]

Because \(\mathcal{S}\) is a \(\sigma\)-algebra on \(X\), we have \(X \in \mathcal{S}\). Hence \(\mu(X)\) belongs to the set above, and so \[ \mu(X) < 1. \]

Choose a real number \(t\) with \(\mu(X) < t < 1\); this is possible because \(\mu(X) < 1\). Then \(t \in [0, 1)\), so by our assumption there exists \(E \in \mathcal{S}\) with \(\mu(E) = t\).

However \(E \subseteq X\) and both sets are in \(\mathcal{S}\), so 2.57(a) gives \[ t = \mu(E) \le \mu(X) < t, \] which is a contradiction.

Thus no such measure space exists.

The essential point is that the set \(\{\mu(E) : E \in \mathcal{S}\}\) always contains the number \(\mu(X)\), which is an upper bound for that entire set. In other words, a bounded set of the form \(\{\mu(E) : E \in \mathcal{S}\}\) must contain its own supremum, and \([0,1)\) does not contain its supremum \(1\).

Define \[ w_k = \mu(\{k\}) \qquad \text{for } k \in \mathbb{Z}^+. \] Because \(\mu\) maps \(2^{\mathbb{Z}^+}\) into \([0, \infty]\), this is a sequence in \([0, \infty]\).

As in 2.55, for a subset \(E \subseteq \mathbb{Z}^+\) the sum \(\sum_{k \in E} w_k\) means the supremum of the finite subsums \(\sum_{k \in D} w_k\) as \(D\) ranges over the finite subsets of \(E\). We must show \(\mu(E)\) equals that supremum for every \(E \subseteq \mathbb{Z}^+\). Note that every subset of \(\mathbb{Z}^+\) lies in the \(\sigma\)-algebra \(2^{\mathbb{Z}^+}\), so \(\mu(E)\) is defined for every \(E\).

First suppose \(E = \emptyset\). Then \(\mu(E) = 0\) by the definition of a measure, and the only finite subset of \(E\) is \(\emptyset\), whose subsum is the empty sum \(0\). So both sides equal \(0\).

Next suppose \(E\) is a nonempty finite set, say \(E = \{k_1, \ldots, k_n\}\) with \(k_1 < \cdots < k_n\). The singletons \(\{k_1\}, \ldots, \{k_n\}\) are disjoint sets in \(2^{\mathbb{Z}^+}\) whose union is \(E\), so finite additivity (which follows from countable additivity, as noted after 2.54, by appending \(\emptyset, \emptyset, \ldots\) to the list) gives \[ \mu(E) = \sum_{j=1}^{n} \mu(\{k_j\}) = \sum_{j=1}^{n} w_{k_j}. \] Since \(E\) itself is the largest finite subset of \(E\) and all the \(w_k\) are nonnegative, the supremum defining \(\sum_{k \in E} w_k\) is attained at \(D = E\) and equals \(\sum_{j=1}^{n} w_{k_j}\). Hence \(\mu(E) = \sum_{k \in E} w_k\).

Finally suppose \(E\) is infinite. Because \(E \subseteq \mathbb{Z}^+\), we may list its elements in increasing order as \(k_1 < k_2 < \cdots\), and then \[ E = \bigcup_{j=1}^{\infty} \{k_j\} \] is a disjoint union of sets in \(2^{\mathbb{Z}^+}\). Countable additivity gives \[ \mu(E) = \sum_{j=1}^{\infty} w_{k_j} = \lim_{n \to \infty} \sum_{j=1}^{n} w_{k_j} = \sup_{n \in \mathbb{Z}^+} \sum_{j=1}^{n} w_{k_j}, \] where the last two equalities hold because the partial sums form an increasing sequence in \([0, \infty]\) (all terms are nonnegative), so the limit of the partial sums equals their supremum. (If some \(w_{k_j} = \infty\) or if the series diverges, both sides equal \(\infty\).)

It remains to see that \[ \sup_{n \in \mathbb{Z}^+} \sum_{j=1}^{n} w_{k_j} = \sup \Big\{ \sum_{k \in D} w_k : D \subseteq E \text{ and } D \text{ is finite} \Big\}. \] The left side is a supremum over the particular finite subsets \(D_n = \{k_1, \ldots, k_n\}\) of \(E\), so it is at most the right side. Conversely, if \(D \subseteq E\) is finite and nonempty, then \(D \subseteq D_n\) for \(n\) large enough (take \(n\) so large that \(D \subseteq \{k_1, \ldots, k_n\}\), which is possible since \(D\) is finite and \(k_1, k_2, \ldots\) lists all of \(E\)); as all \(w_k \ge 0\) this gives \(\sum_{k \in D} w_k \le \sum_{j=1}^{n} w_{k_j}\), which is at most the left side. Hence the right side is at most the left side, and the two suprema are equal.

Combining the displays, \(\mu(E) = \sum_{k \in E} w_k\) for every infinite \(E\) as well, completing the proof.

Define \(w : \mathbb{Z}^+ \to [0, \infty]\) by \(w(k) = 2^{-k}\) and let \[ \mu(E) = \sum_{k \in E} 2^{-k} \qquad \text{for } E \subseteq \mathbb{Z}^+. \] By the third bullet point of 2.55, \(\mu\) is a measure on \((\mathbb{Z}^+, 2^{\mathbb{Z}^+})\). We claim that \(\{\mu(E) : E \subseteq \mathbb{Z}^+\} = [0, 1]\).

One inclusion is easy. For every \(E \subseteq \mathbb{Z}^+\) we have \[ 0 \le \mu(E) \le \mu(\mathbb{Z}^+) = \sum_{k=1}^{\infty} 2^{-k} = 1, \] using 2.57(a) for the middle inequality. Hence \(\{\mu(E) : E \subseteq \mathbb{Z}^+\} \subseteq [0, 1]\).

For the reverse inclusion, let \(t \in [0, 1]\); we produce \(E \subseteq \mathbb{Z}^+\) with \(\mu(E) = t\). If \(t = 1\), take \(E = \mathbb{Z}^+\), which works by the computation above. So assume \(0 \le t < 1\).

Define \(b_1, b_2, \ldots\) in \(\{0, 1\}\) recursively as follows. Having defined \(b_1, \ldots, b_{k-1}\), set \(s_{k-1} = \sum_{j=1}^{k-1} b_j 2^{-j}\) (so \(s_0 = 0\)) and put \[ b_k = \begin{cases} 1 & \text{if } s_{k-1} + 2^{-k} \le t, \\ 0 & \text{otherwise.} \end{cases} \]

We prove by induction on \(k \ge 0\) that \[ s_k \le t < s_k + 2^{-k}. \] For \(k = 0\) this says \(0 \le t < 1\), which holds by assumption. Suppose the inequalities hold for \(k-1\), so \(s_{k-1} \le t < s_{k-1} + 2^{-(k-1)}\). There are two cases.

If \(s_{k-1} + 2^{-k} \le t\), then \(b_k = 1\) and \(s_k = s_{k-1} + 2^{-k}\), so \(s_k \le t\); moreover \[ t < s_{k-1} + 2^{-(k-1)} = s_{k-1} + 2^{-k} + 2^{-k} = s_k + 2^{-k}. \]

If \(s_{k-1} + 2^{-k} > t\), then \(b_k = 0\) and \(s_k = s_{k-1}\), so \(s_k = s_{k-1} \le t\) and \(t < s_{k-1} + 2^{-k} = s_k + 2^{-k}\).

In both cases the inequalities hold for \(k\), completing the induction.

From \(s_k \le t < s_k + 2^{-k}\) we get \(0 \le t - s_k < 2^{-k}\) for every \(k\), hence \(s_k \to t\) as \(k \to \infty\).

Now let \(E = \{k \in \mathbb{Z}^+ : b_k = 1\}\), so that \(s_k = \sum_{j \in E, \, j \le k} 2^{-j}\) for each \(k\); that is, \(s_k\) is the subsum of \(\sum_{k \in E} 2^{-k}\) over the finite set \(E \cap \{1, \ldots, k\}\). We check that \(\mu(E) = \sup_k s_k\). Each \(s_k\) is a subsum of \(\mu(E)\) over a finite subset of \(E\), so \(s_k \le \mu(E)\) and hence \(\sup_k s_k \le \mu(E)\). Conversely, if \(D \subseteq E\) is finite then \(D \subseteq E \cap \{1, \ldots, k\}\) for \(k\) large enough, so \(\sum_{j \in D} 2^{-j} \le s_k \le \sup_k s_k\) because all terms are nonnegative; taking the supremum over such \(D\) gives \(\mu(E) \le \sup_k s_k\). Since \(s_1 \le s_2 \le \cdots\), the supremum equals the limit, and therefore \[ \mu(E) = \sum_{k \in E} 2^{-k} = \lim_{k \to \infty} s_k = t. \]

Therefore every \(t \in [0,1]\) is of the form \(\mu(E)\), and \(\{\mu(E) : E \subseteq \mathbb{Z}^+\} = [0, 1]\).

Let \(X = \mathbb{Z}^+ \times \{1, 2\}\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(X\), and define \(w : X \to [0, \infty]\) by \[ w(k, 1) = 2^{-k}, \qquad w(k, 2) = 3 \qquad \text{for } k \in \mathbb{Z}^+. \] Define \[ \mu(E) = \sum_{x \in E} w(x) \qquad \text{for } E \in \mathcal{S}, \] where as in 2.55 the sum is the supremum of the finite subsums. By the third bullet point of 2.55, \(\mu\) is a measure on \((X, \mathcal{S})\). Informally: \(X\) carries a part of total mass \(1\) whose measures fill out all of \([0,1]\) (the points \((k,1)\), with the weights of Exercise 3), together with infinitely many atoms of mass \(3\) (the points \((k,2)\)).

Given \(E \subseteq X\), write \[ E_1 = \{k \in \mathbb{Z}^+ : (k, 1) \in E\}, \qquad E_2 = \{k \in \mathbb{Z}^+ : (k, 2) \in E\}. \] Then \(E\) is the disjoint union of \(E_1 \times \{1\}\) and \(E_2 \times \{2\}\), both of which lie in \(\mathcal{S}\), so finite additivity gives \[ \mu(E) = s + b, \qquad \text{where } s = \sum_{k \in E_1} 2^{-k} \text{ and } b = \sum_{k \in E_2} 3. \] Here \(b = 3n\) if \(E_2\) is a finite set with exactly \(n\) elements, and \(b = \infty\) if \(E_2\) is infinite (in that case the finite subsums \(3, 6, 9, \ldots\) are unbounded, so their supremum is \(\infty\)).

Exercise 3 above showed that as \(E_1\) ranges over all subsets of \(\mathbb{Z}^+\), the number \(\sum_{k \in E_1} 2^{-k}\) ranges over exactly \([0, 1]\); so \(s\) ranges over exactly \([0,1]\). By the previous paragraph, as \(E_2\) ranges over all subsets of \(\mathbb{Z}^+\) the number \(b\) ranges over exactly \(\{0, 3, 6, 9, \ldots\} \cup \{\infty\}\). Moreover \(E_1\) and \(E_2\) can be chosen independently of each other, since every pair of subsets \((E_1, E_2)\) arises from the set \(E = (E_1 \times \{1\}) \cup (E_2 \times \{2\})\).

Hence \[ \{\mu(E) : E \in \mathcal{S}\} = \{s + b : s \in [0,1], \; b \in \{0, 3, 6, \ldots\} \cup \{\infty\}\}. \] If \(b = \infty\) the sum is \(\infty\) (for every \(s \in [0,1]\)). If \(b = 3k\) for some integer \(k \ge 0\), then as \(s\) runs over \([0,1]\) the sum \(3k + s\) runs over exactly \([3k, 3k+1]\). Therefore \[ \{\mu(E) : E \in \mathcal{S}\} = \{\infty\} \cup \bigcup_{k=0}^{\infty} [3k, 3k+1], \] as required.

Here \(\mathcal{A}\) being a set of disjoint sets means that its elements are pairwise disjoint: if \(A, A^{\prime} \in \mathcal{A}\) and \(A \ne A^{\prime}\), then \(A \cap A^{\prime} = \emptyset\).

For each \(n \in \mathbb{Z}^+\) define \[ \mathcal{A}_n = \Big\{ A \in \mathcal{A} : \mu(A) > \tfrac{1}{n} \Big\}. \]

Step 1: each \(\mathcal{A}_n\) is a finite set, with at most \(n\,\mu(X)\) elements.

Suppose \(A_1, \ldots, A_m\) are \(m\) distinct elements of \(\mathcal{A}_n\). They are pairwise disjoint sets in \(\mathcal{S}\), so their union \(A_1 \cup \cdots \cup A_m\) lies in \(\mathcal{S}\) and finite additivity (which follows from countable additivity, as noted after 2.54) gives \[ \mu(A_1 \cup \cdots \cup A_m) = \sum_{j=1}^{m} \mu(A_j) > \frac{m}{n}. \] On the other hand \(A_1 \cup \cdots \cup A_m \subseteq X\), so 2.57(a) gives \[ \mu(A_1 \cup \cdots \cup A_m) \le \mu(X). \] Combining the two displays yields \(m/n < \mu(X)\), that is, \(m < n\,\mu(X)\). Since \(\mu(X) < \infty\), the number \(n\,\mu(X)\) is a finite real number. Thus every finite list of distinct elements of \(\mathcal{A}_n\) has length less than \(n\,\mu(X)\). If \(\mathcal{A}_n\) were an infinite set, we could choose \(m\) distinct elements of it for some integer \(m \ge n\,\mu(X)\), contradicting the previous sentence. Hence \(\mathcal{A}_n\) is finite.

Step 2: \(\mathcal{A} = \bigcup_{n=1}^{\infty} \mathcal{A}_n\).

Each \(\mathcal{A}_n \subseteq \mathcal{A}\), giving one inclusion. Conversely, if \(A \in \mathcal{A}\) then \(\mu(A) > 0\) by hypothesis, so by the Archimedean property there exists \(n \in \mathbb{Z}^+\) with \(\frac{1}{n} < \mu(A)\), and then \(A \in \mathcal{A}_n\). (Note \(\mu(A)\) could be \(\infty\), which also satisfies \(\mu(A) > \frac{1}{n}\) for every \(n\); in fact \(\mu(A) \le \mu(X) < \infty\) here, but we do not need this.)

Step 3: conclusion.

By Steps 1 and 2, \(\mathcal{A}\) is a countable union of finite sets, and a countable union of countable sets is countable. Therefore \(\mathcal{A}\) is a countable set.

The answer is that \(c = 4\) is the only such number.

Part 1: if such a measure space exists, then \(c = 4\).

Suppose \(c \in [3, \infty)\) and \((X, \mathcal{S}, \mu)\) is a measure space with \[ R := \{\mu(E) : E \in \mathcal{S}\} = [0, 1] \cup [3, c]. \]

First we compute \(\mu(X)\). Every element of \(R\) is at most \(c\), and \(c \in R\) (because \(c \in [3, c]\)). Since \(E \subseteq X\) for every \(E \in \mathcal{S}\), 2.57(a) gives \(\mu(E) \le \mu(X)\) for every \(E \in \mathcal{S}\); taking \(E\) with \(\mu(E) = c\) shows \(c \le \mu(X)\). Also \(\mu(X) \in R\), so \(\mu(X) \le c\). Hence \[ \mu(X) = c < \infty. \]

Next, \(R\) is symmetric about \(c/2\): if \(t \in R\), say \(t = \mu(E)\) with \(E \in \mathcal{S}\), then \(X \setminus E \in \mathcal{S}\), and \(\mu(E) = t \le c < \infty\) since every element of \(R\) is at most \(c\). So 2.57(b), applied to the pair \(E \subseteq X\) of sets in \(\mathcal{S}\) (in the notation of 2.57, with \(D = E\) and with \(X\) playing the role of the larger set), gives \[ \mu(X \setminus E) = \mu(X) - \mu(E) = c - t. \] Thus we have the reflection property \[ t \in R \implies c - t \in R. \]

Apply the reflection property with \(t = 1 \in R\): we get \(c - 1 \in R = [0,1] \cup [3,c]\). Since \(c \ge 3\) we have \(c - 1 \ge 2 > 1\), so \(c - 1 \notin [0,1]\); therefore \(c - 1 \in [3, c]\), which forces \(c - 1 \ge 3\), that is, \[ c \ge 4. \]

Now suppose, seeking a contradiction, that \(c \ge 5\). Then \(3 \le c - 2 \le c\), so \(c - 2 \in [3, c] \subseteq R\). Applying the reflection property with \(t = c - 2\) gives \(c - (c-2) = 2 \in R\). But \(2 \notin [0,1]\) and \(2 \notin [3, c]\), so \(2 \notin R\), a contradiction. Hence \[ 4 \le c < 5. \]

Finally apply the reflection property with \(t = 3 \in R\): we get \(c - 3 \in R\). From \(4 \le c < 5\) we have \(1 \le c - 3 < 2\). The only element of \(R = [0,1] \cup [3,c]\) lying in \([1, 2)\) is \(1\), so \(c - 3 = 1\) and therefore \[ c = 4. \]

Part 2: for \(c = 4\) such a measure space exists.

Let \(X = \{0, 1, 2, 3, \ldots\}\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(X\), and define \(w : X \to [0, \infty]\) by \[ w(0) = 3, \qquad w(k) = 2^{-k} \text{ for } k \in \mathbb{Z}^+. \] Let \(\mu(E) = \sum_{x \in E} w(x)\) for \(E \subseteq X\); by the third bullet point of 2.55 this is a measure on \((X, \mathcal{S})\).

For \(E \subseteq X\) write \(E^{\prime} = E \cap \mathbb{Z}^+\). Splitting \(E\) into the disjoint pieces \(E \cap \{0\}\) and \(E^{\prime}\) and using finite additivity, \[ \mu(E) = \begin{cases} \displaystyle\sum_{k \in E^{\prime}} 2^{-k} & \text{if } 0 \notin E, \\[2mm] 3 + \displaystyle\sum_{k \in E^{\prime}} 2^{-k} & \text{if } 0 \in E. \end{cases} \] By Exercise 3, as \(E^{\prime}\) ranges over all subsets of \(\mathbb{Z}^+\) the quantity \(\sum_{k \in E^{\prime}} 2^{-k}\) ranges over exactly \([0, 1]\), and the choice of \(E^{\prime}\) is independent of whether \(0 \in E\). Hence \[ \{\mu(E) : E \in \mathcal{S}\} = [0, 1] \cup \big(3 + [0,1]\big) = [0, 1] \cup [3, 4], \] which is \([0,1] \cup [3,c]\) with \(c = 4\).

Let \(X = \mathbb{Z}^+ \times \{1, 2\}\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(X\), and define \(w : X \to [0, \infty]\) by \[ w(k, 1) = 2^{-k}, \qquad w(k, 2) = k + 2 \qquad \text{for } k \in \mathbb{Z}^+. \] Define \[ \mu(E) = \sum_{x \in E} w(x) \qquad \text{for } E \in \mathcal{S}, \] the sum again meaning the supremum of the finite subsums, as in 2.55; by the third bullet point of 2.55, \(\mu\) is a measure on \((X, \mathcal{S})\).

The idea: the points \((k,1)\) carry the weights of Exercise 3, giving a part whose measures fill out exactly \([0,1]\); the points \((k,2)\) are atoms of masses \(3, 4, 5, 6, \ldots\), each of which is at least \(3\) (so no value in the forbidden gap \((1,3)\) can be produced) while the integers \(3, 4, 5, \ldots\) are spaced only \(1\) apart (so together with the \([0,1]\) part they sweep out all of \([3, \infty)\)).

Given \(E \subseteq X\), set \[ E_1 = \{k \in \mathbb{Z}^+ : (k,1) \in E\}, \qquad E_2 = \{k \in \mathbb{Z}^+ : (k,2) \in E\}. \] Since \(E\) is the disjoint union of \(E_1 \times \{1\}\) and \(E_2 \times \{2\}\), finite additivity gives \[ \mu(E) = s + b, \qquad \text{where } s = \sum_{k \in E_1} 2^{-k} \text{ and } b = \sum_{k \in E_2} (k+2). \] Note that \(E_1\) and \(E_2\) may be chosen independently: every pair of subsets \((E_1, E_2)\) of \(\mathbb{Z}^+\) arises from \(E = (E_1 \times \{1\}) \cup (E_2 \times \{2\})\).

By Exercise 3, as \(E_1\) ranges over all subsets of \(\mathbb{Z}^+\), the number \(s\) ranges over exactly \([0,1]\). Let \[ B = \Big\{ \sum_{k \in E_2} (k+2) : E_2 \subseteq \mathbb{Z}^+ \Big\} \] be the set of possible values of \(b\). We record three facts about \(B\).

(i) \(B \subseteq \{0\} \cup [3, \infty]\). Indeed, if \(E_2 = \emptyset\) then \(b = 0\). If \(E_2 \ne \emptyset\), pick \(k \in E_2\); then \(\{k\}\) is a finite subset of \(E_2\), so \(b \ge k + 2 \ge 3\).

(ii) Every integer \(n \ge 3\) belongs to \(B\): take \(E_2 = \{n - 2\}\), which is a subset of \(\mathbb{Z}^+\) since \(n - 2 \ge 1\), and then \(b = (n-2) + 2 = n\).

(iii) \(\infty \in B\): take \(E_2 = \mathbb{Z}^+\); the finite subsums \(\sum_{k=1}^{m}(k+2)\) are unbounded, so their supremum is \(\infty\).

Now we show \(\{\mu(E) : E \in \mathcal{S}\} = [0,1] \cup [3, \infty]\).

Containment \(\subseteq\): let \(E \in \mathcal{S}\) and write \(\mu(E) = s + b\) as above, with \(s \in [0,1]\) and \(b \in B\). By (i) either \(b = 0\), in which case \(\mu(E) = s \in [0,1]\); or \(b \ge 3\) (possibly \(b = \infty\)), in which case \(\mu(E) = s + b \ge 3\), so \(\mu(E) \in [3, \infty]\). Either way \(\mu(E) \in [0,1] \cup [3,\infty]\).

Containment \(\supseteq\): let \(t \in [0,1] \cup [3, \infty]\).

If \(t \in [0,1]\), choose \(E_1 \subseteq \mathbb{Z}^+\) with \(\sum_{k \in E_1} 2^{-k} = t\) (possible by Exercise 3) and take \(E_2 = \emptyset\); then \(\mu(E) = t\).

If \(t = \infty\), take \(E_1 = \emptyset\) and \(E_2 = \mathbb{Z}^+\); then \(\mu(E) = \infty\) by (iii).

If \(t \in [3, \infty)\), let \(n = \lfloor t \rfloor\), the greatest integer with \(n \le t\). Then \(n \ge 3\) and \(t - n \in [0, 1)\). Take \(E_2 = \{n-2\}\), so that \(b = n\) by (ii), and choose \(E_1 \subseteq \mathbb{Z}^+\) with \(s = \sum_{k \in E_1} 2^{-k} = t - n\) (possible by Exercise 3, since \(t - n \in [0,1]\)). Then for \(E = (E_1 \times \{1\}) \cup (E_2 \times \{2\})\) we get \[ \mu(E) = (t - n) + n = t. \]

Hence every element of \([0,1] \cup [3, \infty]\) is a value of \(\mu\), and combining the two containments, \[ \{\mu(E) : E \in \mathcal{S}\} = [0, 1] \cup [3, \infty], \] as required.

Take

\begin{equation*} X = \{1, 2, 3, 4\}, \qquad \mathcal{S} = \{E : E \subseteq X\}, \qquad \mathcal{A} = \big\{\{1,2\}, \{2,3\}\big\}, \end{equation*}

and let \(\mu\) and \(\nu\) be the measures on \((X, \mathcal{S})\) defined by

\begin{equation*} \mu = \delta_2 + \delta_4, \qquad \nu = \delta_1 + \delta_3, \end{equation*}

where \(\delta_c\) denotes the Dirac measure at \(c\) (see 2.55). Each of \(\mu\) and \(\nu\) is a measure, being a sum of two measures (Exercise 9 of this section, or the direct verification that a weighted counting measure \(E \mapsto \sum_{x \in E} w(x)\) is a measure, as in 2.55). Explicitly, for \(E \subseteq X\),

\begin{equation*} \mu(E) = \#\big(E \cap \{2,4\}\big), \qquad \nu(E) = \#\big(E \cap \{1,3\}\big). \end{equation*}

First, \(\mathcal{S}\) really is the smallest \(\sigma\)-algebra on \(X\) containing \(\mathcal{A}\). Indeed, let \(\mathcal{T}\) be any \(\sigma\)-algebra on \(X\) with \(\mathcal{A} \subseteq \mathcal{T}\). Then \(\mathcal{T}\) is closed under finite intersections and relative complements (because it is closed under complementation and countable unions), so

\begin{equation*} \{2\} = \{1,2\} \cap \{2,3\} \in \mathcal{T}, \quad \{1\} = \{1,2\} \setminus \{2\} \in \mathcal{T}, \quad \{3\} = \{2,3\} \setminus \{2\} \in \mathcal{T}, \end{equation*}

and hence \(\{4\} = X \setminus (\{1\} \cup \{2\} \cup \{3\}) \in \mathcal{T}\). Thus \(\mathcal{T}\) contains every singleton of \(X\); as \(X\) is finite, every subset of \(X\) is a finite union of singletons, so \(\mathcal{T} = \mathcal{S}\). In particular the smallest \(\sigma\)-algebra containing \(\mathcal{A}\) is \(\mathcal{S}\).

Next, \(\mu\) and \(\nu\) agree on \(\mathcal{A}\):

\begin{equation*} \mu(\{1,2\}) = 1 = \nu(\{1,2\}), \qquad \mu(\{2,3\}) = 1 = \nu(\{2,3\}). \end{equation*}

They also agree on \(X\), with a finite common value:

\begin{equation*} \mu(X) = 2 = \nu(X) < \infty. \end{equation*}

Finally \(\mu \neq \nu\), since

\begin{equation*} \mu(\{1\}) = 0 \neq 1 = \nu(\{1\}). \end{equation*}

This example shows that agreement of two finite measures on a generating family \(\mathcal{A}\) (together with agreement of the total masses) does not force the measures to be equal. The standard uniqueness theorems require in addition that \(\mathcal{A}\) be closed under finite intersections; here that hypothesis fails, because \(\{1,2\} \cap \{2,3\} = \{2\} \notin \mathcal{A}\).

Throughout we use the usual arithmetic of \([0, \infty]\): \(a + \infty = \infty + a = \infty\) for every \(a \in [0, \infty]\). With this convention \([0, \infty]\) is closed under addition, and addition on \([0,\infty]\) is commutative and associative.

Since \(\mu(E) \in [0,\infty]\) and \(\nu(E) \in [0,\infty]\) for each \(E \in \mathcal{S}\), the function \(\mu + \nu\) maps \(\mathcal{S}\) into \([0, \infty]\). Also

\begin{equation*} (\mu + \nu)(\varnothing) = \mu(\varnothing) + \nu(\varnothing) = 0 + 0 = 0 . \end{equation*}

It remains to prove countable additivity. Before doing so we record an elementary fact about series in \([0, \infty]\).

Claim. If \((a_k)\) and \((b_k)\) are sequences in \([0, \infty]\), then

\begin{equation*} \sum_{k=1}^{\infty} (a_k + b_k) = \sum_{k=1}^{\infty} a_k + \sum_{k=1}^{\infty} b_k . \end{equation*}

Here each such series of nonnegative terms is by definition the limit in \([0,\infty]\) of its increasing sequence of partial sums, equivalently the supremum of the partial sums (which exists in \([0,\infty]\) whether or not it is finite).

Proof of Claim. For each \(n \in \mathbf{Z}^{+}\) let

\begin{equation*} A_n = \sum_{k=1}^{n} a_k, \qquad B_n = \sum_{k=1}^{n} b_k, \qquad C_n = \sum_{k=1}^{n}(a_k + b_k). \end{equation*}

Rearranging a finite sum of elements of \([0,\infty]\) gives \(C_n = A_n + B_n\). The sequences \((A_n)\), \((B_n)\), \((C_n)\) are increasing in \([0, \infty]\), with suprema \(A\), \(B\), \(C\) say, and these suprema are the three series in question.

If \(A = \infty\), then for every \(M \in (0,\infty)\) there is \(n\) with \(A_n > M\), hence \(C_n \geq A_n > M\); thus \(C = \infty = A + B\). The case \(B = \infty\) is symmetric. So assume \(A < \infty\) and \(B < \infty\). Then \(C_n = A_n + B_n \leq A + B\) for all \(n\), so \(C \leq A + B\). Conversely, given \(\varepsilon > 0\) choose \(n\) with \(A_n > A - \varepsilon/2\) and \(B_n > B - \varepsilon/2\) (possible, using the same \(n\) for both by monotonicity); then \(C_n = A_n + B_n > A + B - \varepsilon\), so \(C > A + B - \varepsilon\). As \(\varepsilon > 0\) was arbitrary, \(C \geq A + B\). Hence \(C = A + B\), proving the claim.

Now let \(E_1, E_2, \dots\) be a disjoint sequence of sets in \(\mathcal{S}\). Using the countable additivity of \(\mu\) and of \(\nu\) and then the claim with \(a_k = \mu(E_k)\) and \(b_k = \nu(E_k)\), we get

\begin{equation*} \begin{aligned} (\mu + \nu)\Big(\bigcup_{k=1}^{\infty} E_k\Big) &= \mu\Big(\bigcup_{k=1}^{\infty} E_k\Big) + \nu\Big(\bigcup_{k=1}^{\infty} E_k\Big) \\ &= \sum_{k=1}^{\infty} \mu(E_k) + \sum_{k=1}^{\infty} \nu(E_k) \\ &= \sum_{k=1}^{\infty} \big(\mu(E_k) + \nu(E_k)\big) \\ &= \sum_{k=1}^{\infty} (\mu + \nu)(E_k). \end{aligned} \end{equation*}

Thus \(\mu + \nu : \mathcal{S} \to [0, \infty]\) satisfies \((\mu+\nu)(\varnothing) = 0\) and is countably additive, so \(\mu + \nu\) is a measure on \((X, \mathcal{S})\), as required.

The same argument shows more generally that \(c\mu + d\nu\) is a measure for all \(c, d \in [0, \infty)\), and (by an extra application of the interchange of two suprema) that any countable sum \(\sum_{j=1}^{\infty} \mu_j\) of measures on \((X, \mathcal{S})\) is again a measure on \((X, \mathcal{S})\).

Take \(X = \mathbf{Z}^{+}\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(\mathbf{Z}^{+}\), and let \(\mu\) be counting measure on \((X, \mathcal{S})\): thus \(\mu(E) = n\) if \(E\) is a finite set with exactly \(n\) elements, and \(\mu(E) = \infty\) if \(E\) is infinite. As noted in 2.55, counting measure is indeed a measure. Define

\begin{equation*} E_k = \{k, k+1, k+2, \dots\} \quad \text{for } k \in \mathbf{Z}^{+}. \end{equation*}

Then \(E_1 \supseteq E_2 \supseteq \cdots\) is a decreasing sequence of sets in \(\mathcal{S}\).

Each \(E_k\) is an infinite subset of \(\mathbf{Z}^{+}\), so \(\mu(E_k) = \infty\) for every \(k\), and therefore

\begin{equation*} \lim_{k \to \infty} \mu(E_k) = \infty . \end{equation*}

On the other hand

\begin{equation*} \bigcap_{k=1}^{\infty} E_k = \varnothing, \end{equation*}

because if \(n \in \mathbf{Z}^{+}\) then \(n \notin E_{n+1}\), so no positive integer lies in every \(E_k\). Hence

\begin{equation*} \mu\Big(\bigcap_{k=1}^{\infty} E_k\Big) = \mu(\varnothing) = 0 \neq \infty = \lim_{k \to \infty} \mu(E_k), \end{equation*}

as desired.

This example shows that the hypothesis \(\mu(E_1) < \infty\) in 2.60 cannot be deleted. In fact that hypothesis fails as badly as possible here, and necessarily so: if \((E_k)\) is a decreasing sequence in \(\mathcal{S}\) with \(\mu(E_n) < \infty\) for some \(n \in \mathbf{Z}^{+}\), then applying 2.60 to the decreasing sequence \(E_n \supseteq E_{n+1} \supseteq \cdots\) (whose intersection equals \(\bigcap_{k=1}^{\infty} E_k\)) gives

\begin{equation*} \mu\Big(\bigcap_{k=1}^{\infty} E_k\Big) = \lim_{k \to \infty} \mu(E_k), \end{equation*}

where we used that discarding the first \(n - 1\) terms does not change the limit. Thus any counterexample must have \(\mu(E_k) = \infty\) for every \(k\).

The formula is the inclusion-exclusion identity for three sets:

\begin{equation*} \mu(C \cup D \cup E) = \mu( C) + \mu(D) + \mu(E) - \mu(C \cap D) - \mu(C \cap E) - \mu(D \cap E) + \mu(C \cap D \cap E). \end{equation*}

Notation and well-definedness. Write

\begin{equation*} p = \mu(C \cap D), \quad q = \mu(C \cap E), \quad r = \mu(D \cap E), \quad s = \mu(C \cap D \cap E). \end{equation*}

By hypothesis \(p, q, r < \infty\), and since \(C \cap D \cap E \subseteq C \cap D\) we get \(s \leq p < \infty\) from 2.57(a). So all four of \(p, q, r, s\) are (finite) nonnegative real numbers, while \(\mu( C), \mu(D), \mu(E)\) may be \(\infty\). The right side of the displayed formula is therefore unambiguous: it is an element of \([0,\infty]\) from which only finite quantities are subtracted, with the convention \(\infty - t = \infty\) for \(t \in \mathbf{R}\). To avoid any subtraction at all in the proof, we shall establish the equivalent identity

\begin{equation*} \mu(C \cup D \cup E) + p + q + r = \mu( C) + \mu(D) + \mu(E) + s , \end{equation*}

in which every term lies in \([0, \infty]\) and only addition occurs; since \(p, q, r, s\) are finite reals, this identity is equivalent to the displayed formula above.

Step 1. Since \(\mu(C \cap D) = p < \infty\), the result 2.61 applied to \(C\) and \(D\) gives \(\mu(C \cup D) = \mu( C) + \mu(D) - p\), that is,

\begin{equation*} \mu(C \cup D) + p = \mu( C) + \mu(D). \end{equation*}

Step 2. Note that

\begin{equation*} (C \cup D) \cap E = (C \cap E) \cup (D \cap E), \end{equation*}

and that \((C \cap E) \cap (D \cap E) = C \cap D \cap E\), which has measure \(s < \infty\). Hence 2.61 applies to the sets \(C \cap E\) and \(D \cap E\) and yields

\begin{equation*} \mu\big((C \cup D) \cap E\big) = q + r - s . \end{equation*}

In particular \(\mu\big((C \cup D) \cap E\big) \leq q + r < \infty\), and

\begin{equation*} \mu\big((C \cup D) \cap E\big) + s = q + r . \end{equation*}

Step 3. Because \(\mu\big((C \cup D) \cap E\big) < \infty\) by Step 2, the result 2.61 applies to the sets \(C \cup D\) and \(E\) and gives

\begin{equation*} \mu(C \cup D \cup E) = \mu(C \cup D) + \mu(E) - \mu\big((C \cup D) \cap E\big), \end{equation*}

which, since the subtracted quantity is finite, is the same as

\begin{equation*} \mu(C \cup D \cup E) + \mu\big((C \cup D) \cap E\big) = \mu(C \cup D) + \mu(E). \end{equation*}

Step 4. Add \(s\) to both sides of the identity in Step 3 and use Step 2 to replace \(\mu\big((C \cup D) \cap E\big) + s\) by \(q + r\):

\begin{equation*} \mu(C \cup D \cup E) + q + r = \mu(C \cup D) + \mu(E) + s . \end{equation*}

Now add \(p\) to both sides and use Step 1 to replace \(\mu(C \cup D) + p\) by \(\mu( C) + \mu(D)\):

\begin{equation*} \mu(C \cup D \cup E) + p + q + r = \mu( C) + \mu(D) + \mu(E) + s . \end{equation*}

(All rearrangements are legitimate: addition on \([0, \infty]\) is commutative and associative.)

This is exactly the addition-only identity announced above. Finally, since \(p\), \(q\), \(r\), \(s\) are finite real numbers, we may subtract \(p + q + r\) from both sides to conclude

\begin{equation*} \mu(C \cup D \cup E) = \mu( C) + \mu(D) + \mu(E) - \mu(C \cap D) - \mu(C \cap E) - \mu(D \cap E) + \mu(C \cap D \cap E), \end{equation*}

as claimed. (If \(\mu( C)\), \(\mu(D)\), or \(\mu(E)\) equals \(\infty\), both sides equal \(\infty\): the left side because \(\mu(C \cup D \cup E) \geq \mu( C)\) and similarly for \(D\) and \(E\) by 2.57(a), and the right side because a finite quantity is being subtracted from \(\infty\).)

Notation. For a function \(w : X \to [0, \infty]\) and a set \(A \subseteq X\) put

\begin{equation*} \sum_{x \in A} w(x) = \sup\Big\{ \sum_{x \in D} w(x) : D \subseteq A \text{ and } D \text{ is finite} \Big\} \in [0, \infty], \end{equation*}

as in the third bullet of 2.55 (the empty sum is \(0\)). For \(E \in \mathcal{S}\) write \([E \text{ uncountable}]\) for \(1\) if \(E\) is uncountable and \(0\) if \(E\) is countable. Products of elements of \([0,\infty]\) are computed with the usual convention \(0 \cdot \infty = 0\); thus \(c \cdot [E \text{ uncountable}] = 0\) whenever \(E\) is countable, even if \(c = \infty\).

Answer. The measures on \((X, \mathcal{S})\) are exactly the functions \(\mu_{w,c} : \mathcal{S} \to [0,\infty]\) of the form

\begin{equation*} \mu_{w,c}(E) = \sum_{x \in E} w(x) \; + \; c \cdot [E \text{ uncountable}], \end{equation*}

where \(w : X \to [0, \infty]\) is an arbitrary function and \(c \in [0, \infty]\) is an arbitrary constant. In other words: a measure here is a weighted counting measure with weights \(w(x) = \mu(\{x\})\), plus an extra lump of mass \(c\) carried by every uncountable set of \(\mathcal{S}\) (when \(X\) is uncountable, the uncountable sets in \(\mathcal{S}\) are exactly the sets in \(\mathcal{S}\) with countable complement). If \(X\) itself is countable, then \(\mathcal{S}\) is the \(\sigma\)-algebra of all subsets of \(X\), no set is uncountable, and the description collapses to: the measures are exactly \(E \mapsto \sum_{x \in E} w(x)\) with \(w : X \to [0,\infty]\) arbitrary.

We prove three lemmas, then verify each \(\mu_{w,c}\) is a measure, then show every measure is of this form.

Lemma 1 (unordered sums are countably additive). If \(w : X \to [0,\infty]\) and \(A = \bigcup_{k=1}^{\infty} A_k\) with \(A_1, A_2, \dots\) disjoint subsets of \(X\), then \(\sum_{x \in A} w(x) = \sum_{k=1}^{\infty} \sum_{x \in A_k} w(x)\).

Proof. Let \(D \subseteq A\) be finite. Then \(D\) meets only finitely many \(A_k\), say only those with \(k \leq n\), and \(D\) is the disjoint union of the sets \(D \cap A_k\) for \(k \leq n\); hence

\begin{equation*} \sum_{x \in D} w(x) = \sum_{k=1}^{n} \sum_{x \in D \cap A_k} w(x) \leq \sum_{k=1}^{n} \sum_{x \in A_k} w(x) \leq \sum_{k=1}^{\infty} \sum_{x \in A_k} w(x). \end{equation*}

Taking the supremum over such \(D\) gives \(\sum_{x \in A} w(x) \leq \sum_{k=1}^{\infty} \sum_{x \in A_k} w(x)\). Conversely, fix \(n \in \mathbf{Z}^{+}\) and finite sets \(D_k \subseteq A_k\) for \(k = 1, \dots, n\). Then \(D_1 \cup \cdots \cup D_n\) is a finite subset of \(A\), so

\begin{equation*} \sum_{k=1}^{n} \sum_{x \in D_k} w(x) = \sum_{x \in D_1 \cup \cdots \cup D_n} w(x) \leq \sum_{x \in A} w(x). \end{equation*}

Taking suprema over \(D_1\), then \(D_2\), and so on (finitely many independent suprema, each legitimate because addition on \([0,\infty]\) is order preserving and commutes with suprema of increasing families in one variable) gives \(\sum_{k=1}^{n} \sum_{x \in A_k} w(x) \leq \sum_{x \in A} w(x)\). Letting \(n \to \infty\) finishes the proof.

Lemma 2. If \(N \subseteq X\) is uncountable and \(w(x) > 0\) for every \(x \in N\), then \(\sum_{x \in N} w(x) = \infty\).

Proof. We have \(N = \bigcup_{n=1}^{\infty} \{x \in N : w(x) > 1/n\}\). A countable union of countable sets is countable, so some \(\{x \in N : w(x) > 1/n\}\) is uncountable, in particular infinite. For each \(m \in \mathbf{Z}^{+}\) that set therefore contains a finite subset \(D\) with \(m\) elements, and \(\sum_{x \in D} w(x) > m/n\). Hence the supremum defining \(\sum_{x \in N} w(x)\) is \(\infty\).

Lemma 3. Let \(E_1, E_2, \dots\) be a disjoint sequence of sets in \(\mathcal{S}\) and \(E = \bigcup_{k=1}^{\infty} E_k\). Then \(E\) is uncountable if and only if exactly one \(E_k\) is uncountable.

Proof. First, at most one \(E_k\) can be uncountable: if \(E_i\) and \(E_j\) were uncountable with \(i \neq j\), then (being in \(\mathcal{S}\) and uncountable) each has countable complement, and \(E_i \subseteq X \setminus E_j\) because \(E_i \cap E_j = \varnothing\); thus \(E_i\) is a subset of a countable set, contradicting that \(E_i\) is uncountable. Next, if every \(E_k\) is countable then \(E\) is countable, being a countable union of countable sets. Conversely, if some \(E_k\) is uncountable then \(E \supseteq E_k\) is uncountable. Combining these statements gives the lemma.

Step 1: every \(\mu_{w,c}\) is a measure on \((X, \mathcal{S})\).

Clearly \(\mu_{w,c}\) maps \(\mathcal{S}\) into \([0, \infty]\), and \(\mu_{w,c}(\varnothing) = 0 + 0 = 0\) because \(\varnothing\) is countable and the empty sum is \(0\). Let \(E_1, E_2, \dots\) be a disjoint sequence in \(\mathcal{S}\) and \(E = \bigcup_{k=1}^{\infty} E_k\). By Lemma 3,

\begin{equation*} [E \text{ uncountable}] = \sum_{k=1}^{\infty} [E_k \text{ uncountable}], \end{equation*}

since either every term on the right is \(0\) and so is the left side, or exactly one term on the right is \(1\) and the left side is \(1\). Because at most one term of that series is nonzero, we may multiply through by \(c\) termwise with no ambiguity even when \(c = \infty\):

\begin{equation*} c\,[E \text{ uncountable}] = \sum_{k=1}^{\infty} c\,[E_k \text{ uncountable}], \end{equation*}

both sides being \(c\) if some (necessarily unique) \(E_k\) is uncountable and \(0\) otherwise. Combining this with Lemma 1 for the weight part, we get

\begin{equation*} \mu_{w,c}(E) = \sum_{x \in E} w(x) + c\,[E \text{ uncountable}] = \sum_{k=1}^{\infty} \Big( \sum_{x \in E_k} w(x) + c\,[E_k \text{ uncountable}] \Big) = \sum_{k=1}^{\infty} \mu_{w,c}(E_k), \end{equation*}

where the middle equality also uses that two series with terms in \([0,\infty]\) may be added termwise (proved in the solution to Exercise 9 of this section). Hence \(\mu_{w,c}\) is countably additive, so it is a measure.

Step 2: every measure on \((X, \mathcal{S})\) has this form.

Let \(\mu\) be a measure on \((X, \mathcal{S})\) and define

\begin{equation*} w(x) = \mu(\{x\}) \quad \text{for } x \in X, \qquad N = \{x \in X : w(x) > 0\}. \end{equation*}

(Each singleton is countable, hence in \(\mathcal{S}\).)

First, \(\mu(E) = \sum_{x \in E} w(x)\) for every countable \(E \in \mathcal{S}\). Indeed, list the distinct elements of \(E\) as \(x_1, x_2, \dots\) (a finite or infinite list) and apply countable additivity to the disjoint sequence \(\{x_1\}, \{x_2\}, \dots\) (padded with copies of \(\varnothing\) if \(E\) is finite) to get \(\mu(E) = \sum_k w(x_k)\); for nonnegative terms this series equals the unordered sum \(\sum_{x \in E} w(x)\), since its partial sums are exactly the subsums over the finite sets \(\{x_1, \dots, x_n\}\) and these are cofinal among all finite subsets of \(E\).

Second, note that for every \(E \in \mathcal{S}\) and every finite \(D \subseteq E\) we have \(\mu(E) \geq \mu(D) = \sum_{x \in D} w(x)\), using 2.57(a) and finite additivity; taking the supremum over \(D\) gives

\begin{equation*} \mu(E) \geq \sum_{x \in E} w(x) \qquad \text{for all } E \in \mathcal{S}. \end{equation*}

Now distinguish two cases.

Case 1: \(N\) is uncountable. Let \(G \in \mathcal{S}\) be uncountable; then \(X \setminus G\) is countable, so \(G \cap N = N \setminus (X \setminus G)\) is uncountable. By Lemma 2, \(\sum_{x \in G} w(x) \geq \sum_{x \in G \cap N} w(x) = \infty\), and by the inequality just proved \(\mu(G) = \infty\) as well. Hence \(\mu(G) = \sum_{x \in G} w(x) + 0\) for uncountable \(G\), while \(\mu(E) = \sum_{x \in E} w(x) + 0\) for countable \(E\) by the first paragraph of this step. Thus \(\mu = \mu_{w,0}\).

Case 2: \(N\) is countable. Then \(X \setminus N \in \mathcal{S}\); set

\begin{equation*} c = \mu(X \setminus N) \in [0, \infty]. \end{equation*}

Let \(G \in \mathcal{S}\) be uncountable, so \(X \setminus G\) is countable. The sets \(G \cap N\) and \(G \setminus N\) belong to \(\mathcal{S}\) (the first is countable; the second has complement \((X \setminus G) \cup N\), which is countable), and they are disjoint with union \(G\), so

\begin{equation*} \mu(G) = \mu(G \cap N) + \mu(G \setminus N). \end{equation*}

Since \(G \cap N\) is countable, \(\mu(G \cap N) = \sum_{x \in G \cap N} w(x) = \sum_{x \in G} w(x)\), the last equality because \(w\) vanishes off \(N\). For the other term, decompose \(X \setminus N\) as the disjoint union of \(G \setminus N\) and \((X \setminus N) \cap (X \setminus G)\); the latter set is countable (being a subset of \(X \setminus G\)) and \(w\) vanishes on it (being a subset of \(X \setminus N\)), so its measure is \(\sum_{x \in (X \setminus N) \cap (X \setminus G)} w(x) = 0\). Hence

\begin{equation*} c = \mu(X \setminus N) = \mu(G \setminus N) + 0 = \mu(G \setminus N). \end{equation*}

Combining the last three displays gives \(\mu(G) = \sum_{x \in G} w(x) + c\) for every uncountable \(G \in \mathcal{S}\), while \(\mu(E) = \sum_{x \in E} w(x)\) for countable \(E\). Thus \(\mu = \mu_{w,c}\).

In both cases \(\mu\) has the stated form, which completes the description.

Remarks on uniqueness. In the representation \(\mu = \mu_{w,c}\) the weight function is always uniquely determined, since each singleton is countable and hence \(w(x) = \mu_{w,c}(\{x\})\). The constant \(c\), however, need not be. Precisely, \(c\) is uniquely determined by \(\mu\) if and only if \(X\) is uncountable and \(N = \{x \in X : w(x) > 0\}\) is countable. Indeed, if \(X\) is countable then \(\mathcal{S}\) contains no uncountable set, so \(\mu_{w,c} = \mu_{w,c^{\prime}}\) for all \(c, c^{\prime} \in [0,\infty]\); and if \(N\) is uncountable then Lemma 2 (applied as in Case 1) gives \(\sum_{x \in G} w(x) = \infty\) for every uncountable \(G \in \mathcal{S}\), so again \(\mu_{w,c}\) does not depend on \(c\). In both of these degenerate situations one may normalize \(c = 0\). If instead \(X\) is uncountable and \(N\) is countable, then \(X \setminus N\) is an uncountable set in \(\mathcal{S}\) with \(\sum_{x \in X \setminus N} w(x) = 0\), so \(c = \mu(X \setminus N)\) is recovered from \(\mu\).

For example, taking \(w = 0\) and \(c = 3\) recovers the measure of the fourth bullet of 2.55, which assigns \(0\) to countable sets and \(3\) to uncountable sets in \(\mathcal{S}\).

Throughout, a decimal expansion of \(x \in [0,1]\) means a sequence \(d_1, d_2, \ldots\) of digits in \(\{0,1,\ldots,9\}\) with \(x = \sum_{j=1}^{\infty} d_j 10^{-j}\). Let

\begin{equation*} E = \{x \in (0,1) : \text{some decimal expansion of } x \text{ contains one hundred consecutive } 4\text{s}\}. \end{equation*}

A basic observation. For a finite string \(s = (d_1,\ldots,d_m) \in \{0,\ldots,9\}^m\) put

\begin{equation*} a_s = \sum_{j=1}^{m} d_j 10^{-j}, \qquad I_s = [a_s,\, a_s + 10^{-m}]. \end{equation*}

Then a number \(x \in [0,1]\) has a decimal expansion beginning with the string \(s\) if and only if \(x \in I_s\).

Indeed, if \(x = \sum_{j=1}^\infty d_j 10^{-j}\) with \(d_1,\ldots,d_m\) equal to \(s\), then \(x - a_s = \sum_{j > m} d_j 10^{-j} \in [0, 10^{-m}]\), so \(x \in I_s\). Conversely, if \(x \in I_s\), then \(10^{m}(x - a_s) \in [0,1]\), so we may write \(10^{m}(x-a_s) = \sum_{j=1}^{\infty} e_j 10^{-j}\) for some digits \(e_j\) (taking \(e_j = 9\) for all \(j\) in case \(10^m(x-a_s) = 1\)); then \(d_1,\ldots,d_m,e_1,e_2,\ldots\) is a decimal expansion of \(x\) beginning with \(s\).

(a). Let \(F\) denote the string of one hundred 4s. For each \(n \in \mathbb{Z}^{+}\) and each string \(p \in \{0,\ldots,9\}^{n-1}\), the set of \(x \in [0,1]\) having a decimal expansion whose digits in positions \(n, n+1, \ldots, n+99\) all equal 4 and whose first \(n-1\) digits are \(p\) is exactly \(I_{pF}\), where \(pF\) denotes the concatenation of \(p\) with \(F\). A number \(x\) lies in \(E\) exactly when some decimal expansion of \(x\) has one hundred consecutive 4s starting in some position \(n\), so

\begin{equation*} E = (0,1) \cap \bigcup_{n=1}^{\infty} \ \bigcup_{p \in \{0,\ldots,9\}^{n-1}} I_{pF}. \end{equation*}

For each \(n\) there are only \(10^{\,n-1}\) strings \(p\), so the double union is a countable union of closed intervals. Every closed interval is a Borel set and the collection of Borel sets is a \(\sigma\)-algebra, so \(E\) is the intersection of two Borel sets and hence is a Borel set.

(b). We claim \(|E| = 1\); equivalently \(|(0,1) \setminus E| = 0\).

Fix \(N \in \mathbb{Z}^{+}\). Call a string \(s \in \{0,\ldots,9\}^{100N}\) good if none of its \(N\) consecutive blocks of length 100 (positions \(100k+1,\ldots,100k+100\) for \(k = 0,\ldots,N-1\)) consists of one hundred 4s. The number of good strings is \((10^{100} - 1)^{N}\), since each block may be filled in \(10^{100}-1\) ways independently of the others. Let

\begin{equation*} U_N = \bigcup_{s \text{ good}} I_s . \end{equation*}

By subadditivity of outer measure (2.8) and \(|I_s| = 10^{-100N}\) (by 2.14),

\begin{equation*} |U_N| \le (10^{100}-1)^{N} \cdot 10^{-100N} = \Bigl(1 - 10^{-100}\Bigr)^{N}. \end{equation*}

Now suppose \(x \in (0,1) \setminus E\). Choose any decimal expansion \(d_1, d_2, \ldots\) of \(x\) and let \(s\) be the string of its first \(100N\) digits. If \(s\) were not good, then some block of one hundred consecutive positions of this expansion would consist entirely of 4s, so \(x\) would lie in \(E\); hence \(s\) is good. By the basic observation, \(x \in I_s \subseteq U_N\). Therefore \((0,1) \setminus E \subseteq U_N\), and so by 2.5

\begin{equation*} |(0,1) \setminus E| \le \bigl(1 - 10^{-100}\bigr)^{N} \end{equation*}

for every \(N \in \mathbb{Z}^{+}\). Letting \(N \to \infty\) gives \(|(0,1)\setminus E| = 0\).

Finally we compute \(|E|\) using only 2.4, 2.5, 2.8, and 2.14, so that no unproved value of \(|(0,1)|\) is needed. On one hand \(E \subseteq [0,1]\), so \(|E| \le |[0,1]| = 1\) by 2.5 and 2.14. On the other hand

\begin{equation*} [0,1] = E \cup \bigl( (0,1) \setminus E \bigr) \cup \{0,1\}, \end{equation*}

and \(|\{0,1\}| = 0\) by 2.4, so subadditivity (2.8) gives

\begin{equation*} 1 = |[0,1]| \le |E| + |(0,1)\setminus E| + |\{0,1\}| = |E| . \end{equation*}

Hence \(|E| = 1\): the set in (a) has Lebesgue measure \(1\).

We use the set \(V \subseteq [-1,1]\) constructed in the proof of 2.18: \(V\) contains exactly one element of each set \(\widetilde{a} = \{c \in [-1,1] : a - c \in \mathbb{Q}\}\), where \(a\) ranges over \([-1,1]\). Two properties established there will be used:

\begin{equation*} |V| > 0, \qquad \text{and} \qquad (r + V) \cap (r^{\prime} + V) = \varnothing \ \text{ for distinct } r, r^{\prime} \in \mathbb{Q}. \end{equation*}

(The disjointness was verified in the proof of 2.18 for the particular rationals used there, but the argument given uses only that \(r \ne r^{\prime}\) are rational: if \(t = r + v_1 = r^{\prime} + v_2\) with \(v_1, v_2 \in V\), then \(v_1 - v_2 = r^{\prime} - r \in \mathbb{Q}\), so \(v_1\) and \(v_2\) lie in the same set \(\widetilde{a}\), forcing \(v_1 = v_2\) and hence \(r = r^{\prime}\).)

Step 1: every closed subset of \(V\) has outer measure \(0\).

Suppose \(F \subseteq V\) is closed. Let \(r_1, r_2, \ldots\) be a sequence of distinct rational numbers in \([0,1]\). Because \(F \subseteq V\), the sets \(r_1 + F, r_2 + F, \ldots\) are disjoint. Each \(r_k + F\) is a closed subset of \(\mathbb{R}\) (a translate of a closed set is closed), hence a Borel set, and \(r_k + F \subseteq [-1, 2]\). Because outer measure is a measure on the Borel sets (2.68) and outer measure is translation invariant (2.7),

\begin{equation*} 3 = |[-1,2]| \ \ge\ \Bigl| \bigcup_{k=1}^{\infty} (r_k + F) \Bigr| = \sum_{k=1}^{\infty} |r_k + F| = \sum_{k=1}^{\infty} |F| , \end{equation*}

where the first inequality uses that outer measure preserves order (2.5) and 2.14. If \(|F| > 0\) the right side equals \(\infty\), a contradiction. Hence \(|F| = 0\).

Step 2: outer measure scales under dilation.

We claim \(|tB| = t|B|\) for every \(t > 0\) and every \(B \subseteq \mathbb{R}\), where \(tB = \{tb : b \in B\}\). If \(B \subseteq \bigcup_{k} (a_k, b_k)\), then \(tB \subseteq \bigcup_k (ta_k, tb_k)\) and \(\sum_k (tb_k - ta_k) = t \sum_k (b_k - a_k)\); taking the infimum over all such covers in the definition of outer measure (2.1) gives \(|tB| \le t|B|\). Applying this inequality with \(B\) replaced by \(tB\) and \(t\) replaced by \(1/t\) gives \(|B| \le \frac{1}{t}|tB|\), that is, \(t|B| \le |tB|\). Hence \(|tB| = t|B|\).

Step 3: the example.

Let \(c = |V|\), so \(0 < c < \infty\) (indeed \(c \le 2\) since \(V \subseteq [-1,1]\)), and set

\begin{equation*} A = \tfrac{1}{c} V = \Bigl\{ \tfrac{v}{c} : v \in V \Bigr\} . \end{equation*}

Then \(A \subseteq [-\frac{1}{c}, \frac{1}{c}]\), so \(A\) is bounded, and by Step 2,

\begin{equation*} |A| = \tfrac{1}{c}\,|V| = \tfrac{1}{c} \cdot c = 1 . \end{equation*}

Now suppose \(F \subseteq A\) is closed. Then \(cF\) is a closed subset of \(cA = V\), so \(|cF| = 0\) by Step 1, and therefore \(|F| = \frac{1}{c}|cF| = 0\) by Step 2. Thus

\begin{equation*} |F| = 0 = |A| - 1 \end{equation*}

for every closed set \(F \subseteq A\), which is the required inequality.

We first record three general tools, then construct \(A\) from a Vitali set inside \((0,1)\).

Lemma 0 (outer measure of a bounded open interval). If \(a < b\) then \(|(a,b)| = b - a\).

Proof: By 2.5 and 2.14, \(|(a,b)| \le |[a,b]| = b-a\). For the reverse, let \(0 < \varepsilon < (b-a)/2\); then \([a+\varepsilon, b-\varepsilon] \subseteq (a,b)\), so by 2.5 and 2.14 again \(|(a,b)| \ge b - a - 2\varepsilon\). Letting \(\varepsilon \downarrow 0\) gives \(|(a,b)| \ge b-a\).

Lemma 1 (measurable hull). If \(S \subseteq \mathbb{R}\) and \(|S| < \infty\), then there is a Borel set \(B\) with \(S \subseteq B\) and \(|B| = |S|\).

Proof: For each \(k \in \mathbb{Z}^{+}\) the definition of outer measure (2.1) provides open intervals \(I_1, I_2, \ldots\) with \(S \subseteq \bigcup_j I_j\) and \(\sum_j \ell(I_j) < |S| + \frac{1}{k}\). Let \(U_k = \bigcup_j I_j\), an open set containing \(S\) with \(|U_k| \le \sum_j \ell(I_j) < |S| + \frac{1}{k}\). Put \(B = \bigcap_{k=1}^{\infty} U_k\). Then \(B\) is Borel, \(S \subseteq B\), and \(|S| \le |B| \le |U_k| < |S| + \frac{1}{k}\) for every \(k\), so \(|B| = |S|\).

Lemma 2 (splitting by a measurable set). If \(E \subseteq \mathbb{R}\) is Lebesgue measurable and \(S \subseteq \mathbb{R}\), then \(|S| = |S \cap E| + |S \setminus E|\).

Proof: The inequality \(\le\) is subadditivity (2.8). For \(\ge\) we may assume \(|S| < \infty\). Take \(B\) as in Lemma 1. Then \(S \cap E \subseteq B \cap E\) and \(S \setminus E \subseteq B \setminus E\), and \(B \cap E\), \(B \setminus E\) are disjoint Lebesgue measurable sets whose union is \(B\). Because outer measure is a measure on the Lebesgue measurable sets (2.72),

\begin{equation*} |S \cap E| + |S \setminus E| \le |B \cap E| + |B \setminus E| = |B| = |S| . \end{equation*}

Consequence. If \(E_1, \ldots, E_N\) are disjoint Lebesgue measurable sets and \(S \subseteq \mathbb{R}\), then \(|S| \ge \sum_{n=1}^{N} |S \cap E_n|\). This follows by induction on \(N\): Lemma 2 gives \(|S| = |S \cap E_1| + |S \setminus E_1|\), and \(S \setminus E_1\) contains \(S \cap E_n\) for \(n \ge 2\), so the induction hypothesis applies to \(S \setminus E_1\) and \(E_2, \ldots, E_N\).

A Vitali set in \((0,1)\). For \(a \in (0,1)\) let \(\widetilde{a} = \{c \in (0,1) : a - c \in \mathbb{Q}\}\). As in the proof of 2.18, distinct such sets are disjoint and they cover \((0,1)\); let \(V \subseteq (0,1)\) contain exactly one element of each distinct set \(\widetilde{a}\). Exactly as in 2.18, if \(r, r^{\prime}\) are distinct rational numbers then \((r+V) \cap (r^{\prime}+V) = \varnothing\).

If \(a \in (0,1)\) and \(v\) is the unique element of \(V \cap \widetilde{a}\), then \(a - v\) is a rational number in \((-1,1)\); hence, listing \(\mathbb{Q} \cap (-1,1)\) as \(q_1, q_2, \ldots\), we get \((0,1) \subseteq \bigcup_{k} (q_k + V)\). By Lemma 0, 2.5, 2.8, and 2.7,

\begin{equation*} 1 = |(0,1)| \le \sum_{k=1}^{\infty} |q_k + V| = \sum_{k=1}^{\infty} |V| , \end{equation*}

which forces \(|V| > 0\).

Claim 1: every closed set \(F \subseteq V\) has \(|F| = 0\). Let \(r_1, r_2, \ldots\) be distinct rationals in \([0,1]\). The sets \(r_k + F\) are disjoint closed (hence Borel) subsets of \((0,2)\), so by 2.68, 2.7, 2.5, and 2.14,

\begin{equation*} 2 \ge \Bigl| \bigcup_{k=1}^{\infty} (r_k + F) \Bigr| = \sum_{k=1}^{\infty} |r_k+F| = \sum_{k=1}^{\infty}|F| . \end{equation*}

Thus \(|F| = 0\).

Claim 2: \(V\) is not Lebesgue measurable. If it were, then since \(0 < |V| \le 1 < \infty\), condition (b) of 2.71 with \(\varepsilon = |V|/2\) would give a closed set \(F \subseteq V\) with \(|V \setminus F| < |V|/2\); then by 2.8 and Claim 1, \(|V| \le |F| + |V \setminus F| < 0 + |V|/2\), a contradiction.

Claim 3: \(c := |V| + |(0,1)\setminus V| - 1 > 0\). By subadditivity (2.8) and Lemma 0 we have \(|V| + |(0,1)\setminus V| \ge |(0,1)| = 1\), so \(c \ge 0\) (and \(c\) is a real number, since \(|V| \le 1\) and \(|(0,1)\setminus V| \le 1\) by 2.5 and Lemma 0). Suppose \(c = 0\), that is, \(|V| + |(0,1)\setminus V| = 1\). By Lemma 1 choose Borel sets \(B \supseteq V\) and \(C \supseteq (0,1)\setminus V\) with \(|B| = |V|\) and \(|C| = |(0,1)\setminus V|\); replacing \(B\) by \(B \cap (0,1)\) and \(C\) by \(C \cap (0,1)\) changes neither the inclusions nor the outer measures, so we may assume \(B, C \subseteq (0,1)\). Then \(B \cup C = (0,1)\). Since outer measure is a measure on the Borel sets (2.68),

\begin{equation*} |B| + |C| = |B| + |C \setminus B| + |C \cap B| = |B \cup C| + |B \cap C| = 1 + |B \cap C| , \end{equation*}

the last equality because \(B \cup C = (0,1)\) and \(|(0,1)| = 1\) by Lemma 0. (All the quantities here are at most \(1\), so the cancellations are legitimate.) As \(|B| + |C| = 1\), we get \(|B \cap C| = 0\). Now \((0,1) \setminus C \subseteq V\) because \((0,1)\setminus V \subseteq C\), hence

\begin{equation*} V = \bigl( (0,1) \setminus C \bigr) \cup (V \cap C), \end{equation*}

and \(V \cap C \subseteq B \cap C\) has outer measure \(0\) by 2.5, so \(V \cap C\) is Lebesgue measurable (in 2.70 take the Borel subset to be \(\varnothing\)). Since \((0,1)\setminus C\) is Borel and the Lebesgue measurable sets form a \(\sigma\)-algebra (2.72), \(V\) would be Lebesgue measurable, contradicting Claim 2. Hence \(c > 0\).

The construction. Let

\begin{equation*} A = \bigcup_{n=1}^{\infty} (n + V) . \end{equation*}

Since \(V \subseteq (0,1)\), we have \(n + V \subseteq (n, n+1)\), and therefore \(A \cap (n,n+1) = n+V\) for each \(n \in \mathbb{Z}^{+}\).

Let \(G\) be any open set with \(A \subseteq G\), and fix \(n \in \mathbb{Z}^{+}\). Put \(W = G \cap (n,n+1)\), an open set containing \(n+V\), and put \(E = (n,n+1) \setminus W\), a Borel set (hence Lebesgue measurable, by 2.70 with the Borel subset taken to be \(E\) itself) that is disjoint from \(n + V\) because \(n + V \subseteq W\).

First, applying Lemma 2 to \(S = (n+V) \cup E\) and the measurable set \(E\) gives

\begin{equation*} |E| + |n+V| = |S \cap E| + |S \setminus E| = |S| \le |(n,n+1)| = 1 , \end{equation*}

using \(S \cap E = E\) (as \(E \subseteq S\)), \(S \setminus E = n+V\) (as \(E\) and \(n+V\) are disjoint), \(S \subseteq (n,n+1)\), 2.5, and Lemma 0. Hence, by translation invariance (2.7), \(|E| \le 1 - |V|\).

Second, \((n,n+1) \setminus (n+V) \subseteq E \cup \bigl( W \setminus (n+V) \bigr)\), so by 2.8 and 2.7,

\begin{equation*} |(0,1) \setminus V| = \bigl| (n,n+1)\setminus (n+V) \bigr| \le |E| + \bigl| W \setminus (n+V) \bigr| . \end{equation*}

Combining the two displays,

\begin{equation*} \bigl| W \setminus (n+V) \bigr| \ge |(0,1)\setminus V| - (1 - |V|) = c . \end{equation*}

Because \(A \cap (n,n+1) = n+V\) and \(W = G \cap (n,n+1)\), we have \(W \setminus (n+V) = (G \setminus A) \cap (n,n+1)\). Hence

\begin{equation*} \bigl| (G\setminus A) \cap (n,n+1) \bigr| \ge c \qquad \text{for every } n \in \mathbb{Z}^{+} . \end{equation*}

Finally, the intervals \((1,2), (2,3), \ldots, (N, N+1)\) are disjoint Borel sets, so by the Consequence of Lemma 2,

\begin{equation*} |G \setminus A| \ge \sum_{n=1}^{N} \bigl| (G \setminus A) \cap (n,n+1) \bigr| \ge Nc \end{equation*}

for every \(N \in \mathbb{Z}^{+}\). Since \(c > 0\), letting \(N \to \infty\) gives \(|G \setminus A| = \infty\), as desired.

(a). Let \(\mathcal{C}\) be a collection of nontrivial intervals of \(\mathbb{R}\) and let \(U = \bigcup_{I \in \mathcal{C}} I\). For \(I \in \mathcal{C}\) let \(I^{\circ}\) denote the interior of \(I\); because \(I\) is an interval containing at least two points, \(I^{\circ}\) is a nonempty open interval and \(I \setminus I^{\circ}\) consists of at most the two endpoints of \(I\). Put

\begin{equation*} W = \bigcup_{I \in \mathcal{C}} I^{\circ} , \end{equation*}

an open subset of \(\mathbb{R}\) with \(W \subseteq U\).

Step 1: \(W\) is the union of countably many of the sets \(I^{\circ}\). Let \(D\) be the set of pairs \((p,q)\) of rational numbers with \(p < q\) such that \((p,q) \subseteq I^{\circ}\) for at least one \(I \in \mathcal{C}\); for each \((p,q) \in D\) choose one such interval and call it \(I_{p,q} \in \mathcal{C}\). The set \(D\) is countable, being a subset of \(\mathbb{Q} \times \mathbb{Q}\). If \(x \in W\), then \(x \in I^{\circ}\) for some \(I \in \mathcal{C}\), and since \(I^{\circ}\) is open there are rationals \(p < x < q\) with \((p,q) \subseteq I^{\circ}\); thus \((p,q) \in D\) and \(x \in (p,q) \subseteq I_{p,q}\). Hence

\begin{equation*} W \subseteq \bigcup_{(p,q) \in D} I_{p,q} \subseteq U . \end{equation*}

Step 2: \(U \setminus W\) is countable. Suppose \(x \in U \setminus W\). Then \(x \in I\) for some \(I \in \mathcal{C}\), and \(x \notin I^{\circ}\), so \(x\) is an endpoint of the nontrivial interval \(I\). Consequently \(I^{\circ}\) is a nonempty open interval having \(x\) as an endpoint, so there exists \(\delta > 0\) with \((x, x+\delta) \subseteq I^{\circ} \subseteq W\) or \((x-\delta, x) \subseteq I^{\circ} \subseteq W\). Let \(U_{+}\) be the set of \(x \in U \setminus W\) of the first kind and \(U_{-}\) the set of those of the second kind, so \(U \setminus W = U_{+} \cup U_{-}\).

For \(x \in U_{+}\) choose \(\delta_x > 0\) with \((x, x+\delta_x) \subseteq W\) and choose a rational \(r_x \in (x, x + \delta_x)\). The map \(x \mapsto r_x\) is injective on \(U_{+}\): if \(x, y \in U_{+}\) with \(x < y\) and \(r_x = r_y\), then \(x < y < r_y = r_x < x + \delta_x\), so \(y \in (x, x+\delta_x) \subseteq W\), contradicting \(y \notin W\). Hence \(U_{+}\) is countable, and symmetrically so is \(U_{-}\). Thus \(U \setminus W\) is countable.

Step 3. For each \(x \in U \setminus W\) choose \(J_x \in \mathcal{C}\) with \(x \in J_x\). Then

\begin{equation*} \mathcal{D} = \{ I_{p,q} : (p,q) \in D \} \cup \{ J_x : x \in U \setminus W \} \end{equation*}

is a countable subcollection of \(\mathcal{C}\), and

\begin{equation*} U = W \cup (U \setminus W) \subseteq \bigcup_{J \in \mathcal{D}} J \subseteq U . \end{equation*}

Therefore \(U = \bigcup_{J \in \mathcal{D}} J\), the union of a countable subcollection of \(\mathcal{C}\).

(b). Let \(\mathcal{C}\) and \(U\) be as above. By (a) there is a countable subcollection \(\mathcal{D} \subseteq \mathcal{C}\) with \(U = \bigcup_{J \in \mathcal{D}} J\). Every interval \(J\) of \(\mathbb{R}\) is a Borel set. Indeed, let \(a = \inf J \in [-\infty,\infty)\) and \(b = \sup J \in (-\infty,\infty]\) when \(J \ne \varnothing\). The set \(\{x \in \mathbb{R} : a < x < b\}\) is open, hence Borel, and \(J\) differs from it by at most the two points \(a\) and \(b\) (only those that are real and belong to \(J\)). Since every singleton \(\{a\} = \bigcap_{m=1}^{\infty} (a - \frac{1}{m}, a + \frac{1}{m})\) is a countable intersection of open sets and hence Borel, and since the Borel sets form a \(\sigma\)-algebra, \(J\) is Borel; and \(\varnothing\) is Borel. Therefore the countable union \(U = \bigcup_{J \in \mathcal{D}} J\) is a Borel set.

(c). Note first that part (c) drops the word nontrivial: for \(x \in \mathbb{R}\) the singleton \(\{x\} = [x,x]\) is a closed interval, a trivial one. By 2.67 there exists a set \(S \subseteq \mathbb{R}\) with \(|S| < \infty\) that is not a Borel set.

Now let

\begin{equation*} \mathcal{C} = \bigl\{ [x,x] : x \in S \bigr\} , \end{equation*}

a collection of closed intervals of \(\mathbb{R}\). Its union is

\begin{equation*} \bigcup_{x \in S} \{x\} = S , \end{equation*}

which is not a Borel set. This shows that the hypothesis of nontriviality cannot be dropped in (a) or (b): the singletons \(\{x\}\) have empty interior, which is exactly what makes the counting argument of (a) break down.

Suppose \(A \subseteq \mathbb{R}\) is Lebesgue measurable. Because (a) implies (b) in 2.71, for each \(k \in \mathbb{Z}^{+}\) there exists a closed set \(E_k \subseteq A\) with

\begin{equation*} |A \setminus E_k| < \tfrac{1}{k} . \end{equation*}

The sets \(E_k\) produced this way need not be increasing, so define

\begin{equation*} F_k = E_1 \cup E_2 \cup \cdots \cup E_k \qquad \text{for } k \in \mathbb{Z}^{+} . \end{equation*}

Each \(F_k\) is a union of finitely many closed sets and hence is closed. Each \(F_k\) is contained in \(A\) because each \(E_j \subseteq A\). Clearly \(F_1 \subseteq F_2 \subseteq \cdots\), and

\begin{equation*} \bigcup_{k=1}^{\infty} F_k = \bigcup_{k=1}^{\infty} E_k . \end{equation*}

Now fix \(n \in \mathbb{Z}^{+}\). Because \(E_n \subseteq \bigcup_{k=1}^{\infty} F_k\), we have

\begin{equation*} A \setminus \bigcup_{k=1}^{\infty} F_k \subseteq A \setminus E_n , \end{equation*}

and therefore, because outer measure preserves order (2.5),

\begin{equation*} \Bigl| A \setminus \bigcup_{k=1}^{\infty} F_k \Bigr| \le |A \setminus E_n| < \tfrac{1}{n} . \end{equation*}

This holds for every \(n \in \mathbb{Z}^{+}\), so

\begin{equation*} \Bigl| A \setminus \bigcup_{k=1}^{\infty} F_k \Bigr| = 0 , \end{equation*}

which is what we wanted. (This is exactly the argument used to prove that (b) implies (c) in 2.71, with the additional observation that replacing the closed sets by their partial unions makes the sequence increasing without changing the union.)

Suppose \(A\) is Lebesgue measurable, and let \(\varepsilon > 0\).

Because (a) implies (e) in 2.71, there is an open set \(U \supseteq A\) with

\begin{equation*} |U \setminus A| < \tfrac{\varepsilon}{2} . \end{equation*}

By subadditivity of outer measure (2.8), \(|U| \le |A| + |U \setminus A| < \infty\).

Every nonempty open subset of \(\mathbb{R}\) is the union of a countable disjoint collection of nonempty open intervals, namely its connected components; this is the fact invoked in the proof of 2.63, where an arbitrary open subset \(W\) of \(\mathbb{R}\) is written as \(W = \bigcup_{n=1}^{\infty} I_n\) for a sequence of disjoint open intervals. Dispose first of the degenerate case: if \(U = \varnothing\) then \(A \subseteq U\) forces \(A = \varnothing\), and \(G = (0, \varepsilon/3)\) is a union of one bounded open interval with \(|A \setminus G| + |G \setminus A| = |G| \le |[0,\varepsilon/3]| = \varepsilon/3 < \varepsilon\) by 2.5 and 2.14. So assume \(U \ne \varnothing\) and write \(U = \bigcup_{n} I_n\) with \(I_1, I_2, \ldots\) a finite or infinite sequence of disjoint nonempty open intervals; pad a finite list with copies of \(\varnothing\) so that the index set is \(\mathbb{Z}^{+}\).

Each \(I_n\) is bounded: an unbounded nonempty open interval contains closed intervals of arbitrarily large length, so 2.5 and 2.14 would give \(|U| \ge |I_n| = \infty\), contradicting \(|U| < \infty\). Because the \(I_n\) are disjoint Borel sets and outer measure is a measure on the Borel sets (2.68),

\begin{equation*} \sum_{n=1}^{\infty} |I_n| = |U| < \infty . \end{equation*}

Since this series converges, choose \(N \in \mathbb{Z}^{+}\) with

\begin{equation*} \sum_{n > N} |I_n| < \tfrac{\varepsilon}{2}, \end{equation*}

and set \(G = I_1 \cup \cdots \cup I_N\), a union of finitely many disjoint bounded open intervals. (Discarding any empty ones changes nothing.) Then:

• \(G \setminus A \subseteq U \setminus A\), so \(|G \setminus A| \le |U \setminus A| < \frac{\varepsilon}{2}\) by 2.5.

• Since \(A \subseteq U\), we have \(A \setminus G \subseteq U \setminus G = \bigcup_{n > N} I_n\), so by 2.5 and 2.8, \(|A \setminus G| \le \sum_{n>N} |I_n| < \frac{\varepsilon}{2}\).

Adding gives \(|A \setminus G| + |G \setminus A| < \varepsilon\), as required. (Note that this direction did not use \(|A| < \infty\) beyond guaranteeing \(|U| < \infty\), which is where the hypothesis is needed: without it the tail of \(\sum |I_n|\) need not be small.)

Conversely, suppose that for every \(\varepsilon > 0\) such a set \(G\) exists.

Let \(\varepsilon > 0\), and choose \(G\), a union of finitely many disjoint bounded open intervals, with

\begin{equation*} |A \setminus G| + |G \setminus A| < \tfrac{\varepsilon}{2} . \end{equation*}

In particular \(|A \setminus G| < \frac{\varepsilon}{2}\). By the definition of outer measure (2.1) there exist open intervals \(J_1, J_2, \ldots\) with

\begin{equation*} A \setminus G \subseteq \bigcup_{k} J_k \quad \text{and} \quad \sum_{k} \ell(J_k) < \tfrac{\varepsilon}{2} . \end{equation*}

Let \(H = \bigcup_k J_k\). Then \(H\) is open, \(A \setminus G \subseteq H\), and \(|H| \le \sum_k \ell(J_k) < \frac{\varepsilon}{2}\) by 2.8 and 2.14.

Set \(V = G \cup H\). Then \(V\) is open, and

\begin{equation*} A = (A \cap G) \cup (A \setminus G) \subseteq G \cup H = V , \end{equation*}

so \(V \supseteq A\). Moreover

\begin{equation*} V \setminus A = (G \cup H) \setminus A \subseteq (G \setminus A) \cup H , \end{equation*}

so by 2.5 and 2.8,

\begin{equation*} |V \setminus A| \le |G \setminus A| + |H| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

Thus for every \(\varepsilon > 0\) there is an open set \(V \supseteq A\) with \(|V \setminus A| < \varepsilon\); that is, \(A\) satisfies condition (e) of 2.71. Because (e) implies (a) in 2.71, the set \(A\) is Lebesgue measurable.

Suppose \(A \subseteq \mathbb{R}\) is Lebesgue measurable. Because (a) implies (e) in 2.71, for each \(k \in \mathbb{Z}^{+}\) there exists an open set \(U_k \supseteq A\) with

\begin{equation*} |U_k \setminus A| < \tfrac{1}{k} . \end{equation*}

These sets need not be decreasing, so define

\begin{equation*} G_k = U_1 \cap U_2 \cap \cdots \cap U_k \qquad \text{for } k \in \mathbb{Z}^{+} . \end{equation*}

Each \(G_k\) is an intersection of finitely many open sets and hence is open. Each \(G_k\) contains \(A\) because every \(U_j\) contains \(A\). Clearly \(G_1 \supseteq G_2 \supseteq \cdots\), and

\begin{equation*} \bigcap_{k=1}^{\infty} G_k = \bigcap_{k=1}^{\infty} U_k . \end{equation*}

Now fix \(n \in \mathbb{Z}^{+}\). Because \(\bigcap_{k=1}^{\infty} G_k \subseteq U_n\), we have

\begin{equation*} \Bigl( \bigcap_{k=1}^{\infty} G_k \Bigr) \setminus A \subseteq U_n \setminus A , \end{equation*}

and therefore, because outer measure preserves order (2.5),

\begin{equation*} \Bigl| \Bigl( \bigcap_{k=1}^{\infty} G_k \Bigr) \setminus A \Bigr| \le |U_n \setminus A| < \tfrac{1}{n} . \end{equation*}

Since this holds for every \(n \in \mathbb{Z}^{+}\),

\begin{equation*} \Bigl| \Bigl( \bigcap_{k=1}^{\infty} G_k \Bigr) \setminus A \Bigr| = 0 , \end{equation*}

as desired. (This is the argument proving that (e) implies (f) in 2.71, improved by replacing the open sets with their partial intersections, which makes the sequence decreasing without changing the intersection.)

Recall the definition 2.70: a set \(A \subseteq \mathbb{R}\) is Lebesgue measurable if there exists a Borel set \(B \subseteq A\) such that \(|A \setminus B| = 0\). So we need two ingredients: that the Borel sets are translation invariant, and that outer measure is translation invariant (the latter is 2.7).

Step 1: the collection of Borel sets is translation invariant.

Fix \(t \in \mathbb{R}\) and let

\begin{equation*} \mathcal{S} = \{E \subseteq \mathbb{R} : t + E \text{ is a Borel set}\}. \end{equation*}

We claim \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\).

• \(\emptyset \in \mathcal{S}\), because \(t + \emptyset = \emptyset\), which is a Borel set.
• Suppose \(E \in \mathcal{S}\). Because \(x \mapsto t + x\) is a bijection of \(\mathbb{R}\) onto \(\mathbb{R}\), we have \(t + (\mathbb{R} \setminus E) = \mathbb{R} \setminus (t + E)\). The right-hand side is the complement of a Borel set and hence is a Borel set. Thus \(\mathbb{R} \setminus E \in \mathcal{S}\).
• Suppose \(E_1, E_2, \ldots \in \mathcal{S}\). Then

\begin{equation*} t + \bigcup_{k=1}^{\infty} E_k = \bigcup_{k=1}^{\infty} (t + E_k), \end{equation*}

which is a countable union of Borel sets and hence is a Borel set. Thus \(\bigcup_{k=1}^{\infty} E_k \in \mathcal{S}\).

Hence \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). Furthermore, if \(G \subseteq \mathbb{R}\) is open then \(t + G\) is open (translation is a homeomorphism of \(\mathbb{R}\); concretely, if \(x \in t + G\) then \(x - t \in G\), so \((x - t - \delta, x - t + \delta) \subseteq G\) for some \(\delta > 0\), and then \((x - \delta, x + \delta) \subseteq t + G\)), and every open set is a Borel set. Thus \(\mathcal{S}\) contains all open subsets of \(\mathbb{R}\).

Because the collection of Borel sets is by definition (2.29) the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing all the open sets, and \(\mathcal{S}\) is a \(\sigma\)-algebra containing all the open sets, we conclude that every Borel set belongs to \(\mathcal{S}\). In other words, if \(B\) is a Borel set and \(t \in \mathbb{R}\), then \(t + B\) is a Borel set.

Step 2: conclusion.

Suppose \(A \subseteq \mathbb{R}\) is Lebesgue measurable and \(t \in \mathbb{R}\). By 2.70 there exists a Borel set \(B \subseteq A\) with \(|A \setminus B| = 0\).

By Step 1, \(t + B\) is a Borel set. Because \(B \subseteq A\) we have \(t + B \subseteq t + A\).

Because \(x \mapsto t + x\) is injective,

\begin{equation*} (t + A) \setminus (t + B) = t + (A \setminus B). \end{equation*}

Indeed, \(x \in (t+A) \setminus (t+B)\) if and only if \(x - t \in A\) and \(x - t \notin B\), i.e. if and only if \(x - t \in A \setminus B\).

Hence, by the translation invariance of outer measure (2.7),

\begin{equation*} \bigl| (t + A) \setminus (t + B) \bigr| = \bigl| t + (A \setminus B) \bigr| = |A \setminus B| = 0 . \end{equation*}

Thus \(t + B\) is a Borel set contained in \(t + A\) whose complement in \(t + A\) has outer measure \(0\). By 2.70, \(t + A\) is Lebesgue measurable.

The proof parallels the one for translation (Exercise 8 in this section), using in place of 2.7 the dilation formula \(|tA| = |t| \, |A|\) from Exercise 2 in Section 2A (with the convention \(0 \cdot \infty = 0\)).

The degenerate case \(t = 0\). If \(A = \emptyset\) then \(0A = \emptyset\); otherwise \(0A = \{0\}\). In either case \(0A\) is a closed set, hence a Borel set, hence Lebesgue measurable. (Alternatively: \(|0A| = 0\) by 2.3, and every set of outer measure \(0\) is Lebesgue measurable, taking \(B = \emptyset\) in 2.70.) So assume from now on that \(t \neq 0\).

Step 1: the collection of Borel sets is dilation invariant.

Fix \(t \in \mathbb{R}\) with \(t \neq 0\) and let

\begin{equation*} \mathcal{S} = \{E \subseteq \mathbb{R} : tE \text{ is a Borel set}\}. \end{equation*}

We claim \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). Note first that because \(t \neq 0\), the map \(x \mapsto tx\) is a bijection of \(\mathbb{R}\) onto \(\mathbb{R}\), with inverse \(x \mapsto \frac{1}{t}x\).

• \(\emptyset \in \mathcal{S}\), because \(t\emptyset = \emptyset\) is a Borel set.
• Suppose \(E \in \mathcal{S}\). By the bijectivity just noted, \(t(\mathbb{R} \setminus E) = \mathbb{R} \setminus tE\), which is a Borel set. Thus \(\mathbb{R} \setminus E \in \mathcal{S}\).
• Suppose \(E_1, E_2, \ldots \in \mathcal{S}\). Then

\begin{equation*} t \bigcup_{k=1}^{\infty} E_k = \bigcup_{k=1}^{\infty} tE_k , \end{equation*}

a countable union of Borel sets, hence a Borel set. Thus \(\bigcup_{k=1}^{\infty} E_k \in \mathcal{S}\).

Hence \(\mathcal{S}\) is a \(\sigma\)-algebra on \(\mathbb{R}\). If \(G \subseteq \mathbb{R}\) is open, then \(tG\) is open: if \(x \in tG\) then \(\frac{x}{t} \in G\), so \(\bigl(\frac{x}{t} - \delta, \frac{x}{t} + \delta\bigr) \subseteq G\) for some \(\delta > 0\), and then the interval of radius \(|t|\delta\) centered at \(x\) is contained in \(tG\). Thus \(\mathcal{S}\) contains every open subset of \(\mathbb{R}\).

Because the collection of Borel sets is the smallest \(\sigma\)-algebra on \(\mathbb{R}\) containing the open sets (2.29), we conclude that every Borel set is in \(\mathcal{S}\). In other words, if \(B\) is a Borel set and \(t \neq 0\), then \(tB\) is a Borel set.

Step 2: conclusion.

Suppose \(A \subseteq \mathbb{R}\) is Lebesgue measurable and \(t \neq 0\). By 2.70 there is a Borel set \(B \subseteq A\) with \(|A \setminus B| = 0\).

By Step 1, \(tB\) is a Borel set, and clearly \(tB \subseteq tA\). Because \(x \mapsto tx\) is injective,

\begin{equation*} tA \setminus tB = t(A \setminus B); \end{equation*}

indeed \(x \in tA \setminus tB\) if and only if \(\frac{x}{t} \in A\) and \(\frac{x}{t} \notin B\), i.e. if and only if \(\frac{x}{t} \in A \setminus B\).

By Exercise 2 in Section 2A,

\begin{equation*} |tA \setminus tB| = |t(A \setminus B)| = |t| \, |A \setminus B| = |t| \cdot 0 = 0 . \end{equation*}

Thus \(tB\) is a Borel set contained in \(tA\) with \(|tA \setminus tB| = 0\), so \(tA\) is Lebesgue measurable by 2.70.

Remark on the cited dilation formula. For completeness, here is the short proof of \(|tA| = |t| \, |A|\) for \(t \neq 0\). If \(I_1, I_2, \ldots\) are open intervals whose union contains \(A\), then \(tI_1, tI_2, \ldots\) are open intervals whose union contains \(tA\), and \(\ell(tI_k) = |t| \, \ell(I_k)\). Hence \(|tA| \leq \sum_{k=1}^{\infty} |t| \, \ell(I_k) = |t| \sum_{k=1}^{\infty} \ell(I_k)\); taking the infimum over all such covers gives \(|tA| \leq |t| \, |A|\). Applying this inequality to \(\frac{1}{t}\) and the set \(tA\) gives \(|A| = \bigl|\frac{1}{t}(tA)\bigr| \leq \frac{1}{|t|} |tA|\), i.e. \(|t| \, |A| \leq |tA|\). Combining the two inequalities gives equality.

Note that no measurability is assumed of \(A\); the point of the exercise is that a single measurable set already splits outer measure additively off any disjoint companion.

The inequality \(|A \cup B| \leq |A| + |B|\) is immediate from countable subadditivity of outer measure (2.8), applied to the sequence \(A, B, \emptyset, \emptyset, \ldots\). So the whole content is the reverse inequality.

A lemma: every set has a Borel hull.

We first record the following, which we use again in Exercise 12 of this section.

Lemma. If \(S \subseteq \mathbb{R}\), then there exists a Borel set \(E\) with \(S \subseteq E\) and \(|E| = |S|\).

Proof of Lemma. If \(|S| = \infty\), take \(E = \mathbb{R}\); then \(S \subseteq E\) and \(|E| = \infty = |S|\), since \(|S| \leq |E|\) by 2.5. So suppose \(|S| < \infty\) and let \(n \in \mathbb{Z}^+\). Because \(|S|\) is defined in 2.2 as an infimum, there exist open intervals \(I_1, I_2, \ldots\) with \(S \subseteq \bigcup_{k=1}^{\infty} I_k\) and

\begin{equation*} \sum_{k=1}^{\infty} \ell(I_k) < |S| + \tfrac{1}{n}. \end{equation*}

Let \(G_n = \bigcup_{k=1}^{\infty} I_k\), an open set containing \(S\). Because \(|I| \leq \ell(I)\) for every open interval \(I\) (take the cover \(I, \emptyset, \emptyset, \ldots\) in 2.2), countable subadditivity (2.8) gives

\begin{equation*} |G_n| \leq \sum_{k=1}^{\infty} |I_k| \leq \sum_{k=1}^{\infty} \ell(I_k) < |S| + \tfrac{1}{n}. \end{equation*}

Now let \(E = \bigcap_{n=1}^{\infty} G_n\). Each \(G_n\) is open, hence Borel, so \(E\) is Borel; and \(S \subseteq E\) because \(S \subseteq G_n\) for every \(n\). By 2.5 we have \(|E| \leq |G_n| < |S| + \frac{1}{n}\) for every \(n \in \mathbb{Z}^+\), hence \(|E| \leq |S|\). The reverse inequality \(|S| \leq |E|\) holds by 2.5. Thus \(|E| = |S|\), proving the Lemma.

Proof of the exercise.

Suppose \(A, B \subseteq \mathbb{R}\) are disjoint and \(B\) is Lebesgue measurable.

Because \(B\) is Lebesgue measurable, 2.70 provides a Borel set \(D \subseteq B\) with \(|B \setminus D| = 0\). Then

\begin{equation*} |B| = |D \cup (B \setminus D)| \leq |D| + |B \setminus D| = |D| \leq |B|, \end{equation*}

using 2.8 for the first inequality and 2.5 for the last. Hence \(|D| = |B|\).

By the Lemma, choose a Borel set \(E\) with \(A \cup B \subseteq E\) and \(|E| = |A \cup B|\).

Since \(D \subseteq B \subseteq E\), the sets \(D\) and \(E \setminus D\) are disjoint Borel sets whose union is \(E\). Because outer measure is a measure on the Borel sets (2.68), it is in particular finitely additive there, so

\begin{equation*} |E| = |D| + |E \setminus D|. \end{equation*}

Now observe that \(A \subseteq E \setminus D\): indeed \(A \subseteq A \cup B \subseteq E\), and \(A \cap D \subseteq A \cap B = \emptyset\) because \(D \subseteq B\) and \(A, B\) are disjoint. Hence \(|A| \leq |E \setminus D|\) by 2.5.

Putting these together,

\begin{equation*} |A \cup B| = |E| = |D| + |E \setminus D| \geq |B| + |A| . \end{equation*}

Combined with the subadditivity inequality \(|A \cup B| \leq |A| + |B|\) noted at the start, this gives

\begin{equation*} |A \cup B| = |A| + |B| , \end{equation*}

as desired. (All the arithmetic above takes place in \([0, \infty]\), so no finiteness assumptions are needed.)

This strengthens the existence of a nonmeasurable set (which follows from 2.18 and 2.72): every set of positive outer measure, no matter how small or how strangely placed, contains a nonmeasurable subset. As in 2.18, the Axiom of Choice is used.

A Vitali set with representatives in \([0,1]\).

Define a relation on \(\mathbb{R}\) by declaring \(x\) and \(y\) equivalent when \(x - y \in \mathbb{Q}\). This is an equivalence relation (reflexivity, symmetry, and transitivity all follow from the fact that \(\mathbb{Q}\) is closed under negation and addition). Write \(\tilde{x} = x + \mathbb{Q}\) for the equivalence class of \(x\); distinct classes are disjoint.

Each class \(\tilde{x}\) meets \([0,1]\): choose a rational \(q\) with \(-x \leq q \leq -x + 1\), which exists because \(\mathbb{Q}\) is dense in \(\mathbb{R}\); then \(x + q \in \tilde{x} \cap [0,1]\).

The sets \(\tilde{x} \cap [0,1]\), as \(\tilde{x}\) ranges over the distinct equivalence classes, are disjoint and (by the previous paragraph) nonempty. So the Axiom of Choice provides a set \(V\) containing exactly one element of \(\tilde{x} \cap [0,1]\) for each class \(\tilde{x}\). Then \(V \subseteq [0,1]\), and because \(V \cap \tilde{x} = V \cap (\tilde{x} \cap [0,1])\), the set \(V\) contains exactly one element of each equivalence class.

Two properties of \(V\):

• (i) \(\mathbb{R} = \bigcup_{r \in \mathbb{Q}} (r + V)\). Indeed, given \(x \in \mathbb{R}\), let \(v\) be the unique element of \(V \cap \tilde{x}\). Then \(x - v \in \mathbb{Q}\) and \(x = (x - v) + v \in (x - v) + V\).
• (ii) The sets \(r + V\), for \(r \in \mathbb{Q}\), are pairwise disjoint. Indeed, if \(t \in (r + V) \cap (s + V)\) with \(r, s \in \mathbb{Q}\), write \(t = r + v_1 = s + v_2\) with \(v_1, v_2 \in V\). Then \(v_1 - v_2 = s - r \in \mathbb{Q}\), so \(v_1\) and \(v_2\) lie in the same equivalence class; because \(V\) contains exactly one element of each class, \(v_1 = v_2\), and therefore \(r = s\).

Proof of the exercise.

Suppose \(A \subseteq \mathbb{R}\) with \(|A| > 0\), and suppose toward a contradiction that every subset of \(A\) is Lebesgue measurable.

For \(r \in \mathbb{Q}\) let

\begin{equation*} A_r = A \cap (r + V). \end{equation*}

Each \(A_r\) is a subset of \(A\), hence is Lebesgue measurable by our assumption. By (i) we have \(A = \bigcup_{r \in \mathbb{Q}} A_r\), and by (ii) this union is disjoint. The set \(\mathbb{Q}\) is countable, so this is a countable disjoint union of Lebesgue measurable sets. Because outer measure is a measure on the \(\sigma\)-algebra of Lebesgue measurable sets (2.72), it is countably additive there, so

\begin{equation*} 0 < |A| = \sum_{r \in \mathbb{Q}} |A_r| . \end{equation*}

Hence there exists \(r \in \mathbb{Q}\) with \(|A_r| > 0\). Fix such an \(r\) and set

\begin{equation*} W = (-r) + A_r . \end{equation*}

By Exercise 8 in this section, \(W\) is Lebesgue measurable, and by the translation invariance of outer measure (2.7), \(|W| = |A_r| > 0\). Moreover \(A_r \subseteq r + V\) gives \(W \subseteq V \subseteq [0,1]\).

Now let \(q_1, q_2, \ldots\) be a sequence of distinct rational numbers in \([0,1]\) (for instance an enumeration of \(\mathbb{Q} \cap [0,1]\)). For each \(k \in \mathbb{Z}^+\):

• \(q_k + W\) is Lebesgue measurable (Exercise 8) and \(|q_k + W| = |W|\) (2.7);
• \(q_k + W \subseteq q_k + V\), so by (ii) the sets \(q_1 + W, q_2 + W, \ldots\) are pairwise disjoint;
• \(q_k + W \subseteq [0,1] + [0,1] \subseteq [0,2]\), since \(W \subseteq [0,1]\) and \(q_k \in [0,1]\).

Therefore \(\bigcup_{k=1}^{\infty} (q_k + W) \subseteq [0,2]\), and by countable additivity of outer measure on the Lebesgue measurable sets (2.72), together with 2.5 and 2.14,

\begin{equation*} \sum_{k=1}^{\infty} |W| \;=\; \sum_{k=1}^{\infty} |q_k + W| \;=\; \Bigl| \bigcup_{k=1}^{\infty} (q_k + W) \Bigr| \;\leq\; |[0,2]| \;=\; 2 . \end{equation*}

But \(|W| > 0\), so the left side of the display above equals \(\infty\), a contradiction.

Hence our assumption was false: some subset of \(A\) is not Lebesgue measurable.

Throughout we use that \(|(b,c)| = c - b\) (Exercise 6 in Section 2A). For completeness: \((b,c) \subseteq [b,c]\) gives \(|(b,c)| \leq |[b,c]| = c - b\) by 2.5 and 2.14, while \([b,c] = (b,c) \cup \{b,c\}\) gives \(c - b = |[b,c]| \leq |(b,c)| + |\{b,c\}| = |(b,c)|\) by 2.8 and 2.3.

Suppose \(A\) is Lebesgue measurable.

The sets \((b,c) \setminus A\) and \(A\) are disjoint, and \(A\) is Lebesgue measurable. By Exercise 10 in this section (applied with the measurable set in the role of \(B\)),

\begin{equation*} \bigl| \bigl( (b,c) \setminus A \bigr) \cup A \bigr| = |(b,c) \setminus A| + |A| . \end{equation*}

Because \(A \subseteq (b,c)\), the union on the left equals \((b,c)\). Hence

\begin{equation*} |A| + |(b,c) \setminus A| = |(b,c)| = c - b . \end{equation*}

Suppose \(|A| + |(b,c) \setminus A| = c - b\).

Both \(A\) and \((b,c) \setminus A\) are subsets of \((b,c)\), so by 2.5 each of \(|A|\) and \(|(b,c) \setminus A|\) is at most \(c - b\); in particular both are finite, so the arithmetic below is legitimate.

By the Lemma proved in Exercise 10 of this section, there is a Borel set \(E_0\) with \((b,c) \setminus A \subseteq E_0\) and \(|E_0| = |(b,c) \setminus A|\). Replace \(E_0\) by

\begin{equation*} E = E_0 \cap (b,c), \end{equation*}

which is again Borel and still contains \((b,c) \setminus A\); and since

\begin{equation*} |(b,c) \setminus A| \leq |E| \leq |E_0| = |(b,c) \setminus A| \end{equation*}

by 2.5, we still have \(|E| = |(b,c) \setminus A|\). Thus \(E\) is a Borel subset of \((b,c)\) containing \((b,c) \setminus A\) with \(|E| = |(b,c) \setminus A|\).

Now let

\begin{equation*} B = (b,c) \setminus E . \end{equation*}

Then \(B\) is a Borel set. Moreover \(B \subseteq A\): if \(x \in B\) then \(x \in (b,c)\) and \(x \notin E\); since \((b,c) \setminus A \subseteq E\), we get \(x \notin (b,c) \setminus A\), and hence \(x \in A\).

The sets \(B\) and \(E\) are disjoint Borel sets whose union is \((b,c)\). Because outer measure is a measure on the Borel sets (2.68), it is finitely additive there, so

\begin{equation*} |B| + |E| = |(b,c)| = c - b . \end{equation*}

Since \(|E| = |(b,c) \setminus A|\) is finite, we may subtract and use the hypothesis:

\begin{equation*} |B| = c - b - |(b,c) \setminus A| = |A| . \end{equation*}

Finally, \(A \setminus B\) and \(B\) are disjoint and \(B\) is Lebesgue measurable (it is a Borel set). By Exercise 10 in this section,

\begin{equation*} \bigl| (A \setminus B) \cup B \bigr| = |A \setminus B| + |B| . \end{equation*}

Because \(B \subseteq A\), the union on the left equals \(A\). Hence

\begin{equation*} |A| = |A \setminus B| + |B| = |A \setminus B| + |A| . \end{equation*}

Since \(|A| \leq c - b < \infty\), we may cancel \(|A|\) to conclude \(|A \setminus B| = 0\).

Thus \(B\) is a Borel set with \(B \subseteq A\) and \(|A \setminus B| = 0\), which by definition 2.70 says exactly that \(A\) is Lebesgue measurable.

For each \(n \in \mathbb{Z}^+\) set

\begin{equation*} A_n = (-n,n) \cap A . \end{equation*}

Then \(A_n \subseteq (-n,n)\), and

\begin{equation*} (-n,n) \setminus A_n = (-n,n) \setminus A , \end{equation*}

because removing \(A\) from \((-n,n)\) is the same as removing the part of \(A\) that lies inside \((-n,n)\). So the displayed condition in the exercise says precisely

\begin{equation*} |A_n| + |(-n,n) \setminus A_n| = 2n = n - (-n) \end{equation*}

for every \(n \in \mathbb{Z}^+\). Exercise 12 in this section (with \(b = -n\) and \(c = n\)) says that this condition holds if and only if \(A_n\) is Lebesgue measurable. So the exercise reduces to showing:

\begin{equation*} A \text{ is Lebesgue measurable} \iff A_n \text{ is Lebesgue measurable for every } n \in \mathbb{Z}^+ . \end{equation*}

Suppose \(A\) is Lebesgue measurable.

The collection \(\mathcal{L}\) of Lebesgue measurable sets is a \(\sigma\)-algebra on \(\mathbb{R}\) (2.72). Each interval \((-n,n)\) is open, hence a Borel set, hence Lebesgue measurable. Because a \(\sigma\)-algebra is closed under intersection (2.25), \(A_n = (-n,n) \cap A\) is Lebesgue measurable for every \(n \in \mathbb{Z}^+\). By the forward direction of Exercise 12 applied to \(A_n \subseteq (-n,n)\),

\begin{equation*} |(-n,n) \cap A| + |(-n,n) \setminus A| = |A_n| + |(-n,n) \setminus A_n| = 2n \end{equation*}

for every \(n \in \mathbb{Z}^+\).

Suppose the equality holds for every \(n \in \mathbb{Z}^+\).

Fix \(n \in \mathbb{Z}^+\). As noted above, the hypothesis says \(|A_n| + |(-n,n) \setminus A_n| = n - (-n)\), and \(A_n \subseteq (-n,n)\). By the reverse direction of Exercise 12, \(A_n\) is Lebesgue measurable.

Finally,

\begin{equation*} A = \bigcup_{n=1}^{\infty} A_n , \end{equation*}

because every real number lies in \((-n,n)\) for all sufficiently large \(n\), so every element of \(A\) lies in some \(A_n\); and each \(A_n \subseteq A\). Since \(\mathcal{L}\) is a \(\sigma\)-algebra (2.72) it is closed under countable unions, so \(A\) is Lebesgue measurable.

By 2.75, the Cantor set \(C\) is the set of numbers in \([0,1]\) that have a base \(3\) representation containing only \(0\)s and \(2\)s. So it suffices to exhibit such a representation for each of the two numbers. We do this by running the usual greedy base \(3\) algorithm (repeatedly multiply by \(3\); the integer part is the next digit, the fractional part is the new remainder) and then verifying the resulting series.

The number \(\frac{1}{4}\).

Running the algorithm:

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{1}{4} &= \tfrac{3}{4} = 0 + \tfrac{3}{4}, &&\text{first digit } 0, \text{ remainder } \tfrac{3}{4};\\ 3 \cdot \tfrac{3}{4} &= \tfrac{9}{4} = 2 + \tfrac{1}{4}, &&\text{second digit } 2, \text{ remainder } \tfrac{1}{4}. \end{aligned} \end{equation*}

The remainder \(\frac{1}{4}\) has reappeared, so the block \(02\) repeats forever, suggesting

\begin{equation*} \tfrac{1}{4} = 0.020202\ldots_3 . \end{equation*}

Verification of this claim directly from the definition of a base \(3\) representation:

\begin{equation*} \sum_{k=1}^{\infty} \frac{2}{3^{2k}} = 2 \sum_{k=1}^{\infty} \Bigl(\frac{1}{9}\Bigr)^{k} = 2 \cdot \frac{\frac{1}{9}}{1 - \frac{1}{9}} = 2 \cdot \frac{1}{8} = \frac{1}{4}. \end{equation*}

Thus \(\frac{1}{4} \in [0,1]\) has a base \(3\) representation whose digits are all \(0\)s and \(2\)s. By 2.75, \(\frac{1}{4} \in C\).

The number \(\frac{9}{13}\).

Running the algorithm:

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{9}{13} &= \tfrac{27}{13} = 2 + \tfrac{1}{13}, &&\text{first digit } 2, \text{ remainder } \tfrac{1}{13};\\ 3 \cdot \tfrac{1}{13} &= \tfrac{3}{13} = 0 + \tfrac{3}{13}, &&\text{second digit } 0, \text{ remainder } \tfrac{3}{13};\\ 3 \cdot \tfrac{3}{13} &= \tfrac{9}{13} = 0 + \tfrac{9}{13}, &&\text{third digit } 0, \text{ remainder } \tfrac{9}{13}. \end{aligned} \end{equation*}

The original remainder \(\frac{9}{13}\) has reappeared, so the block \(200\) repeats forever, suggesting

\begin{equation*} \tfrac{9}{13} = 0.200200200\ldots_3 . \end{equation*}

Verification directly from the definition:

\begin{equation*} \sum_{k=0}^{\infty} \frac{2}{3^{3k+1}} = \frac{2}{3} \sum_{k=0}^{\infty} \Bigl(\frac{1}{27}\Bigr)^{k} = \frac{2}{3} \cdot \frac{1}{1 - \frac{1}{27}} = \frac{2}{3} \cdot \frac{27}{26} = \frac{18}{26} = \frac{9}{13}. \end{equation*}

Thus \(\frac{9}{13} \in [0,1]\) has a base \(3\) representation whose digits are all \(0\)s and \(2\)s. By 2.75, \(\frac{9}{13} \in C\).

Remark. As the book notes just after 2.75, neither of these numbers is an endpoint of any of the intervals removed in the construction of \(C\). Indeed, every such endpoint is of the form \(\frac{m}{3^n}\) for some integers \(m\) and \(n\) (each \(G_n\) is a union of intervals whose endpoints have denominator \(3^n\)), and neither \(\frac{1}{4}\) nor \(\frac{9}{13}\) can be written that way, since in lowest terms their denominators \(4\) and \(13\) are not powers of \(3\). So these are genuine examples of Cantor set elements that the construction does not hand you directly.

By 2.75, the Cantor set \(C\) is the set of numbers in \([0,1]\) having a base \(3\) representation whose digits are all \(0\) or \(2\). So it suffices to show that every base \(3\) representation of \(\frac{13}{17}\) contains a digit equal to \(1\).

Step 1: only one base 3 representation.

Suppose \(x \in [0,1]\) has two different base \(3\) representations, say

\begin{equation*} x = \sum_{k=1}^{\infty} \frac{a_k}{3^k} = \sum_{k=1}^{\infty} \frac{b_k}{3^k}, \end{equation*}

with all \(a_k, b_k \in \{0,1,2\}\), and let \(n\) be the smallest index with \(a_n \neq b_n\); say \(a_n < b_n\). Then

\begin{equation*} 0 = \sum_{k=n}^{\infty} \frac{b_k - a_k}{3^k} \geq \frac{b_n - a_n}{3^n} - \sum_{k=n+1}^{\infty} \frac{2}{3^k} = \frac{b_n - a_n}{3^n} - \frac{1}{3^n} \geq 0 , \end{equation*}

because \(b_n - a_n \geq 1\). Hence equality holds throughout, which forces \(b_n - a_n = 1\) and \(b_k - a_k = -2\) for all \(k > n\), i.e. \(b_k = 0\) and \(a_k = 2\) for all \(k > n\). In particular

\begin{equation*} x = \sum_{k=1}^{n} \frac{b_k}{3^k} = \frac{m}{3^n} \quad \text{where } m = \sum_{k=1}^{n} b_k 3^{n-k} \in \mathbb{Z}. \end{equation*}

Thus a number with two base \(3\) representations must be of the form \(m/3^n\).

Now if \(\frac{13}{17} = \frac{m}{3^n}\) for integers \(m, n\), then \(13 \cdot 3^n = 17 m\), so \(17\) divides \(13 \cdot 3^n\). But \(17\) is prime and divides neither \(13\) nor \(3\), a contradiction. Hence \(\frac{13}{17}\) has exactly one base \(3\) representation.

Step 2: that representation contains a 1.

Run the standard digit algorithm: writing \(x_0 = \frac{13}{17}\) and \(a_k = \lfloor 3 x_{k-1} \rfloor\), \(x_k = 3x_{k-1} - a_k\), we get

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{13}{17} &= \tfrac{39}{17} = 2 + \tfrac{5}{17}, &&a_1 = 2,\\ 3 \cdot \tfrac{5}{17} &= \tfrac{15}{17} = 0 + \tfrac{15}{17}, &&a_2 = 0,\\ 3 \cdot \tfrac{15}{17} &= \tfrac{45}{17} = 2 + \tfrac{11}{17}, &&a_3 = 2,\\ 3 \cdot \tfrac{11}{17} &= \tfrac{33}{17} = 1 + \tfrac{16}{17}, &&a_4 = 1 . \end{aligned} \end{equation*}

Each \(x_k\) produced lies in \([0,1)\), so this algorithm does produce a base \(3\) representation of \(\frac{13}{17}\); explicitly

\begin{equation*} \frac{13}{17} = \frac{2}{3} + \frac{0}{3^2} + \frac{2}{3^3} + \frac{1}{3^4} + \frac{1}{3^4}\cdot\frac{16}{17}, \end{equation*}

which one checks directly: \(\frac{61}{81} + \frac{16}{1377} = \frac{1037 + 16}{1377} = \frac{1053}{1377} = \frac{13}{17}\).

So the unique base \(3\) representation of \(\frac{13}{17}\) is \(0.2021\ldots_3\), and its fourth digit is \(1\). By Step 1 there is no other representation, so no base \(3\) representation of \(\frac{13}{17}\) consists only of \(0\)s and \(2\)s. By 2.75, \(\frac{13}{17} \notin C\).

Concretely, \(\frac{13}{17}\) lies in the open interval \(\left(\frac{61}{81}, \frac{62}{81}\right)\), one of the eight intervals making up \(G_4\) (see Exercise 16).

By 2.74, \(G_4\) is the union of the middle-third open intervals of the intervals making up \([0,1] \setminus (G_1 \cup G_2 \cup G_3)\). So first we identify those intervals.

From 2.74 we have \(G_1 = \left(\frac{1}{3},\frac{2}{3}\right)\), \(G_2 = \left(\frac{1}{9},\frac{2}{9}\right) \cup \left(\frac{7}{9},\frac{8}{9}\right)\), and

\begin{equation*} G_3 = \left(\tfrac{1}{27},\tfrac{2}{27}\right) \cup \left(\tfrac{7}{27},\tfrac{8}{27}\right) \cup \left(\tfrac{19}{27},\tfrac{20}{27}\right) \cup \left(\tfrac{25}{27},\tfrac{26}{27}\right), \end{equation*}

and \([0,1] \setminus (G_1 \cup G_2) = \left[0,\frac{1}{9}\right] \cup \left[\frac{2}{9},\frac{1}{3}\right] \cup \left[\frac{2}{3},\frac{7}{9}\right] \cup \left[\frac{8}{9},1\right]\). Deleting from each of these four closed intervals of length \(\frac{1}{9}\) its middle third (which is the corresponding interval of \(G_3\)) leaves eight closed intervals of length \(\frac{1}{27}\):

\begin{equation*} \left[0,\tfrac{1}{27}\right],\ \left[\tfrac{2}{27},\tfrac{3}{27}\right],\ \left[\tfrac{6}{27},\tfrac{7}{27}\right],\ \left[\tfrac{8}{27},\tfrac{9}{27}\right],\ \left[\tfrac{18}{27},\tfrac{19}{27}\right],\ \left[\tfrac{20}{27},\tfrac{21}{27}\right],\ \left[\tfrac{24}{27},\tfrac{25}{27}\right],\ \left[\tfrac{26}{27},1\right]. \end{equation*}

These are exactly the intervals of \([0,1] \setminus (G_1 \cup G_2 \cup G_3)\).

The middle third of \(\left[a, a + \frac{1}{27}\right]\) is \(\left(a + \frac{1}{81}, a + \frac{2}{81}\right)\). Writing each left endpoint above as \(\frac{3j}{81}\) with \(j \in \{0,2,6,8,18,20,24,26\}\), the middle third is \(\left(\frac{3j+1}{81}, \frac{3j+2}{81}\right)\). Hence

\begin{equation*} \begin{aligned} G_4 = &\left(\tfrac{1}{81},\tfrac{2}{81}\right) \cup \left(\tfrac{7}{81},\tfrac{8}{81}\right) \cup \left(\tfrac{19}{81},\tfrac{20}{81}\right) \cup \left(\tfrac{25}{81},\tfrac{26}{81}\right)\\ &\cup \left(\tfrac{55}{81},\tfrac{56}{81}\right) \cup \left(\tfrac{61}{81},\tfrac{62}{81}\right) \cup \left(\tfrac{73}{81},\tfrac{74}{81}\right) \cup \left(\tfrac{79}{81},\tfrac{80}{81}\right). \end{aligned} \end{equation*}

As a check, use the base \(3\) description preceding 2.75: there \(G_1 \cup \cdots \cup G_n\) is described as the set of numbers in \([0,1]\) all of whose base \(3\) representations have a \(1\) among the first \(n\) digits, so \(G_4\) consists of those numbers in \([0,1]\) all of whose base \(3\) representations have their first \(1\) in the fourth digit. Such numbers are precisely those in an interval \(\left(\frac{k}{81}, \frac{k+1}{81}\right)\) with \(k = (d_1 d_2 d_3 1)_3 = 27 d_1 + 9 d_2 + 3 d_3 + 1\) and \(d_1, d_2, d_3 \in \{0,2\}\): a point strictly inside such an interval has every base \(3\) representation beginning \(0.d_1 d_2 d_3 1\), since the set of numbers admitting a representation with that prefix is exactly the closed interval \(\left[\frac{k}{81}, \frac{k+1}{81}\right]\). The eight choices of \((d_1,d_2,d_3)\) give

\begin{equation*} k = 1,\ 7,\ 19,\ 25,\ 55,\ 61,\ 73,\ 79, \end{equation*}

in agreement with the list above. Note also that \(|G_4| = 8 \cdot \frac{1}{81} = \frac{8}{81} = \frac{2^{4-1}}{3^4}\), as asserted in the proof of 2.76.

Write \(S = \left\{\frac{1}{2}x + \frac{1}{2}y : x, y \in C\right\}\).

The inclusion \(S \subseteq [0,1]\) is immediate: if \(x, y \in C\) then \(x, y \in [0,1]\) by 2.74, and \([0,1]\) is convex, so \(\frac{1}{2}x + \frac{1}{2}y \in [0,1]\).

For the reverse inclusion, let \(t \in [0,1]\). If \(t = 1\), take \(x = y = 1 \in C\) (indeed \(1 = 0.222\ldots_3\), so \(1 \in C\) by 2.75). So assume \(t \in [0,1)\) and choose a base \(3\) representation

\begin{equation*} t = \sum_{k=1}^{\infty} \frac{t_k}{3^k}, \qquad t_k \in \{0,1,2\}. \end{equation*}

Define digits \(a_k, b_k \in \{0,2\}\) by

\begin{equation*} (a_k, b_k) = \begin{cases} (0,0) & \text{if } t_k = 0,\\ (0,2) & \text{if } t_k = 1,\\ (2,2) & \text{if } t_k = 2. \end{cases} \end{equation*}

In every case \(a_k + b_k = 2 t_k\), and \(a_k, b_k \in \{0,2\}\).

Now set

\begin{equation*} x = \sum_{k=1}^{\infty} \frac{a_k}{3^k}, \qquad y = \sum_{k=1}^{\infty} \frac{b_k}{3^k}. \end{equation*}

Both series converge (they are dominated by \(\sum 2/3^k = 1\)), and both displays exhibit a base \(3\) representation of \(x\) and of \(y\) using only the digits \(0\) and \(2\). Hence \(x \in C\) and \(y \in C\) by 2.75.

Finally, since both series converge absolutely we may add them term by term:

\begin{equation*} \frac{1}{2}x + \frac{1}{2}y = \sum_{k=1}^{\infty} \frac{1}{2} \cdot \frac{a_k + b_k}{3^k} = \sum_{k=1}^{\infty} \frac{t_k}{3^k} = t . \end{equation*}

Thus \(t \in S\). As \(t \in [0,1]\) was arbitrary, \([0,1] \subseteq S\), and therefore \(S = [0,1]\).

This is the counterexample announced in the discussion introducing the Cantor set (just before 2.73): the sum of two sets of Lebesgue measure \(0\) can have positive Lebesgue measure. Indeed, put \(A = \frac{1}{2}C\). Scaling a cover of \(C\) by open intervals scales the sum of the lengths by \(\frac{1}{2}\), so outer measure satisfies \(|A| = \frac{1}{2}|C| = 0\) by 2.76(b); yet \(S = A + A = [0,1]\), so \(|A + A| = 1\).

Let \(C\) denote the Cantor set. It suffices to prove that \(C\) has no isolated points, meaning: for every \(x \in C\) and every \(\varepsilon > 0\), the interval \((x - \varepsilon, x + \varepsilon)\) contains infinitely many elements of \(C\). Granting that, let \(I\) be an open interval of \(\mathbb{R}\) that contains at least one element of \(C\), say \(x \in I \cap C\). Because \(I\) is open, there exists \(\varepsilon > 0\) with \((x-\varepsilon, x+\varepsilon) \subseteq I\); then \(I \cap C \supseteq (x-\varepsilon,x+\varepsilon) \cap C\) is infinite. So \(I \cap C\) is either empty or infinite, which is the assertion of the exercise.

Now fix \(x \in C\) and \(\varepsilon > 0\). By 2.75 there is a base \(3\) representation

\begin{equation*} x = \sum_{k=1}^{\infty} \frac{a_k}{3^k}, \qquad a_k \in \{0,2\} \text{ for every } k . \end{equation*}

For each positive integer \(n\) define a new digit string by flipping the \(n\)-th digit:

\begin{equation*} a_k^{(n)} = \begin{cases} a_k & \text{if } k \neq n,\\ 2 - a_n & \text{if } k = n, \end{cases} \qquad\text{and}\qquad x^{(n)} = \sum_{k=1}^{\infty} \frac{a_k^{(n)}}{3^k}. \end{equation*}

Since \(a_n \in \{0,2\}\) we have \(2 - a_n \in \{0,2\}\), so every digit \(a_k^{(n)}\) lies in \(\{0,2\}\). The series converges and its sum lies in \([0,1]\), so the display above is a base \(3\) representation of \(x^{(n)}\) using only \(0\)s and \(2\)s. By 2.75, \(x^{(n)} \in C\).

The two strings differ only in the \(n\)-th digit, so

\begin{equation*} \left|x - x^{(n)}\right| = \frac{|a_n - (2 - a_n)|}{3^n} = \frac{2}{3^n} . \end{equation*}

In particular \(x^{(n)} \neq x\) for every \(n\), and the numbers \(\left|x - x^{(n)}\right| = \frac{2}{3^n}\) are pairwise distinct for distinct \(n\); hence \(x^{(1)}, x^{(2)}, x^{(3)}, \ldots\) are pairwise distinct elements of \(C\), all different from \(x\).

Choose \(N\) with \(\frac{2}{3^N} < \varepsilon\). Then for every \(n \geq N\) we have \(\left|x - x^{(n)}\right| = \frac{2}{3^n} \leq \frac{2}{3^N} < \varepsilon\), so \(x^{(n)} \in (x-\varepsilon, x+\varepsilon)\). Thus \((x - \varepsilon, x + \varepsilon)\) contains the infinitely many distinct points \(x^{(N)}, x^{(N+1)}, x^{(N+2)}, \ldots\) of \(C\), as required.

Combined with 2.76(a), this says that \(C\) is a perfect set: closed with no isolated points. (Consistently with 2.76(c), the conclusion is not that \(C\) contains an interval; the infinitely many nearby Cantor set points found above are separated from \(x\) by gaps belonging to the removed sets \(G_n\).)

The answer is \(\int_0^1 \Lambda = \frac{1}{2}\).

By 2.79 the Cantor function \(\Lambda\) is continuous on \([0,1]\), hence Riemann integrable there; being bounded and continuous it is also Lebesgue measurable, and the two integrals agree, so the computation below settles the value under either reading.

The whole point is the following symmetry.

Claim: \(\Lambda(1-x) = 1 - \Lambda(x)\) for every \(x \in [0,1]\).

Case 1: \(x \in C\). By 2.75 write \(x = \sum_{k=1}^{\infty} a_k 3^{-k}\) with every \(a_k \in \{0,2\}\). This representation is the unique base \(3\) representation of \(x\) using only \(0\)s and \(2\)s: by Step 1 of the solution to Exercise 15, if a number has two base \(3\) representations then in one of them some digit is followed by all \(2\)s while in the other that digit is increased by \(1\) and followed by all \(0\)s; a digit \(d \in \{0,2\}\) increased by \(1\) becomes \(1\) or \(3\), neither of which is an admissible digit in \(\{0,2\}\), so at most one of the two representations consists of \(0\)s and \(2\)s. Since \(\sum_{k=1}^{\infty} 2 \cdot 3^{-k} = 1\),

\begin{equation*} 1 - x = \sum_{k=1}^{\infty} \frac{2 - a_k}{3^k}, \end{equation*}

and every digit \(2 - a_k\) lies in \(\{0,2\}\). So \(1 - x \in C\) and the display is its \(\{0,2\}\)-representation. By 2.77, with \(e_k = a_k/2 \in \{0,1\}\),

\begin{equation*} \Lambda(x) = \sum_{k=1}^{\infty} \frac{e_k}{2^k}, \qquad \Lambda(1-x) = \sum_{k=1}^{\infty} \frac{1 - e_k}{2^k} = \sum_{k=1}^{\infty} \frac{1}{2^k} - \Lambda(x) = 1 - \Lambda(x). \end{equation*}

Case 2: \(x \in [0,1] \setminus C\). By 2.74 we have \([0,1] \setminus C = \bigcup_{n=1}^{\infty} G_n\), so \(x\) lies in some \(G_n\). We first pin down the intervals composing \(G_n\).

Subclaim: for every \(n \geq 1\), the set \([0,1] \setminus (G_1 \cup \cdots \cup G_n)\) is the union of the \(2^n\) closed intervals \(\left[c, c + \frac{1}{3^n}\right]\) with \(c = \sum_{i=1}^{n} \frac{d_i}{3^i}\) and \(d_1, \ldots, d_n \in \{0,2\}\); and \(G_n\) is the union of the \(2^{n-1}\) open intervals \(\left(a, a + \frac{1}{3^n}\right)\) with

\begin{equation*} a = \sum_{i=1}^{n-1} \frac{d_i}{3^i} + \frac{1}{3^n}, \qquad d_1, \ldots, d_{n-1} \in \{0,2\} . \end{equation*}

Proof of the subclaim, by induction on \(n\), where the statement at level \(0\) is read as the assertion that \([0,1]\) is the union of the single closed interval \(\left[0, 1\right]\) attached to the empty digit string (true by inspection). For \(n = 1\): by 2.74, \(G_1 = \left(\frac{1}{3}, \frac{2}{3}\right)\), which is the single interval \(\left(a, a + \frac{1}{3}\right)\) with \(a = \frac{1}{3}\) (an empty sum plus \(\frac{1}{3}\)), and \([0,1] \setminus G_1 = \left[0,\frac{1}{3}\right] \cup \left[\frac{2}{3},1\right]\) is the union of the two intervals attached to \(d_1 = 0\) and \(d_1 = 2\). Now let \(n \geq 2\) and suppose the description of \([0,1] \setminus (G_1 \cup \cdots \cup G_{n-1})\) holds for \(n - 1\), so that set is the union of the intervals \(\left[c, c + \frac{1}{3^{n-1}}\right]\) with \(c = \sum_{i=1}^{n-1} d_i 3^{-i}\), \(d_i \in \{0,2\}\). By 2.74, \(G_n\) is the union of the middle-third open intervals of these, and the middle third of \(\left[c, c + \frac{1}{3^{n-1}}\right]\) is \(\left(c + \frac{1}{3^n},\, c + \frac{2}{3^n}\right)\), which is exactly the interval \(\left(a, a + \frac{1}{3^n}\right)\) with \(a = c + \frac{1}{3^n}\). Removing it leaves the two intervals \(\left[c, c + \frac{1}{3^n}\right]\) and \(\left[c + \frac{2}{3^n}, c + \frac{3}{3^n}\right]\), which are the intervals attached to the digit strings \((d_1, \ldots, d_{n-1}, 0)\) and \((d_1, \ldots, d_{n-1}, 2)\). This proves the subclaim.

So fix the interval \((a,b)\) of \(G_n\) containing \(x\), where \(a\) is as in the subclaim, \(b = a + \frac{1}{3^n}\), and put \(e_i = d_i/2\). Using \(\frac{1}{3^n} = \sum_{k > n} \frac{2}{3^k}\) we get the two \(\{0,2\}\)-representations

\begin{equation*} a = 0.d_1 \ldots d_{n-1} 0\,2\,2\,2\ldots_3, \qquad b = 0.d_1 \ldots d_{n-1} 2\,0\,0\,0\ldots_3 , \end{equation*}

so \(a, b \in C\) and, by 2.77,

\begin{equation*} \Lambda(a) = \sum_{i=1}^{n-1}\frac{e_i}{2^i} + \sum_{k>n}\frac{1}{2^k} = \sum_{i=1}^{n-1}\frac{e_i}{2^i} + \frac{1}{2^n} = \Lambda(b). \end{equation*}

Next, for \(x \in (a,b)\) every base \(3\) representation of \(x\) begins \(0.d_1 \ldots d_{n-1} 1\). Indeed, the set of numbers admitting a representation with a given prefix \(c_1 \ldots c_n\) is exactly the closed interval \(\left[\sum_{i \leq n} c_i 3^{-i},\, \sum_{i \leq n} c_i 3^{-i} + 3^{-n}\right]\), since the tails \(\sum_{k>n} c_k 3^{-k}\) range over all of \(\left[0, 3^{-n}\right]\). Reading a prefix \(c_1 \ldots c_n\) as the base \(3\) integer \(k = \sum_{i \leq n} c_i 3^{n-i} \in \{0, 1, \ldots, 3^n - 1\}\), that interval is \(\left[\frac{k}{3^n}, \frac{k+1}{3^n}\right]\); distinct prefixes give distinct \(k\) and hence intervals with disjoint interiors. The prefix \(d_1 \ldots d_{n-1} 1\) has \(\frac{k}{3^n} = a\), so it gives exactly \([a,b]\), and a point of the open interval \((a,b)\) lies in the interior of \([a,b]\) and therefore in no other prefix interval. Truncating after that first \(1\) and halving the earlier \(2\)s, 2.77 gives \(\Lambda(x) = \sum_{i<n} e_i 2^{-i} + 2^{-n}\), which is the value \(\Lambda(a) = \Lambda(b)\) computed above. Hence \(\Lambda\) is constant on \([a,b]\), with value \(\Lambda(a)\).

Now \(1 - x \in (1-b, 1-a)\), and \(1-a, 1-b \in C\) by Case 1. Since \(\Lambda\) is increasing (2.79),

\begin{equation*} \Lambda(1-b) \leq \Lambda(1-x) \leq \Lambda(1-a), \end{equation*}

while Case 1 gives \(\Lambda(1-a) = 1 - \Lambda(a)\) and \(\Lambda(1-b) = 1 - \Lambda(b) = 1 - \Lambda(a)\). The two outer bounds coincide, so \(\Lambda(1-x) = 1 - \Lambda(a) = 1 - \Lambda(x)\). This proves the claim.

Given the claim, substitute \(u = 1 - x\) in the Riemann integral:

\begin{equation*} \int_0^1 \Lambda(x)\,dx = \int_0^1 \Lambda(1-u)\,du = \int_0^1 \bigl(1 - \Lambda(u)\bigr)\,du = 1 - \int_0^1 \Lambda(u)\,du . \end{equation*}

Writing \(I\) for the common value of these integrals gives \(I = 1 - I\), so \(I = \frac{1}{2}\).

As a check, compute the integral by summing \(\Lambda\) over the intervals it is constant on, whose lengths total \(1\) by the proof of 2.76(b). At stage \(n\) there are \(2^{n-1}\) removed intervals, each of length \(3^{-n}\); by the formula above the value of \(\Lambda\) on the one indexed by \((d_1,\ldots,d_{n-1})\) is \(\frac{2m+1}{2^n}\), where \(m = \sum_{i<n} e_i 2^{n-1-i}\) runs bijectively over \(0, 1, \ldots, 2^{n-1}-1\). Those values sum to \(\sum_{m=0}^{2^{n-1}-1} \frac{2m+1}{2^n} = \frac{(2^{n-1})^2}{2^n} = 2^{n-2}\), so stage \(n\) contributes \(2^{n-2} 3^{-n}\) and the total is

\begin{equation*} \sum_{n=1}^{\infty} \frac{2^{n-2}}{3^n} = \frac{1}{4}\sum_{n=1}^{\infty}\left(\frac{2}{3}\right)^{n} = \frac{1}{4} \cdot 2 = \frac{1}{2}, \end{equation*}

agreeing with the answer above.

(a) First find the base \(3\) representation of \(\frac{9}{13}\) by the usual digit algorithm:

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{9}{13} &= \tfrac{27}{13} = 2 + \tfrac{1}{13}, &&\text{digit } 2,\\ 3 \cdot \tfrac{1}{13} &= \tfrac{3}{13} = 0 + \tfrac{3}{13}, &&\text{digit } 0,\\ 3 \cdot \tfrac{3}{13} &= \tfrac{9}{13} = 0 + \tfrac{9}{13}, &&\text{digit } 0, \end{aligned} \end{equation*}

and now the remainder \(\frac{9}{13}\) has returned, so the digits repeat with period \(3\):

\begin{equation*} \frac{9}{13} = 0.\overline{200}_3 . \end{equation*}

This is confirmed directly by summing the geometric series: \(\sum_{j=0}^{\infty} \frac{2}{3^{3j+1}} = \frac{2/3}{1 - 1/27} = \frac{2/3 \cdot 27}{26} = \frac{18}{26} = \frac{9}{13}\).

All digits are \(0\) or \(2\), so \(\frac{9}{13} \in C\) by 2.75 (this is the fact asserted in Exercise 14). By 2.77, \(\Lambda\!\left(\frac{9}{13}\right)\) is obtained by replacing each \(2\) by \(1\) and reading the result in base \(2\):

\begin{equation*} \Lambda\!\left(\frac{9}{13}\right) = 0.\overline{100}_2 = \sum_{j=0}^{\infty} \frac{1}{2^{3j+1}} = \frac{1/2}{1 - 1/8} = \frac{4}{7}. \end{equation*}

As a sanity check, \(\frac{2}{3} < \frac{9}{13} < \frac{7}{9}\) (numerically \(0.667 < 0.692 < 0.778\)), and by Example 2.78 \(\Lambda\!\left(\frac{2}{3}\right) = \frac{1}{2}\) while \(\Lambda\) equals \(\frac{3}{4}\) on \(\left(\frac{7}{9},\frac{8}{9}\right)\). Since \(\Lambda\) is increasing (2.79), the value must lie between \(\frac{1}{2}\) and \(\frac{3}{4}\), and indeed \(\frac{4}{7} \approx 0.571\) does.

(b) Here \(0.93\) is a decimal, so we must evaluate \(\Lambda\) at \(\frac{93}{100}\). Its base \(3\) digits:

\begin{equation*} \begin{aligned} 3 \cdot \tfrac{93}{100} &= \tfrac{279}{100} = 2 + \tfrac{79}{100}, &&\text{digit } 2,\\ 3 \cdot \tfrac{79}{100} &= \tfrac{237}{100} = 2 + \tfrac{37}{100}, &&\text{digit } 2,\\ 3 \cdot \tfrac{37}{100} &= \tfrac{111}{100} = 1 + \tfrac{11}{100}, &&\text{digit } 1 . \end{aligned} \end{equation*}

So \(\frac{93}{100} = 0.221\ldots_3\), with a \(1\) appearing in the third digit. Since \(\frac{93}{100}\) is not of the form \(m/3^k\) (that would force \(100 \mid 93 \cdot 3^k\), impossible as \(\gcd(93 \cdot 3^k, 100)\) is \(1\)), by Step 1 of Exercise 15 this base \(3\) representation is unique, and it contains a \(1\); hence \(\frac{93}{100} \notin C\) by 2.75.

By 2.77, truncate after the first \(1\) and replace each \(2\) before that \(1\) by \(1\): the string \(0.221\) becomes \(0.111\), read in base \(2\). Therefore

\begin{equation*} \Lambda(0.93) = 0.111_2 = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}. \end{equation*}

Equivalently and as a check: \(0.221_3 = \frac{18 + 6 + 1}{27} = \frac{25}{27}\) and \(\frac{26}{27} \approx 0.96296\), so \(\frac{25}{27} < 0.93 < \frac{26}{27}\), i.e. \(0.93\) lies in the interval \(\left(\frac{25}{27},\frac{26}{27}\right)\) of \(G_3\) listed in 2.74. The graph in Example 2.78 shows \(\Lambda\) taking the value \(\frac{7}{8}\) on that rightmost third-stage interval.

The answers are \(\Lambda^{-1}\bigl(\{\frac{1}{3}\}\bigr) = \left\{\frac{1}{4}\right\}\) and \(\Lambda^{-1}\bigl(\{\frac{5}{16}\}\bigr) = \left[\frac{19}{81}, \frac{20}{81}\right]\).

Set up two bookkeeping facts about 2.77, splitting \([0,1]\) into \(C\) and its complement.

Fact 1 (points of \(C\)). By 2.75 each \(x \in C\) has a base \(3\) representation \(x = \sum_{k} a_k 3^{-k}\) with all \(a_k \in \{0,2\}\), and (as shown in the solution to Exercise 19) that representation is unique. Putting \(e_k = a_k/2 \in \{0,1\}\), definition 2.77 gives \(\Lambda(x) = \sum_{k} e_k 2^{-k}\). Conversely, any string \((e_k)\) of \(0\)s and \(1\)s arises this way from \(x = \sum_k 2 e_k 3^{-k} \in C\). Hence \(x \mapsto (e_k)\) is a bijection between \(C\) and the set of binary digit strings, and under it \(\Lambda(x)\) is the number that the string represents in base \(2\). So for \(y \in [0,1]\), the preimages of \(y\) lying in \(C\) correspond exactly to the base \(2\) representations of \(y\).

Fact 2 (points off \(C\)). By the subclaim in Case 2 of the solution to Exercise 19, together with \([0,1] \setminus C = \bigcup_{n} G_n\) (2.74) and the disjointness of \(G_1, G_2, \ldots\) noted in the proof of 2.76, the set \([0,1] \setminus C\) is the disjoint union, over \(n \geq 1\) and over \((d_1,\ldots,d_{n-1}) \in \{0,2\}^{n-1}\), of the open intervals \((a,b)\) with \(a = \sum_{i<n} d_i 3^{-i} + 3^{-n}\) and \(b = a + 3^{-n}\); and on such an interval \(\Lambda\) has the constant value

\begin{equation*} \sum_{i=1}^{n-1} \frac{e_i}{2^i} + \frac{1}{2^n} = 0.e_1\ldots e_{n-1}1_2 = \frac{2m+1}{2^n}, \qquad e_i = \tfrac{d_i}{2},\ \ m = \sum_{i<n} e_i 2^{n-1-i}. \end{equation*}

As \((d_1,\ldots,d_{n-1})\) ranges over \(\{0,2\}^{n-1}\), the integer \(m\) ranges over \(0,1,\ldots,2^{n-1}-1\) bijectively. So the values \(\Lambda\) takes off \(C\) are exactly the numbers \((2m+1)/2^n\), i.e. the dyadic rationals in \((0,1)\); and such a value determines \(n\) (the \(2\)-adic denominator) and then \(m\), hence determines \((d_1,\ldots,d_{n-1})\) and the interval uniquely.

Fact 3 (base \(2\) representations). The argument of Step 1 in the solution to Exercise 15 works verbatim with \(3\) replaced by \(2\) and the digit bound \(2\) replaced by \(1\): if \(y \in [0,1]\) has two base \(2\) representations, then \(y = m/2^n\) for some integers \(m, n\), and the two representations are \(0.b_1 \ldots b_{n-1} b_n 0 0 0 \ldots_2\) and \(0.b_1 \ldots b_{n-1} (b_n - 1) 1 1 1 \ldots_2\) for a single index \(n\). Conversely, if \(y \in (0,1)\) is dyadic, write \(y = m/2^n\) with \(m\) odd and \(0 < m < 2^n\); then \(y = 0.b_1 \ldots b_n 0 0 0 \ldots_2\) with \(b_n = 1\), and since \(\sum_{k > n} 2^{-k} = 2^{-n}\) we also have \(y = 0.b_1 \ldots b_{n-1} 0 1 1 1 \ldots_2\). So a dyadic rational in \((0,1)\) has exactly two base \(2\) representations, and every other number in \([0,1]\) has exactly one.

(a) \(y = \frac{1}{3}\). If \(\frac{1}{3} = \frac{m}{2^n}\) then \(2^n = 3m\), impossible since \(3 \nmid 2^n\). So \(\frac{1}{3}\) is not a dyadic rational, and by Fact 2 no point outside \(C\) is mapped to it. By Fact 3, \(\frac{1}{3}\) has exactly one base \(2\) representation. That representation is \(0.\overline{01}_2\), since

\begin{equation*} 0.\overline{01}_2 = \sum_{j=1}^{\infty} \frac{1}{4^{j}} = \frac{1/4}{1 - 1/4} = \frac{1}{3}. \end{equation*}

By Fact 1 the one and only preimage is the point of \(C\) whose \(\{0,2\}\)-representation is \(0.\overline{02}_3\), namely

\begin{equation*} x = \sum_{j=1}^{\infty} \frac{2}{9^{j}} = \frac{2/9}{1-1/9} = \frac{2}{8} = \frac{1}{4}. \end{equation*}

Hence \(\Lambda^{-1}\bigl(\{\tfrac{1}{3}\}\bigr) = \left\{\frac{1}{4}\right\}\). (Note \(\frac{1}{4} \in C\), as asserted in Exercise 14.)

(b) \(y = \frac{5}{16}\). This is dyadic, so both cases contribute.

Preimages off \(C\): write \(\frac{5}{16} = \frac{2m+1}{2^n}\); this forces \(n = 4\) and \(2m+1 = 5\), so \(m = 2\), whose \(3\)-digit binary form is \(e_1e_2e_3 = 010\), i.e. \((d_1,d_2,d_3) = (0,2,0)\). By Fact 2 the corresponding interval is \((a,b)\) with

\begin{equation*} a = \frac{0}{3} + \frac{2}{9} + \frac{0}{27} + \frac{1}{81} = \frac{18 + 1}{81} = \frac{19}{81}, \qquad b = a + \frac{1}{81} = \frac{20}{81}, \end{equation*}

which is indeed one of the eight intervals of \(G_4\) found in Exercise 16. So \(\Lambda^{-1}\bigl(\{\tfrac{5}{16}\}\bigr) \setminus C = \left(\frac{19}{81}, \frac{20}{81}\right)\).

Preimages in \(C\): \(\frac{5}{16} = \frac{1}{4} + \frac{1}{16} = 0.0101_2\) is dyadic, so by Fact 3 it has exactly two base \(2\) representations,

\begin{equation*} 0.0101\,000\ldots_2 \qquad\text{and}\qquad 0.0100\,111\ldots_2 . \end{equation*}

By Fact 1 these give the two Cantor set preimages

\begin{equation*} 0.0202\,000\ldots_3 = \frac{2}{9} + \frac{2}{81} = \frac{18+2}{81} = \frac{20}{81}, \end{equation*}

\begin{equation*} 0.0200\,222\ldots_3 = \frac{2}{9} + \sum_{k=5}^{\infty}\frac{2}{3^k} = \frac{18}{81} + \frac{1}{81} = \frac{19}{81}. \end{equation*}

(Both are in \(C\), being the endpoints of the removed interval above, as in Case 2 of Exercise 19.)

Combining,

\begin{equation*} \Lambda^{-1}\left(\left\{\tfrac{5}{16}\right\}\right) = \left\{\tfrac{19}{81}\right\} \cup \left(\tfrac{19}{81}, \tfrac{20}{81}\right) \cup \left\{\tfrac{20}{81}\right\} = \left[\frac{19}{81}, \frac{20}{81}\right]. \end{equation*}

The contrast between (a) and (b) is exactly the dichotomy of Facts 1 and 2: \(\Lambda\) is injective at a value \(y\) that is not a dyadic rational, while at each dyadic rational in \((0,1)\) the level set is a full closed interval, one of the intervals on which the Cantor function is flat.

Here \(C\) is the Cantor set and \(\Lambda\) is the Cantor function (2.77). Everything rests on the classical characterization of rationality in terms of digit strings, which we prove first in the exact form needed.

Throughout, fix an integer \(b \ge 2\). By a base \(b\) representation of \(t \in [0,1]\) we mean a sequence \(d_1, d_2, \ldots\) of digits from \(\{0, 1, \ldots, b-1\}\) with

\begin{equation*} t = \sum_{j=1}^{\infty} \frac{d_j}{b^{j}} . \end{equation*}

(The number \(1\) is represented by the constant string \(b-1\); every \(t \in [0,1)\) has one or two representations, as the next lemma shows.)

Lemma 1 (near-uniqueness). If \((c_j)\) and \((c_j^{\prime})\) are base \(b\) representations of the same number \(t\) and \((c_j) \ne (c_j^{\prime})\), then, letting \(n\) be the first index where they differ and assuming without loss of generality that \(c_n > c_n^{\prime}\), we have

\begin{equation*} c_n = c_n^{\prime} + 1, \qquad c_j = 0 \ \text{ and } \ c_j^{\prime} = b - 1 \ \text{ for all } j > n . \end{equation*}

In particular, each of the two strings is eventually constant.

Proof. Since the strings agree before index \(n\),

\begin{equation*} 0 = \sum_{j=1}^{\infty} \frac{c_j - c_j^{\prime}}{b^{j}} = \frac{c_n - c_n^{\prime}}{b^{n}} + \sum_{j > n} \frac{c_j - c_j^{\prime}}{b^{j}} \ge \frac{1}{b^{n}} - \sum_{j > n} \frac{b-1}{b^{j}} = \frac{1}{b^{n}} - \frac{1}{b^{n}} = 0 , \end{equation*}

where the inequality used \(c_n - c_n^{\prime} \ge 1\) and \(c_j - c_j^{\prime} \ge -(b-1)\). Thus every inequality used is an equality, forcing \(c_n - c_n^{\prime} = 1\) and \(c_j - c_j^{\prime} = -(b-1)\), that is \(c_j = 0\) and \(c_j^{\prime} = b-1\), for every \(j > n\). \(\square\)

Lemma 2. Let \((d_j)\) be a base \(b\) representation of \(t \in [0,1]\). Then \(t\) is rational if and only if the sequence \((d_j)\) is eventually periodic (that is, there exist \(N \ge 0\) and \(k \ge 1\) with \(d_{j+k} = d_j\) for all \(j > N\)).

Proof. Suppose first that \((d_j)\) is eventually periodic, with \(d_{j+k} = d_j\) for all \(j > N\). Put

\begin{equation*} z = \sum_{i=1}^{\infty} \frac{d_{N+i}}{b^{i}} , \end{equation*}

so that \(t = \sum_{j \le N} d_j b^{-j} + b^{-N} z\). The string \((d_{N+i})_{i \ge 1}\) is periodic with period \(k\), hence

\begin{equation*} b^{k} z = \sum_{i=1}^{\infty} d_{N+i} b^{\,k-i} = m + \sum_{i=1}^{\infty} \frac{d_{N+k+i}}{b^{i}} = m + z , \end{equation*}

where \(m = \sum_{i=1}^{k} d_{N+i} b^{\,k-i}\) is an integer. Thus \(z = m/(b^{k}-1)\) is rational, and therefore \(t\) is rational.

Conversely, suppose \(t = p/q\) with \(p, q\) integers, \(q \ge 1\). For \(n \ge 0\) let

\begin{equation*} s_n = \sum_{i=1}^{\infty} \frac{d_{n+i}}{b^{i}} = b^{n} t - \sum_{j=1}^{n} d_j b^{\,n-j} . \end{equation*}

Then \(s_n \in [0,1]\), and \(q s_n = b^{n} p - q \sum_{j \le n} d_j b^{\,n-j}\) is an integer. Hence \(q s_n \in \{0, 1, \ldots, q\}\), so the numbers \(s_0, s_1, s_2, \ldots\) take at most \(q+1\) distinct values. By the pigeonhole principle there are \(n < m\) with \(s_n = s_m\). Now \((d_{n+i})_{i \ge 1}\) and \((d_{m+i})_{i \ge 1}\) are two base \(b\) representations of the same number \(s_n = s_m\). If they are equal, then \(d_{j+(m-n)} = d_j\) for every \(j > n\) and \((d_j)\) is eventually periodic. If they are not equal, Lemma 1 says that each of them is eventually constant, so \((d_j)\) is eventually constant and in particular eventually periodic. \(\square\)

(a) Let \(x \in [0,1]\) be rational. There are two cases, according to the two bullet points in the definition 2.77 of \(\Lambda\).

Case 1: \(x \notin C\). By 2.75, a number in \([0,1]\) lies in \(C\) exactly when it has a base \(3\) representation using only the digits \(0\) and \(2\); so every base \(3\) representation of \(x\) contains a digit \(1\). The definition of \(\Lambda\) takes such a representation, truncates it immediately after the first \(1\), replaces each \(2\) occurring before that \(1\) by a \(1\), and reads the resulting finite string \(e_1 e_2 \cdots e_n\) in base \(2\). Hence

\begin{equation*} \Lambda(x) = \sum_{j=1}^{n} \frac{e_j}{2^{j}} , \end{equation*}

a finite sum of rationals, so \(\Lambda(x)\) is rational (indeed a dyadic rational). Note that this case did not use the rationality of \(x\): \(\Lambda\) is rational at every point of \([0,1] \setminus C\), which is unsurprising since \(\Lambda\) is constant on each interval removed in the construction of \(C\).

Case 2: \(x \in C\). Let \(d_1 d_2 \cdots\) be the base \(3\) representation of \(x\) using only \(0\)s and \(2\)s, which is the one used in the definition of \(\Lambda\). Since \(x\) is rational, Lemma 2 (with \(b = 3\)) shows that \((d_j)\) is eventually periodic. By definition,

\begin{equation*} \Lambda(x) = \sum_{j=1}^{\infty} \frac{d_j/2}{2^{j}} , \end{equation*}

that is, \((d_j / 2)_{j \ge 1}\) is a base \(2\) representation of \(\Lambda(x)\). Replacing each digit by half of it does not disturb eventual periodicity, so \((d_j/2)\) is eventually periodic, and Lemma 2 (with \(b = 2\)) shows that \(\Lambda(x)\) is rational.

In both cases \(\Lambda(x)\) is rational, as desired.

(b) Suppose \(x \in C\) and \(\Lambda(x)\) is rational. Let \(d_1 d_2 \cdots\) be the base \(3\) representation of \(x\) consisting only of \(0\)s and \(2\)s; this representation is unique, as asserted in 2.77, because two distinct representations of the same number must, by Lemma 1, differ by exactly \(1\) at the first index where they differ, whereas two digits from \(\{0,2\}\) differ by \(0\) or \(2\).

By the definition of \(\Lambda\), the sequence \((d_j/2)_{j \ge 1}\), whose terms lie in \(\{0,1\}\), is a base \(2\) representation of the number \(\Lambda(x)\). Since \(\Lambda(x)\) is rational, Lemma 2 applied with \(b = 2\) to this particular representation shows that \((d_j/2)\) is eventually periodic. Multiplying every term by \(2\), the sequence \((d_j)\) is eventually periodic. Now Lemma 2 applied with \(b = 3\) shows that

\begin{equation*} x = \sum_{j=1}^{\infty} \frac{d_j}{3^{j}} \end{equation*}

is rational, as desired.

Note that the hypothesis \(x \in C\) cannot be dropped in (b): \(\Lambda\) takes the rational value \(\tfrac12\) at every point of the interval \(\left(\tfrac13, \tfrac23\right)\) by 2.78, and that interval contains irrational numbers.

We construct such an \(f\) explicitly out of decimal digits. The idea: build a function whose value at \(x\) depends only on the tail of the decimal expansion of \(x\), and which already takes all its values on every set of numbers sharing a fixed initial block of digits. Since every nonempty open interval contains such a set, the function is onto on every interval.

The digit map. For \(x \in \mathbb{R}\) let \(\langle x \rangle = x - \lfloor x \rfloor \in [0,1)\) denote the fractional part. Every \(t \in [0,1)\) has exactly one decimal representation \(t = \sum_{j \ge 1} d_j 10^{-j}\) with digits \(d_j \in \{0, 1, \ldots, 9\}\) that is not eventually equal to \(9\). For existence, set \(a_j = \lfloor 10^{j} t \rfloor\) and \(d_j = a_j - 10 a_{j-1}\) for \(j \ge 1\); from \(a_{j-1} \le 10^{j-1} t < a_{j-1} + 1\) we get \(10 a_{j-1} \le a_j \le 10 a_{j-1} + 9\), so each \(d_j\) is a digit, the sums telescope to \(\sum_{j=1}^{n} d_j 10^{-j} = a_n 10^{-n}\) (using \(a_0 = 0\)), and \(a_n 10^{-n} \le t < a_n 10^{-n} + 10^{-n}\) forces these partial sums to converge to \(t\); moreover \(d_j = 9\) for all \(j > n\) would give \(t = a_n 10^{-n} + 10^{-n}\), contradicting that same inequality. Uniqueness follows from Lemma 1 of Exercise 22 with \(b = 10\): of two distinct representations of one number, one is eventually \(0\) and the other eventually \(9\), so at most one avoids being eventually \(9\). Write \(d_j(x)\) for the \(j\)-th digit of this canonical representation of \(\langle x \rangle\), and define \(\varphi : \mathbb{R} \to [0,9]\) by

\begin{equation*} \varphi(x) = \limsup_{n \to \infty} \frac{d_1(x) + d_2(x) + \cdots + d_n(x)}{n} . \end{equation*}

The averages lie in \([0,9]\), so \(\varphi\) is well defined with values in \([0,9]\).

Composing with a surjection. Define \(h : [0,9] \to \mathbb{R}\) by

\begin{equation*} h(t) = \tan\!\left( \pi \left( \frac{t}{9} - \frac12 \right) \right) \ \text{ for } t \in (0,9), \qquad h(0) = h(9) = 0 . \end{equation*}

As \(t\) runs over \((0,9)\), the quantity \(\pi(t/9 - 1/2)\) runs over \(\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)\), on which \(\tan\) is a bijection onto \(\mathbb{R}\). Hence \(h\) maps \([0,9]\) onto \(\mathbb{R}\). Set

\begin{equation*} f = h \circ \varphi : \mathbb{R} \to \mathbb{R} . \end{equation*}

We claim \(f(I) = \mathbb{R}\) for every nonempty open interval \(I\). Because \(f(I) \subseteq \mathbb{R}\) and \(h([0,9]) = \mathbb{R}\), it suffices to prove that \(\varphi(I) = [0,9]\).

Step 1: every interval contains a full decimal cylinder. Let \(I\) be a nonempty open interval. Every nonempty open interval contains a nonempty open subinterval lying strictly between two consecutive integers, so we may choose an integer \(m\) and real numbers \(u, v\) with

\begin{equation*} m < u < v < m+1 \qquad \text{and} \qquad (u,v) \subseteq I ; \end{equation*}

then \(\lfloor x \rfloor = m\), and hence \(\langle x \rangle = x - m\), for every \(x \in (u,v)\). Choose \(k \in \mathbb{Z}^{+}\) with \(10^{-k} < (v-u)/2\) and set

\begin{equation*} M = \left\lfloor (u-m) 10^{k} \right\rfloor + 1 , \end{equation*}

an integer satisfying \(M \ge 1\) and \((u-m) 10^{k} < M \le (u-m) 10^{k} + 1\). Dividing by \(10^{k}\) turns these into

\begin{equation*} u - m < \frac{M}{10^{k}} \quad\text{and}\quad \frac{M+1}{10^{k}} \le (u-m) + \frac{2}{10^{k}} < (u-m) + (v-u) = v - m . \end{equation*}

Note the first inequality is strict, which is what will place our points in the open interval. Also \((M+1) 10^{-k} < v - m < 1\), so \(M + 1 < 10^{k}\) and therefore \(1 \le M < 10^{k}\). Let \(c_1 c_2 \cdots c_k\) be the digits of \(M\) written with exactly \(k\) digits, allowing leading zeros, so that \(M 10^{-k} = 0.c_1 c_2 \cdots c_k\). Now let \(e_1, e_2, \ldots\) be any digit sequence that is not eventually equal to \(9\), and put

\begin{equation*} x = m + \frac{M}{10^{k}} + \frac{1}{10^{k}} \sum_{i=1}^{\infty} \frac{e_i}{10^{i}} . \end{equation*}

Because \(\sum_{i \ge 1} e_i 10^{-i} \in [0,1]\), the number \(\langle x \rangle = x - m\) lies in \(\left[ M 10^{-k}, (M+1) 10^{-k} \right]\), and this closed interval is contained in \((u-m, v-m)\) precisely because the two inequalities displayed above are \(u - m < M 10^{-k}\) and \((M+1) 10^{-k} < v-m\). Hence \(x \in (u,v) \subseteq I\). Moreover the canonical decimal digits of \(\langle x \rangle\) are exactly \(c_1, \ldots, c_k, e_1, e_2, \ldots\): this string is a decimal representation of \(\langle x \rangle\), and it is not eventually \(9\) (because \((e_i)\) is not), so it is the canonical one.

A fixed finite prefix does not affect the limit superior of the averages, so

\begin{equation*} \varphi(x) = \limsup_{n \to \infty} \frac{e_1 + \cdots + e_n}{n} . \end{equation*}

Indeed, write \(K = c_1 + \cdots + c_k\) and \(S_j = e_1 + \cdots + e_j\), so that for \(n > k\) the average of the first \(n\) digits of \(\langle x \rangle\) equals \((K + S_{n-k})/n\). Substituting \(j = n-k\) and subtracting,

\begin{equation*} \frac{K + S_j}{j+k} - \frac{S_j}{j} = \frac{K}{j+k} - \frac{k\, S_j}{j (j+k)} , \end{equation*}

and since \(0 \le S_j \le 9j\) the right side has absolute value at most \(\frac{K}{j+k} + \frac{9k}{j+k}\), which tends to \(0\) as \(j \to \infty\). Two sequences whose difference tends to \(0\) have the same limit superior, so \(\varphi(x) = \limsup_{j \to \infty} S_j / j\), as claimed.

Step 2: the tail averages realize every value in \([0,9]\). Fix \(t \in [0,9]\). We produce a digit sequence \((e_i)\), not eventually equal to \(9\), with \(\limsup_n (e_1 + \cdots + e_n)/n = t\).

If \(t = 9\), let \(e_i = 0\) when \(i\) is a power of \(2\) and \(e_i = 9\) otherwise. The number of powers of \(2\) in \(\{1, \ldots, n\}\) is at most \(1 + \log_2 n\), so

\begin{equation*} 9 \ge \frac{e_1 + \cdots + e_n}{n} \ge \frac{9n - 9(1 + \log_2 n)}{n} \longrightarrow 9 , \end{equation*}

whence the limit, and so the limit superior, is \(9\). This sequence has infinitely many \(0\)s, so it is not eventually \(9\).

If \(t \in [0,9)\), define \((e_i)\) greedily: having chosen \(e_1, \ldots, e_{n-1}\), with \(S_{n-1} = e_1 + \cdots + e_{n-1}\) (and \(S_0 = 0\)), put

\begin{equation*} e_n = \begin{cases} 9 & \text{if } \dfrac{S_{n-1} + 9}{n} \le t, \\[4pt] 0 & \text{otherwise.} \end{cases} \end{equation*}

Write \(A_n = S_n / n\). We check \(A_n \le t\) for all \(n \ge 1\) by induction. If \(e_n = 9\) then \(A_n = (S_{n-1}+9)/n \le t\) by the defining rule. If \(e_n = 0\) then \(A_n = S_{n-1}/n \le S_{n-1}/(n-1) = A_{n-1} \le t\) for \(n \ge 2\), while for \(n = 1\) we get \(A_1 = 0 \le t\). So \(A_n \le t\) always, and therefore \(\limsup_n A_n \le t\).

Next, \(e_n = 0\) for infinitely many \(n\): otherwise \(e_n = 9\) for all \(n\) beyond some point, forcing \(A_n \to 9\), contradicting \(A_n \le t < 9\). Whenever \(e_n = 0\) we have \((S_{n-1} + 9)/n > t\), hence

\begin{equation*} A_n = \frac{S_{n-1}}{n} > t - \frac{9}{n} . \end{equation*}

Letting \(n \to \infty\) through the infinitely many indices with \(e_n = 0\) gives \(\limsup_n A_n \ge t\). Therefore \(\limsup_n A_n = t\). Finally, since \(e_n = 0\) infinitely often, \((e_n)\) is not eventually equal to \(9\).

Conclusion. Combining the two steps: given a nonempty open interval \(I\) and any \(t \in [0,9]\), Step 2 supplies a legal digit tail \((e_i)\) with tail average limsup equal to \(t\), and Step 1 converts it into a point \(x \in I\) with \(\varphi(x) = t\). Thus \(\varphi(I) = [0,9]\) and hence

\begin{equation*} f(I) = h(\varphi(I)) = h([0,9]) = \mathbb{R} \end{equation*}

for every nonempty open interval \(I\), as required.

Two remarks. First, such an \(f\) is necessarily discontinuous at every point, since it is unbounded on every interval. Second, and more relevant to this book, the function \(f\) constructed above is nevertheless Borel measurable: each digit function \(d_j\) is constant on each interval of a countable partition of \(\mathbb{R}\) into intervals, hence is Borel measurable; therefore each average \((d_1 + \cdots + d_n)/n\) is Borel measurable, and \(\varphi\), being the limit superior of a sequence of Borel measurable functions, is Borel measurable, since \(\limsup\) is an infimum of suprema and both operations preserve measurability by 2.53. Since \(h\) is Borel measurable (it is continuous on \((0,9)\) and constant on the two-point set \(\{0,9\}\)), the composition \(f = h \circ \varphi\) is Borel measurable by 2.44. So wild behavior of this kind is entirely compatible with measurability.

Write \(\mu_{*}(A)\) for the inner measure of \(A\) and, as in the book, \(|A|\) for the outer measure of \(A\). Note that the empty set is a closed bounded subset of every \(A\), so the supremum is over a nonempty collection and \(\mu_{*}(A) \in [0, \infty]\) is always defined. Note also that every closed subset of \(\mathbb{R}\) is a Borel set and hence Lebesgue measurable, so \(|F|\) in the definition is the Lebesgue measure of \(F\).

(a) Let \(A \subseteq \mathbb{R}\) be Lebesgue measurable.

The inequality \(\mu_{*}(A) \le |A|\). This holds for every set \(A\), measurable or not: if \(F \subseteq A\) then \(|F| \le |A|\) because outer measure preserves order (2.5). Taking the supremum over all closed bounded \(F \subseteq A\) gives \(\mu_{*}(A) \le |A|\).

The inequality \(\mu_{*}(A) \ge |A|\). Let \(\varepsilon > 0\). Because \(A\) is Lebesgue measurable, 2.71 (the implication (a) \(\Rightarrow\) (b)) provides a closed set \(F \subseteq A\) with

\begin{equation*} |A \setminus F| < \varepsilon . \end{equation*}

The set \(F\) need not be bounded, so we truncate it. For \(n \in \mathbb{Z}^{+}\) put

\begin{equation*} F_n = F \cap [-n, n] . \end{equation*}

Each \(F_n\) is closed (an intersection of two closed sets) and bounded, and \(F_n \subseteq F \subseteq A\); hence \(|F_n| \le \mu_{*}(A)\) for every \(n\). The sets \(F_1 \subseteq F_2 \subseteq \cdots\) are Lebesgue measurable with union \(F\), so by 2.59 (the measure of an increasing union, applied to Lebesgue measure, which is a measure on the \(\sigma\)-algebra of Lebesgue measurable sets by 2.72)

\begin{equation*} \lim_{n \to \infty} |F_n| = |F| . \end{equation*}

Consequently \(\sup_n |F_n| = |F|\) and therefore

\begin{equation*} \mu_{*}(A) \ge |F| . \end{equation*}

Now \(A = F \cup (A \setminus F)\), so subadditivity of outer measure (2.8) gives

\begin{equation*} |A| \le |F| + |A \setminus F| < |F| + \varepsilon . \end{equation*}

If \(|A| = \infty\), this forces \(|F| = \infty\) and hence \(\mu_{*}(A) = \infty = |A|\), and we are done. If \(|A| < \infty\), we obtain

\begin{equation*} \mu_{*}(A) \ge |F| > |A| - \varepsilon . \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, \(\mu_{*}(A) \ge |A|\).

Combining the two inequalities, \(\mu_{*}(A) = |A|\) whenever \(A\) is Lebesgue measurable. In particular \(\mu_{*}([0,1)) = |[0,1)| = 1\), a fact used below.

(b) Inner measure does satisfy \(\mu_{*}(\varnothing) = 0\), so to show that it is not a measure on the \(\sigma\)-algebra of all subsets of \(\mathbb{R}\) we must exhibit a sequence of disjoint sets whose inner measures do not add up to the inner measure of their union. We use the Vitali-type construction that appears in the discussion following 2.72.

Two preliminary observations. First, \(\mu_{*}\) is translation invariant: if \(t \in \mathbb{R}\) and \(A \subseteq \mathbb{R}\), then \(F \mapsto t + F\) is a bijection from the closed bounded subsets of \(A\) onto the closed bounded subsets of \(t + A\), and \(|t + F| = |F|\) by 2.7; hence \(\mu_{*}(t + A) = \mu_{*}(A)\). Second, \(\mu_{*}\) is monotone: \(A \subseteq B\) implies \(\mu_{*}(A) \le \mu_{*}(B)\), since every closed bounded subset of \(A\) is one of \(B\).

Rotation of \([0,1)\). For \(r \in [0,1)\) and \(S \subseteq [0,1)\) define

\begin{equation*} S \oplus r = \bigl( (S \cap [0, 1-r)) + r \bigr) \cup \bigl( (S \cap [1-r, 1)) + r - 1 \bigr) \subseteq [0,1) , \end{equation*}

so \(x \mapsto x \oplus r\) (addition modulo \(1\)) is a bijection of \([0,1)\) onto itself. Note that if \(S\) is Lebesgue measurable then so is \(S \oplus r\), and, since the two pieces above are disjoint, additivity of Lebesgue measure (2.72) together with translation invariance of outer measure (2.7) gives

\begin{equation*} |S \oplus r| = |S \cap [0,1-r)| + |S \cap [1-r,1)| = |S| . \end{equation*}

The Vitali set. Define an equivalence relation on \([0,1)\) by \(x \sim y\) if and only if \(x - y \in \mathbb{Q}\). By the axiom of choice, let \(V \subseteq [0,1)\) contain exactly one element of each equivalence class. Let \(r_1, r_2, r_3, \ldots\) be an enumeration of the countably infinite set \(\mathbb{Q} \cap [0,1)\), and set

\begin{equation*} V_k = V \oplus r_k \qquad (k \in \mathbb{Z}^{+}) . \end{equation*}

The sets \(V_1, V_2, \ldots\) are disjoint with union \([0,1)\). Suppose \(x \in V_j \cap V_k\). Then \(x = v \oplus r_j = w \oplus r_k\) for some \(v, w \in V\); since \(x - (v + r_j) \in \{0, -1\}\) and likewise \(x - (w + r_k) \in \{0, -1\}\), we get \(v - w \in \mathbb{Q}\), so \(v \sim w\) and therefore \(v = w\) by the choice of \(V\). Then \(v \oplus r_j = v \oplus r_k\), and since \(r \mapsto v \oplus r\) is injective on \([0,1)\) we get \(r_j = r_k\), i.e. \(j = k\). For the union, let \(x \in [0,1)\) and let \(v \in V\) be the representative of the class of \(x\). Then \(x - v \in \mathbb{Q} \cap (-1, 1)\); put \(r = x - v\) if \(x \ge v\) and \(r = x - v + 1\) otherwise, so \(r \in \mathbb{Q} \cap [0,1)\) and \(x = v \oplus r\). Thus \(x \in V_k\) for the index \(k\) with \(r_k = r\).

Each \(V_k\) has inner measure \(0\). Since rotation by \(r_k\) maps each equivalence class onto itself (because \(x \oplus r_k \sim x\) for rational \(r_k\)) and is injective, the set \(V_k\), like \(V\), contains exactly one element of each equivalence class. So it suffices to prove the following: if \(S \subseteq [0,1)\) contains at most one element of each equivalence class, then \(\mu_{*}(S) = 0\).

Let \(F \subseteq S\) be closed and bounded. Then \(F\) is a Borel set, hence Lebesgue measurable. The sets \(F \oplus r\), for \(r \in \mathbb{Q} \cap [0,1)\), are pairwise disjoint: if \(x \oplus r = y \oplus r^{\prime}\) with \(x, y \in F\) and \(r \ne r^{\prime}\), then \(x - y \in \mathbb{Q}\), so \(x \sim y\); also \(x \ne y\), since \(x = y\) would give \(x \oplus r = x \oplus r^{\prime}\) and hence \(r = r^{\prime}\); but \(x, y \in F \subseteq S\) and \(S\) contains at most one element of each class, a contradiction. Each \(F \oplus r\) is a Lebesgue measurable subset of \([0,1)\) with \(|F \oplus r| = |F|\), by the computation above. Therefore, by countable additivity of Lebesgue measure (2.72) and monotonicity (2.5),

\begin{equation*} \sum_{r \in \mathbb{Q} \cap [0,1)} |F| = \sum_{r \in \mathbb{Q} \cap [0,1)} |F \oplus r| = \Bigl| \bigcup_{r \in \mathbb{Q} \cap [0,1)} (F \oplus r) \Bigr| \le |[0,1)| = 1 . \end{equation*}

The left side is a sum of countably infinitely many copies of the same number \(|F|\), so it equals \(\infty\) unless \(|F| = 0\). Hence \(|F| = 0\). As \(F\) was an arbitrary closed bounded subset of \(S\), we conclude \(\mu_{*}(S) = 0\), and in particular

\begin{equation*} \mu_{*}(V_k) = 0 \qquad \text{for every } k \in \mathbb{Z}^{+} . \end{equation*}

Conclusion. The sets \(V_1, V_2, \ldots\) are disjoint subsets of \(\mathbb{R}\), all belonging to the \(\sigma\)-algebra of all subsets of \(\mathbb{R}\), and their union is \([0,1)\). By part (a),

\begin{equation*} \mu_{*}\Bigl( \bigcup_{k=1}^{\infty} V_k \Bigr) = \mu_{*}\bigl( [0,1) \bigr) = \bigl| [0,1) \bigr| = 1 , \end{equation*}

whereas

\begin{equation*} \sum_{k=1}^{\infty} \mu_{*}(V_k) = \sum_{k=1}^{\infty} 0 = 0 . \end{equation*}

Thus \(\mu_{*}\) fails to be countably additive, so inner measure is not a measure on the \(\sigma\)-algebra of all subsets of \(\mathbb{R}\).

Contrast this with part (a): restricted to the \(\sigma\)-algebra of Lebesgue measurable sets, \(\mu_{*}\) coincides with Lebesgue measure and hence is a measure there. The failure in (b) is caused precisely by the nonmeasurable sets, which can have positive outer measure while containing no closed bounded set of positive measure.

If \(X = \emptyset\) the assertion is vacuous (every condition of the form “for all \(x \in X\)” holds), so assume \(X = \{x_1, \dots, x_N\}\) with \(N \in \mathbb{Z}^+\).

Suppose \(f_1, f_2, \dots\) is a sequence of functions from \(X\) to \(\mathbb{R}\) converging pointwise on \(X\) to \(f : X \to \mathbb{R}\). Let \(\varepsilon > 0\). By the definition of pointwise convergence (2.82), for each \(j \in \{1, \dots, N\}\) there exists \(n_j \in \mathbb{Z}^+\) such that

\begin{equation*} |f_k(x_j) - f(x_j)| < \varepsilon \quad \text{for all integers } k \ge n_j . \end{equation*}

Because \(\{n_1, \dots, n_N\}\) is a finite set of positive integers, it has a maximum. Let

\begin{equation*} n = \max\{n_1, \dots, n_N\} \in \mathbb{Z}^+ . \end{equation*}

Now suppose \(k \ge n\) is an integer and \(x \in X\). Then \(x = x_j\) for some \(j \in \{1, \dots, N\}\), and \(k \ge n \ge n_j\), so \(|f_k(x) - f(x)| = |f_k(x_j) - f(x_j)| < \varepsilon\).

Thus for every \(\varepsilon > 0\) we have produced a single \(n \in \mathbb{Z}^+\) that works simultaneously for all \(x \in X\). By 2.82, \(f_1, f_2, \dots\) converges uniformly on \(X\) to \(f\).

The point is that pointwise convergence differs from uniform convergence only in the order of the quantifiers: pointwise convergence supplies a threshold \(n\) depending on both \(\varepsilon\) and \(x\), while uniform convergence demands a threshold depending only on \(\varepsilon\). When \(X\) is finite, the supremum over \(x \in X\) of these thresholds is attained and is itself a positive integer, so the two notions coincide. (The argument fails for infinite \(X\) because a supremum of infinitely many positive integers may be \(\infty\); see Exercise 2 in this section.)

For each \(k \in \mathbb{Z}^+\) define \(f_k : \mathbb{Z}^+ \to \mathbb{R}\) by

\begin{equation*} f_k(m) = \begin{cases} 1 & \text{if } m = k, \\ 0 & \text{if } m \neq k, \end{cases} \end{equation*}

in other words \(f_k = \chi_{\{k\}}\).

Pointwise convergence: fix \(m \in \mathbb{Z}^+\). For every integer \(k > m\) we have \(k \neq m\) and hence \(f_k(m) = 0\). Thus \(\lim_{k \to \infty} f_k(m) = 0\). Since \(m\) was arbitrary, \(f_1, f_2, \dots\) converges pointwise on \(\mathbb{Z}^+\) to the function \(f\) that is identically \(0\).

Failure of uniform convergence: if \(f_1, f_2, \dots\) converged uniformly on \(\mathbb{Z}^+\), the uniform limit would have to agree with the pointwise limit \(f = 0\) (uniform convergence implies pointwise convergence). Take \(\varepsilon = \tfrac12\). For every \(n \in \mathbb{Z}^+\) the integer \(k = n\) satisfies \(k \ge n\), and the point \(x = n \in \mathbb{Z}^+\) satisfies

\begin{equation*} |f_k(x) - f(x)| = |f_n(n) - 0| = 1 > \tfrac12 . \end{equation*}

So no \(n \in \mathbb{Z}^+\) has the property required in 2.82 for \(\varepsilon = \tfrac12\). Hence \(f_1, f_2, \dots\) does not converge uniformly on \(\mathbb{Z}^+\).

Equivalently, \(\sup_{m \in \mathbb{Z}^+} |f_k(m) - f(m)| = 1\) for every \(k\), so the sequence cannot converge uniformly to \(f\).

Note how this example is exactly the failure of the argument in Exercise 1: for the point \(m\), the threshold needed is \(n_m = m + 1\), and \(\sup_m n_m = \infty\).

For each \(k \in \mathbb{Z}^+\) define \(f_k : [0,1] \to \mathbb{R}\) by

\begin{equation*} f_k(x) = \begin{cases} k^2 x & \text{if } 0 \le x \le \frac{1}{k}, \\ \dfrac{1}{x} & \text{if } \frac{1}{k} < x \le 1. \end{cases} \end{equation*}

Thus the graph of \(f_k\) rises linearly from \((0,0)\) to \(\bigl(\frac1k, k\bigr)\) and then follows the hyperbola \(y = 1/x\) out to \((1,1)\).

Each \(f_k\) is continuous: it is continuous on \([0, \frac1k]\) and on \([\frac1k, 1]\) (on the latter it is \(1/x\), continuous since \(x \ge \frac1k > 0\)), and the two formulas agree at the junction, since \(k^2 \cdot \frac1k = k = \frac{1}{1/k}\). A function that is continuous on each of two closed intervals whose union is \([0,1]\) and which agree at the shared endpoint is continuous on \([0,1]\).

Define \(f : [0,1] \to \mathbb{R}\) by

\begin{equation*} f(x) = \begin{cases} 0 & \text{if } x = 0, \\ \dfrac{1}{x} & \text{if } 0 < x \le 1. \end{cases} \end{equation*}

Note that \(f\) takes values in \(\mathbb{R}\) (it is real-valued at every point of \([0,1]\), including \(0\)).

The sequence converges pointwise to \(f\): if \(x = 0\), then \(f_k(0) = k^2 \cdot 0 = 0 = f(0)\) for every \(k\), so \(\lim_{k \to \infty} f_k(0) = f(0)\). If \(0 < x \le 1\), choose \(n \in \mathbb{Z}^+\) with \(n > 1/x\); then for every integer \(k \ge n\) we have \(\frac1k \le \frac1n < x\), hence \(f_k(x) = \frac1x = f(x)\). Thus \(\lim_{k \to \infty} f_k(x) = f(x)\) (the sequence is eventually constant at each such \(x\)).

Finally, \(f\) is not bounded: for each \(k \in \mathbb{Z}^+\) the point \(x = \frac1k \in [0,1]\) satisfies \(f(x) = k\), so \(\sup_{x \in [0,1]} |f(x)| = \infty\).

This also gives another example for Exercise 2’s phenomenon: the convergence here is certainly not uniform, since a uniform limit of bounded functions is bounded, while each \(f_k\) is bounded (by \(k\)) and \(f\) is not.

The statement is true. This is the uniform-continuity analogue of 2.84 (which says that a uniform limit of functions continuous at a point is continuous at that point), and it is proved by the same \(\varepsilon/3\) device, with the key observation that the \(\delta\) produced for \(f_n\) does not depend on the point.

Recall that \(g : A \to \mathbb{R}\) is uniformly continuous on \(A\) if for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|g(x) - g(y)| < \varepsilon\) for all \(x, y \in A\) with \(|x - y| < \delta\).

Suppose \(\varepsilon > 0\). Because \(f_1, f_2, \dots\) converges uniformly on \(A\) to \(f\), there exists \(n \in \mathbb{Z}^+\) such that

\begin{equation*} |f_n(z) - f(z)| < \frac{\varepsilon}{3} \quad \text{for all } z \in A . \end{equation*}

Because \(f_n\) is uniformly continuous on \(A\), there exists \(\delta > 0\) such that

\begin{equation*} |f_n(x) - f_n(y)| < \frac{\varepsilon}{3} \quad \text{for all } x, y \in A \text{ with } |x - y| < \delta . \end{equation*}

Now suppose \(x, y \in A\) and \(|x - y| < \delta\). Then

\begin{equation*} \begin{aligned} |f(x) - f(y)| &\le |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)| \\ &< \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} \\ &= \varepsilon . \end{aligned} \end{equation*}

Thus for every \(\varepsilon > 0\) we have produced \(\delta > 0\) that works for all pairs of points of \(A\) simultaneously. Hence \(f\) is uniformly continuous on \(A\).

Two remarks on where the hypotheses are used. Uniform convergence is essential: the sequence of Exercise 3 above consists of functions each of which is uniformly continuous on \([0,1]\) (each \(f_k\) is continuous on the compact interval \([0,1]\), hence uniformly continuous there), yet the pointwise limit \(f\) is not even bounded, so certainly not uniformly continuous. And note that nothing about \(A\) was needed: the argument works verbatim for functions on any metric space, since the only property of \(A\) used is that \(|x-y|\) makes sense.

Let \(X = \mathbb{R}\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}\), and let \(\mu\) be Lebesgue measure on \(\mathcal{S}\) (so \(\mu(E) = |E|\), the outer measure of \(E\)). Then \(\mu(X) = \infty\).

For each \(k \in \mathbb{Z}^+\) define \(f_k : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f_k = \chi_{[k,\, k+1]} . \end{equation*}

Each \(f_k\) is \(\mathcal{S}\)-measurable, since \([k, k+1]\) is a Borel set (the characteristic function of a set \(E\) is \(\mathcal{S}\)-measurable if and only if \(E \in \mathcal{S}\)).

The sequence converges pointwise on \(\mathbb{R}\) to the function \(f\) that is identically \(0\): given \(x \in \mathbb{R}\), every integer \(k > x\) satisfies \(x \notin [k, k+1]\), so \(f_k(x) = 0\) for all such \(k\).

Now take \(\varepsilon = 1\) and suppose \(E \in \mathcal{S}\) satisfies \(\mu(\mathbb{R} \setminus E) < 1\). Fix any \(k \in \mathbb{Z}^+\). Since \([k, k+1] \setminus E \subseteq \mathbb{R} \setminus E\), monotonicity of outer measure gives

\begin{equation*} \bigl| [k, k+1] \setminus E \bigr| \le \bigl| \mathbb{R} \setminus E \bigr| < 1 = \bigl| [k, k+1] \bigr| . \end{equation*}

Hence \([k, k+1] \setminus E \neq [k, k+1]\), which means \([k, k+1] \cap E \neq \emptyset\). Choose \(x_k \in [k, k+1] \cap E\). Then

\begin{equation*} |f_k(x_k) - f(x_k)| = 1 . \end{equation*}

Consequently, for the value \(\varepsilon^{\prime} = \tfrac12\) in the definition of uniform convergence there is no \(n \in \mathbb{Z}^+\) that works: whatever \(n\) is chosen, the integer \(k = n\) and the point \(x_n \in E\) give \(|f_n(x_n) - f(x_n)| = 1 > \tfrac12\). So \(f_1, f_2, \dots\) does not converge uniformly to \(f\) on \(E\).

Since \(E\) was an arbitrary member of \(\mathcal{S}\) with \(\mu(\mathbb{R} \setminus E) < 1\), there is no set \(E \in \mathcal{S}\) with \(\mu(X \setminus E) < \varepsilon\) on which the convergence is uniform. Thus the conclusion of Egorov’s Theorem (2.85) fails for \(\varepsilon = 1\), even though every other hypothesis of 2.85 holds. The hypothesis \(\mu(X) < \infty\) therefore cannot be dropped.

(It is instructive to see exactly where the proof of 2.85 breaks. There 2.59 supplies \(\lim_{m \to \infty} \mu(A_{m,n}) = \mu(X)\), and finiteness of \(\mu(X)\) is what allows this to be turned into the choice of \(m_n\) with \(\mu(X) - \mu(A_{m_n,n}) < \varepsilon/2^n\), and then into \(\mu(X \setminus A_{m_n,n}) < \varepsilon/2^n\) by 2.57(b). Both steps are meaningless when \(\mu(X) = \infty\). Indeed, in the example above

\begin{equation*} A_{m,n} = \bigcap_{k \ge m} \bigl\{ x \in \mathbb{R} : |f_k(x)| < \tfrac1n \bigr\} = \mathbb{R} \setminus \bigcup_{k \ge m} [k, k+1] = (-\infty, m), \end{equation*}

so \(\mu(A_{m,n}) = \infty = \mu(X)\) for every \(m\) and \(n\), while \(\mu(\mathbb{R} \setminus A_{m,n}) = \mu([m,\infty)) = \infty\) for every \(m\) and \(n\). The sets that the proof would select are never of small complementary measure.)

The proof follows the proof of Egorov’s Theorem (2.85), with the sets \(\{x : |f_k(x) - f(x)| < \frac1n\}\) replaced by the sets \(\{x : f_k(x) > n\}\).

Suppose \(\varepsilon > 0\). Temporarily fix \(n \in \mathbb{Z}^+\). For \(m \in \mathbb{Z}^+\) let

\begin{equation*} A_{m,n} = \bigcap_{k = m}^{\infty} \{ x \in X : f_k(x) > n \} . \end{equation*}

Each set \(\{x \in X : f_k(x) > n\}\) equals \(f_k^{-1}\bigl((n, \infty)\bigr)\), which belongs to \(\mathcal{S}\) because \(f_k\) is \(\mathcal{S}\)-measurable and \((n, \infty)\) is a Borel subset of \(\mathbb{R}\). Hence each \(A_{m,n} \in \mathcal{S}\), being a countable intersection of elements of the \(\sigma\)-algebra \(\mathcal{S}\).

Clearly \(A_{1,n} \subseteq A_{2,n} \subseteq \cdots\), because intersecting over fewer indices gives a larger set. Moreover

\begin{equation*} \bigcup_{m = 1}^{\infty} A_{m,n} = X . \end{equation*}

Indeed, the inclusion \(\subseteq\) is trivial; for the reverse, let \(x \in X\). Since \(\lim_{k \to \infty} f_k(x) = \infty\), there exists \(m \in \mathbb{Z}^+\) such that \(f_k(x) > n\) for all integers \(k \ge m\), which says exactly that \(x \in A_{m,n}\).

By 2.59 (measure of an increasing union), \(\lim_{m \to \infty} \mu(A_{m,n}) = \mu(X)\). Since \(\mu(X) < \infty\), there exists \(m_n \in \mathbb{Z}^+\) such that

\begin{equation*} \mu(X) - \mu(A_{m_n, n}) < \frac{\varepsilon}{2^n} . \end{equation*}

Because \(A_{m_n,n} \subseteq X\) and \(\mu(X) < \infty\), 2.57(b) gives \(\mu(X \setminus A_{m_n,n}) = \mu(X) - \mu(A_{m_n,n}) < \varepsilon/2^n\) (the subtraction is legitimate precisely because \(\mu(A_{m_n,n}) \le \mu(X) < \infty\)).

Now let

\begin{equation*} E = \bigcap_{n = 1}^{\infty} A_{m_n, n} . \end{equation*}

Then \(E \in \mathcal{S}\), and by countable subadditivity (2.58),

\begin{equation*} \mu(X \setminus E) = \mu\Bigl( \bigcup_{n=1}^{\infty} (X \setminus A_{m_n, n}) \Bigr) \le \sum_{n=1}^{\infty} \mu(X \setminus A_{m_n, n}) < \sum_{n=1}^{\infty} \frac{\varepsilon}{2^n} = \varepsilon . \end{equation*}

It remains to verify that \(f_1, f_2, \dots\) converges uniformly to \(\infty\) on \(E\). Suppose \(t > 0\). Choose \(n \in \mathbb{Z}^+\) with \(n \ge t\), and set \(n_0 = m_n\). Because \(E \subseteq A_{m_n, n}\), every \(x \in E\) lies in \(A_{m_n,n}\), which means

\begin{equation*} f_k(x) > n \ge t \quad \text{for all integers } k \ge m_n = n_0 \text{ and all } x \in E . \end{equation*}

Thus for every \(t > 0\) there exists \(n_0 \in \mathbb{Z}^+\) with \(f_k(x) > t\) for all integers \(k \ge n_0\) and all \(x \in E\), which is exactly the assertion that \(f_1, f_2, \dots\) converges uniformly to \(\infty\) on \(E\).

As with Egorov’s Theorem itself, the hypothesis \(\mu(X) < \infty\) cannot be omitted. Let \(X = \mathbb{R}\) with \(\mathcal{S}\) the Borel sets and \(\mu\) Lebesgue measure, and for \(k \in \mathbb{Z}^+\) define the continuous (hence Borel measurable) function

\begin{equation*} f_k(x) = \max\{ k - |x|,\, 0 \} . \end{equation*}

For each fixed \(x \in \mathbb{R}\) we have \(f_k(x) = k - |x|\) once \(k > |x|\), so \(\lim_{k \to \infty} f_k(x) = \infty\). Now take \(\varepsilon = 1\) and let \(E \in \mathcal{S}\) satisfy \(\mu(\mathbb{R} \setminus E) < 1\). Exactly as in Exercise 5, for each \(k \in \mathbb{Z}^+\) the inequality \(\bigl| [k, k+1] \setminus E \bigr| \le \bigl| \mathbb{R} \setminus E \bigr| < 1 = \bigl| [k, k+1] \bigr|\) forces \(E \cap [k, k+1] \neq \emptyset\); choose \(x_k\) in that intersection. Since \(|x_k| \ge k\), we get \(f_k(x_k) = 0\). Hence for \(t = 1\) there is no \(n \in \mathbb{Z}^+\) with \(f_k(x) > 1\) for all \(k \ge n\) and all \(x \in E\), because \(k = n\) and \(x = x_n \in E\) give \(f_n(x_n) = 0\). So \(f_1, f_2, \dots\) converges uniformly to \(\infty\) on no such \(E\), and the conclusion fails for \(\varepsilon = 1\).

First note that \(g\) is well defined and real-valued: for each \(x \in F\) the sequence \(g_1(x) \le g_2(x) \le \cdots\) is increasing and bounded above (its supremum is finite by hypothesis), so it converges to

\begin{equation*} g(x) = \sup\{g_1(x), g_2(x), \dots\} \in \mathbb{R} . \end{equation*}

In particular \(g_k(x) \le g(x)\) for every \(k \in \mathbb{Z}^+\) and every \(x \in F\).

If \(F = \emptyset\) both statements hold vacuously, so assume \(F \neq \emptyset\).

One direction requires neither monotonicity nor any hypothesis on \(F\). Suppose \(g_1, g_2, \dots\) converges uniformly on \(F\) to \(g\). Each \(g_k\) is continuous at each point \(b \in F\), so 2.84 (uniform limit of continuous functions is continuous) shows that \(g\) is continuous at each \(b \in F\). Hence \(g\) is continuous on \(F\).

For the other direction, suppose \(g\) is continuous on \(F\). For each \(k \in \mathbb{Z}^+\) define

\begin{equation*} h_k = g - g_k . \end{equation*}

Then each \(h_k\) is a continuous real-valued function on \(F\) (a difference of continuous functions), and

• \(h_k(x) \ge 0\) for all \(x \in F\), because \(g_k(x) \le g(x)\);
• \(h_1(x) \ge h_2(x) \ge \cdots\) for all \(x \in F\), because \(g_k(x) \le g_{k+1}(x)\);
• \(\lim_{k \to \infty} h_k(x) = 0\) for each \(x \in F\), by the definition of \(g\).

Proving that \(g_1, g_2, \dots\) converges uniformly to \(g\) on \(F\) is the same as proving that for every \(\varepsilon > 0\) there exists \(n \in \mathbb{Z}^+\) with \(h_k(x) < \varepsilon\) for all integers \(k \ge n\) and all \(x \in F\); and by the monotonicity of \((h_k)\) in \(k\) it suffices to find, for each \(\varepsilon > 0\), a single \(n\) with \(h_n(x) < \varepsilon\) for all \(x \in F\), since then \(0 \le h_k(x) \le h_n(x) < \varepsilon\) for all \(k \ge n\).

Suppose, seeking a contradiction, that this fails for some \(\varepsilon > 0\). Then for every \(n \in \mathbb{Z}^+\) there exists \(x_n \in F\) with

\begin{equation*} h_n(x_n) \ge \varepsilon . \end{equation*}

The sequence \(x_1, x_2, \dots\) lies in the bounded set \(F\), so by the Bolzano-Weierstrass Theorem it has a convergent subsequence \(x_{n_1}, x_{n_2}, \dots\) with \(n_1 < n_2 < \cdots\), say \(\lim_{j \to \infty} x_{n_j} = x\). Because \(F\) is closed and each \(x_{n_j} \in F\), we have \(x \in F\).

Now fix \(m \in \mathbb{Z}^+\). For every \(j\) large enough that \(n_j \ge m\), monotonicity in the index gives

\begin{equation*} h_m(x_{n_j}) \ge h_{n_j}(x_{n_j}) \ge \varepsilon . \end{equation*}

Since \(h_m\) is continuous at \(x\) and \(x_{n_j} \to x\) with all \(x_{n_j} \in F\), letting \(j \to \infty\) yields

\begin{equation*} h_m(x) = \lim_{j \to \infty} h_m(x_{n_j}) \ge \varepsilon . \end{equation*}

(Continuity of \(h_m\) at \(x\) relative to \(F\) is exactly what licenses passing to the limit here.)

But \(m \in \mathbb{Z}^+\) was arbitrary, so \(h_m(x) \ge \varepsilon\) for every \(m\), contradicting \(\lim_{m \to \infty} h_m(x) = 0\). This contradiction shows that for every \(\varepsilon > 0\) there is an \(n \in \mathbb{Z}^+\) with \(h_n(x) < \varepsilon\) for all \(x \in F\), and hence, as observed above, \(h_k(x) < \varepsilon\) for all \(k \ge n\) and all \(x \in F\). Therefore \(g_1, g_2, \dots\) converges uniformly on \(F\) to \(g\).

Each of the remaining hypotheses is genuinely used in the direction just proved.

Monotonicity of the sequence cannot be dropped. On \(F = [0,1]\) let \(g_k\) be the continuous piecewise-linear function with \(g_k(0) = 0\), \(g_k\bigl(\frac{1}{2k}\bigr) = 1\), \(g_k\bigl(\frac1k\bigr) = 0\), and \(g_k(x) = 0\) for \(x \ge \frac1k\). Then \(g_1, g_2, \dots\) converges pointwise on \([0,1]\) to the continuous function \(g \equiv 0\), yet \(\sup_{x \in [0,1]} |g_k(x) - g(x)| = 1\) for every \(k\), so the convergence is not uniform. This sequence is of course not increasing.

Closedness of \(F\) cannot be dropped. On \(F = (0,1]\), which is bounded but not closed, the functions \(g_k(x) = \min\{kx, 1\}\) are continuous, increasing in \(k\), and bounded above by \(1\); they converge pointwise to the continuous function \(g \equiv 1\). But \(g_k\bigl(\frac{1}{2k}\bigr) = \frac12\), so \(\sup_{x \in (0,1]} |g(x) - g_k(x)| \ge \frac12\) for every \(k\) and the convergence is not uniform. Boundedness cannot be dropped either: on the closed but unbounded set \(F = [0, \infty)\) the functions \(g_k(x) = \max\{1 - \frac{x}{k},\, 0\}\) are continuous, increasing in \(k\), and bounded above by \(1\), and they converge pointwise to \(g \equiv 1\); but \(g_k(x) = 0\) for \(x \ge k\), so \(\sup_{x \ge 0} |g(x) - g_k(x)| = 1\) for every \(k\).

Finally, note that continuity of \(g\) is a genuine hypothesis rather than an automatic consequence of the other assumptions: on \(F = [0,1]\) the functions \(g_k(x) = 1 - x^k\) are continuous, satisfy \(g_1(x) \le g_2(x) \le \cdots\) (since \(x^k\) is nonincreasing in \(k\) for \(x \in [0,1]\)), and have \(\sup\{g_1(x), g_2(x), \dots\} \le 1 < \infty\) for each \(x\); yet their pointwise limit equals \(1\) on \([0,1)\) and \(0\) at \(x = 1\), so \(g\) is not continuous. By the theorem just proved, the convergence is therefore not uniform, as one checks directly from \(g_k\bigl(2^{-1/k}\bigr) = \tfrac12\) while \(g\bigl(2^{-1/k}\bigr) = 1\).

The point is that \(\mu\) concentrates almost all of its mass on a finite set, and pointwise convergence on a finite set is automatically uniform.

Let \(\varepsilon > 0\). Because \(2^n \to \infty\), there exists \(N \in \mathbb{Z}^+\) with \(\frac{1}{2^N} < \varepsilon\). Put

\begin{equation*} E = \{1, 2, \ldots, N\}. \end{equation*}

Then \(\mathbb{Z}^+ \setminus E = \{N+1, N+2, \ldots\}\), and summing the geometric series gives

\begin{equation*} \mu(\mathbb{Z}^+ \setminus E) = \sum_{n=N+1}^{\infty} \frac{1}{2^n} = \frac{1}{2^N} < \varepsilon. \end{equation*}

Note that \(E\) was chosen using only \(\varepsilon\); no sequence of functions was involved.

Now suppose \(f_1, f_2, \ldots\) is any sequence of functions from \(\mathbb{Z}^+\) to \(\mathbb{R}\) that converges pointwise on \(\mathbb{Z}^+\), say to \(f : \mathbb{Z}^+ \to \mathbb{R}\). We show \(f_k \to f\) uniformly on \(E\).

Let \(\delta > 0\). For each \(n \in \{1, \ldots, N\}\), pointwise convergence at \(n\) provides \(M_n \in \mathbb{Z}^+\) such that

\begin{equation*} |f_k(n) - f(n)| < \delta \quad \text{for all } k \geq M_n . \end{equation*}

Let \(M = \max\{M_1, \ldots, M_N\}\), which is a well-defined element of \(\mathbb{Z}^+\) because the maximum is taken over a finite set. If \(k \geq M\) and \(n \in E\), then \(k \geq M_n\) and hence \(|f_k(n) - f(n)| < \delta\). Thus

\begin{equation*} \sup_{n \in E} |f_k(n) - f(n)| \leq \delta \quad \text{for all } k \geq M . \end{equation*}

Because \(\delta > 0\) was arbitrary, \(f_1, f_2, \ldots\) converges uniformly to \(f\) on \(E\).

Hence for each \(\varepsilon > 0\) the single set \(E = \{1, \ldots, N\}\) works simultaneously for every pointwise convergent sequence, as desired.

Write \(F = F_1 \cup \cdots \cup F_n\). Continuity of \(g\) means continuity of \(g\) as a function on its domain \(F\): for each \(x \in F\) and each \(\varepsilon > 0\) there is \(\delta > 0\) such that \(|g(y) - g(x)| < \varepsilon\) for all \(y \in F\) with \(|y - x| < \delta\).

Fix \(x \in F\) and \(\varepsilon > 0\). Because \(F_1, \ldots, F_n\) cover \(F\), there is \(j \in \{1, \ldots, n\}\) with \(x \in F_j\). Let

\begin{equation*} C = \bigcup_{k \neq j} F_k . \end{equation*}

Being a finite union of closed sets, \(C\) is closed. Because the sets \(F_1, \ldots, F_n\) are disjoint and \(x \in F_j\), we have \(x \notin C\). Hence \(\mathbb{R} \setminus C\) is an open set containing \(x\), so there exists \(\delta_1 > 0\) such that

\begin{equation*} (x - \delta_1, x + \delta_1) \cap C = \varnothing . \end{equation*}

(If \(n = 1\) then \(C = \varnothing\) and any \(\delta_1 > 0\) works.)

Because \(g|_{F_j}\) is continuous at \(x\), there exists \(\delta_2 > 0\) such that

\begin{equation*} |g(y) - g(x)| < \varepsilon \quad \text{for all } y \in F_j \text{ with } |y - x| < \delta_2 . \end{equation*}

Let \(\delta = \min\{\delta_1, \delta_2\} > 0\). Suppose \(y \in F\) and \(|y - x| < \delta\). Then \(|y - x| < \delta_1\), so \(y \notin C\); since \(y \in F = F_j \cup C\), this forces \(y \in F_j\). Also \(|y - x| < \delta_2\), so \(|g(y) - g(x)| < \varepsilon\).

Thus \(g\) is continuous at \(x\). As \(x \in F\) was arbitrary, \(g\) is a continuous function.

Remarks on the hypotheses. Disjointness made the argument easy but is not actually needed. Suppose \(F_1, \ldots, F_n\) are closed but possibly overlapping and \(g|_{F_k}\) is continuous for each \(k\) (which forces the restrictions to agree on overlaps, since they are all restrictions of the single function \(g\)). Fix \(x \in F\) and \(\varepsilon > 0\), let \(P = \{k : x \in F_k\}\), which is nonempty, and let \(C = \bigcup_{k \notin P} F_k\), a closed set with \(x \notin C\); choose \(\delta_1 > 0\) with \((x - \delta_1, x + \delta_1) \cap C = \varnothing\). For each \(k \in P\), continuity of \(g|_{F_k}\) at \(x\) gives \(\eta_k > 0\) such that \(|g(y) - g(x)| < \varepsilon\) whenever \(y \in F_k\) and \(|y - x| < \eta_k\). Let \(\delta = \min\bigl(\{\delta_1\} \cup \{\eta_k : k \in P\}\bigr) > 0\), a minimum over a finite set. If \(y \in F\) and \(|y - x| < \delta\), then \(y \notin C\), so \(y \in F_k\) for some \(k \in P\), and \(|y - x| < \eta_k\) gives \(|g(y) - g(x)| < \varepsilon\). So \(g\) is continuous. This overlapping version is what is usually called the pasting lemma for finitely many closed sets. Closedness, however, is essential: with \(F_1 = (-\infty, 0)\) and \(F_2 = [0, \infty)\) the function \(g = \chi_{F_2}\) restricts to a continuous function on each piece yet is not continuous on \(\mathbb{R}\). Finiteness is essential too: let \(q_1, q_2, \ldots\) enumerate \(\mathbb{Q}\) and take \(F_k = \{q_k\}\), a disjoint sequence of closed sets with union \(\mathbb{Q}\). Every restriction \(g|_{F_k}\) is continuous, because a function on a one-point set is trivially continuous. But the function \(g : \mathbb{Q} \to \mathbb{R}\) defined by \(g(q_1) = 1\) and \(g(q) = 0\) for \(q \neq q_1\) is not continuous at \(q_1\): every interval about \(q_1\) contains rationals other than \(q_1\), at which \(g\) takes the value \(0\).

We prove the contrapositive: if \(F\) is not closed, then some continuous function from \(F\) to \(\mathbb{R}\) has no continuous extension to \(\mathbb{R}\).

So suppose \(F\) is not a closed subset of \(\mathbb{R}\). Then \(F\) does not contain all of its limit points, so there exists \(b \in \mathbb{R}\) such that \(b \notin F\) but \(b\) is a limit point of \(F\); equivalently, there is a sequence \(x_1, x_2, \ldots\) of elements of \(F\) with \(\lim_{j \to \infty} x_j = b\) and \(x_j \neq b\) for every \(j\).

Define \(g : F \to \mathbb{R}\) by

\begin{equation*} g(x) = \frac{1}{x - b}. \end{equation*}

This makes sense because \(b \notin F\), so \(x - b \neq 0\) for every \(x \in F\). The function \(x \mapsto \frac{1}{x-b}\) is continuous on \(\mathbb{R} \setminus \{b\}\), and \(F \subseteq \mathbb{R} \setminus \{b\}\); hence \(g\), being the restriction of a continuous function to a subset of its domain, is a continuous function from \(F\) to \(\mathbb{R}\).

Suppose, for contradiction, that there is a continuous function \(h : \mathbb{R} \to \mathbb{R}\) with \(h|_F = g\). Continuity of \(h\) at \(b\) and \(x_j \to b\) give

\begin{equation*} \lim_{j \to \infty} h(x_j) = h(b) \in \mathbb{R}. \end{equation*}

In particular the sequence \(h(x_1), h(x_2), \ldots\) is bounded, being convergent in \(\mathbb{R}\). But \(x_j \in F\), so

\begin{equation*} |h(x_j)| = |g(x_j)| = \frac{1}{|x_j - b|} \to \infty \end{equation*}

as \(j \to \infty\), because \(0 < |x_j - b| \to 0\). Thus \(h(x_1), h(x_2), \ldots\) is unbounded, a contradiction.

Hence \(g\) admits no continuous extension to \(\mathbb{R}\), contradicting the hypothesis on \(F\). Therefore \(F\) is closed.

The converse is also true (though not asked): by the Tietze extension theorem, every continuous real-valued function on a closed subset of \(\mathbb{R}\) extends continuously to \(\mathbb{R}\). For \(F \subseteq \mathbb{R}\) closed this is elementary, since \(\mathbb{R} \setminus F\) is a countable disjoint union of open intervals and one simply extends affinely across each bounded gap and constantly across each unbounded one.

The statement is true. Restricting attention to bounded functions costs nothing: the unbounded function \(x \mapsto \frac{1}{x-b}\) used in Exercise 10 can be replaced by a bounded function that oscillates rather than blows up.

We prove the contrapositive: if \(F\) is not closed, then some bounded continuous function from \(F\) to \(\mathbb{R}\) has no continuous extension to \(\mathbb{R}\).

Suppose \(F\) is not closed. As in Exercise 10, there is a point \(b \notin F\) that is a limit point of \(F\); because \(b \notin F\), this means \(F \cap (b - \delta, b + \delta) \neq \varnothing\) for every \(\delta > 0\).

We first fix a side. We claim that either \(F \cap (b, b + \delta) \neq \varnothing\) for every \(\delta > 0\), or \(F \cap (b - \delta, b) \neq \varnothing\) for every \(\delta > 0\). Indeed, if both alternatives failed, there would exist \(\delta_1 > 0\) with \(F \cap (b, b + \delta_1) = \varnothing\) and \(\delta_2 > 0\) with \(F \cap (b - \delta_2, b) = \varnothing\); then \(\delta = \min\{\delta_1, \delta_2\}\) would satisfy \(F \cap (b - \delta, b + \delta) = \varnothing\) (using \(b \notin F\) to discard the point \(b\) itself), contradicting the previous paragraph.

Suppose the second alternative holds. Let \(-F = \{-x : x \in F\}\). The map \(x \mapsto -x\) is a homeomorphism of \(\mathbb{R}\) onto itself carrying \(F\) onto \(-F\) and \(b\) to \(-b\), so \(-b \notin -F\) and \((-F) \cap (-b, -b + \delta) = -\bigl(F \cap (b - \delta, b)\bigr) \neq \varnothing\) for every \(\delta > 0\): the first alternative holds for \(-F\) at \(-b\). Moreover, if \(g\) is a bounded continuous function on \(-F\) admitting no continuous extension to \(\mathbb{R}\), then \(x \mapsto g(-x)\) is a bounded continuous function on \(F\) admitting no continuous extension to \(\mathbb{R}\), since a continuous extension \(h\) of the latter would make \(x \mapsto h(-x)\) a continuous extension of the former. So it suffices to treat the first alternative, and we assume from now on that

\begin{equation*} F \cap (b, b + \delta) \neq \varnothing \quad \text{for every } \delta > 0 . \end{equation*}

Construct recursively a strictly decreasing sequence in \(F\) converging to \(b\): choose \(x_1 \in F \cap (b, b+1)\), and having chosen \(x_j\), choose

\begin{equation*} x_{j+1} \in F \cap \Bigl(b, \ b + \tfrac{x_j - b}{2}\Bigr), \end{equation*}

which is nonempty by the displayed property. Then \(b < x_{j+1} < x_j\) for all \(j\) and \(0 < x_j - b \le 2^{1-j}\), so \(x_j \to b\).

Now set

\begin{equation*} a_j = \begin{cases} 1 & \text{if } j \text{ is odd},\\ 0 & \text{if } j \text{ is even},\end{cases} \end{equation*}

and define \(\varphi : \mathbb{R} \setminus \{b\} \to [0,1]\) by

\begin{equation*} \varphi(x) = \begin{cases} 0 & \text{if } x < b,\\ a_{j+1} + (a_j - a_{j+1}) \dfrac{x - x_{j+1}}{x_j - x_{j+1}} & \text{if } x \in [x_{j+1}, x_j] \text{ for some } j \in \mathbb{Z}^+,\\ 1 & \text{if } x \geq x_1 . \end{cases} \end{equation*}

In words: \(\varphi\) is \(0\) to the left of \(b\), and on \((b, x_1]\) it is the piecewise linear function whose graph connects the points \((x_{j+1}, a_{j+1})\) and \((x_j, a_j)\) by line segments.

This definition is legitimate. First, the intervals \([x_{j+1}, x_j]\), \(j \in \mathbb{Z}^+\), together with \([x_1, \infty)\) cover \((b, \infty)\): given \(x > b\), the set \(\{i \in \mathbb{Z}^+ : x \le x_i\}\) is finite because \(x_i \to b < x\), so either it is empty (and then \(x > x_1\)) or it has a largest element \(j\), in which case \(x_{j+1} < x \le x_j\). Second, consecutive formulas agree where the intervals overlap: at \(x = x_j\) the \(j\)-th formula and the \((j-1)\)-st formula both give \(\varphi(x_j) = a_j\), and at \(x = x_1\) the first formula gives \(a_1 = 1\). So \(\varphi\) is a well-defined function with values in \([0,1]\), and

\begin{equation*} \varphi(x_j) = a_j \quad \text{for every } j \in \mathbb{Z}^+ . \end{equation*}

Also \(\varphi\) is continuous on \(\mathbb{R} \setminus \{b\}\). Indeed, \(\varphi\) is constant on the open set \((-\infty, b)\), hence continuous there. And if \(x > b\), choose \(j\) with \(x_j < x\); on the closed half-line \([x_j, \infty)\) the function \(\varphi\) is built from the finitely many closed pieces \([x_j, x_{j-1}], \ldots, [x_2, x_1], [x_1, \infty)\), on each of which it is affine, with matching values at the shared endpoints, so \(\varphi|_{[x_j, \infty)}\) is continuous (this is the pasting principle for finitely many closed sets discussed in Exercise 9). Since \((x_j, \infty)\) is an open neighborhood of \(x\) on which \(\varphi\) agrees with this continuous function, \(\varphi\) is continuous at \(x\).

Now define \(g = \varphi|_F\). Because \(b \notin F\) we have \(F \subseteq \mathbb{R} \setminus \{b\}\), so \(g\) is a continuous function from \(F\) to \(\mathbb{R}\), and \(0 \le g \le 1\), so \(g\) is bounded.

Suppose \(h : \mathbb{R} \to \mathbb{R}\) were continuous with \(h|_F = g\). Since \(x_j \in F\) and \(x_j \to b\), continuity of \(h\) at \(b\) gives

\begin{equation*} h(b) = \lim_{j \to \infty} h(x_j) = \lim_{j \to \infty} g(x_j) = \lim_{j \to \infty} a_j . \end{equation*}

But \(a_1, a_2, \ldots\) is the sequence \(1, 0, 1, 0, \ldots\), which does not converge. This contradiction shows that \(g\) has no continuous extension to \(\mathbb{R}\).

Therefore, if every bounded continuous function from \(F\) to \(\mathbb{R}\) extends to a continuous function on \(\mathbb{R}\), then \(F\) must be closed.

Luzin’s Theorem says a Borel measurable function is continuous when restricted to the complement of a set of small measure, and one might hope to shrink that exceptional set to measure \(0\). This exercise shows that hope is false.

The obvious first guess, \(f = \chi_{\mathbb{Q}}\), does not work: take \(B = \mathbb{R} \setminus \mathbb{Q}\); then \(|\mathbb{R} \setminus B| = |\mathbb{Q}| = 0\) and \(f|_B\) is identically \(0\), hence continuous. What we need is a Borel set \(A\) that, together with its complement, has positive measure in every interval, so that no set of full measure can separate \(A\) from its complement locally.

Step 1: a closed nowhere dense set of positive measure inside any interval.

Claim: if \(J\) is a nonempty bounded open interval, there is a closed set \(K \subseteq J\) with empty interior and \(|K| > 0\).

Write \(J = (c,d)\) and set \(c^{\prime} = c + \frac{d-c}{4}\), \(d^{\prime} = d - \frac{d-c}{4}\), so \([c^{\prime}, d^{\prime}] \subseteq J\) and \(d^{\prime} - c^{\prime} = \frac{d-c}{2} > 0\). Let \(q_1, q_2, \ldots\) be an enumeration of \(\mathbb{Q}\), let \(\varepsilon = \frac{d^{\prime}-c^{\prime}}{2}\), and let

\begin{equation*} U = \bigcup_{j=1}^{\infty} \Bigl( q_j - \frac{\varepsilon}{2^{j+1}}, \ q_j + \frac{\varepsilon}{2^{j+1}} \Bigr). \end{equation*}

Then \(U\) is open, \(\mathbb{Q} \subseteq U\), and by countable subadditivity of outer measure (2.8) together with 2.14,

\begin{equation*} |U| \leq \sum_{j=1}^{\infty} \frac{\varepsilon}{2^{j}} = \varepsilon . \end{equation*}

Let \(K = [c^{\prime}, d^{\prime}] \setminus U\). Then \(K\) is closed (it is the intersection of the closed sets \([c^{\prime},d^{\prime}]\) and \(\mathbb{R} \setminus U\)) and \(K \subseteq J\). Since \(K\) contains no rational number and every nonempty open interval contains a rational number, \(K\) has empty interior. Finally \([c^{\prime},d^{\prime}] \subseteq K \cup U\), so subadditivity and 2.14 give

\begin{equation*} d^{\prime} - c^{\prime} \leq |K| + |U| \leq |K| + \varepsilon = |K| + \frac{d^{\prime}-c^{\prime}}{2}, \end{equation*}

hence \(|K| \geq \frac{d^{\prime}-c^{\prime}}{2} > 0\). This proves the claim.

Step 2: a finite union of closed sets with empty interior has empty interior.

It suffices to do two sets, and then induct. Suppose \(E\) and \(G\) are closed with empty interior and \(V \subseteq E \cup G\) with \(V\) open and nonempty. Then \(V \setminus E\) is open; it is nonempty, since \(V \subseteq E\) would contradict \(E\) having empty interior. But \(V \setminus E \subseteq G\), contradicting \(G\) having empty interior. So \(E \cup G\) has empty interior; being a union of two closed sets it is closed.

Step 3: construction of \(A\).

Let \(I_1, I_2, \ldots\) be an enumeration of all open intervals with rational endpoints (a countable collection). We construct recursively closed sets \(K_1, L_1, K_2, L_2, \ldots\), each with empty interior and positive measure, and all pairwise disjoint, with \(K_n \cup L_n \subseteq I_n\).

Suppose \(n \in \mathbb{Z}^+\) and \(K_1, L_1, \ldots, K_{n-1}, L_{n-1}\) have been chosen (for \(n = 1\) nothing has been chosen). Let

\begin{equation*} C = K_1 \cup L_1 \cup \cdots \cup K_{n-1} \cup L_{n-1}, \end{equation*}

a finite union of closed sets with empty interior, hence closed with empty interior by Step 2. Therefore \(I_n \setminus C = I_n \cap (\mathbb{R} \setminus C)\) is open and nonempty (it is nonempty because otherwise the nonempty open set \(I_n\) would be contained in \(C\)). Pick \(y \in I_n \setminus C\). Since \(I_n \setminus C\) is open, there is \(r > 0\) with \((y - r, y + r) \subseteq I_n \setminus C\); set

\begin{equation*} J = \Bigl( y - \tfrac{r}{2}, \ y \Bigr), \qquad J^{\prime} = \Bigl( y, \ y + \tfrac{r}{2} \Bigr), \end{equation*}

which are disjoint nonempty bounded open intervals contained in \(I_n \setminus C\). By Step 1, choose closed sets with empty interior \(K_n \subseteq J\) and \(L_n \subseteq J^{\prime}\) with \(|K_n| > 0\) and \(|L_n| > 0\). By construction \(K_n\) and \(L_n\) are disjoint from each other and from all previously chosen sets, and they are contained in \(I_n\). The recursion continues, producing a pairwise disjoint family.

Now define

\begin{equation*} A = \bigcup_{n=1}^{\infty} K_n , \qquad f = \chi_A . \end{equation*}

Then \(A\) is Borel, being a countable union of closed sets, and \(L_n \cap A = \varnothing\) for every \(n\) by disjointness. The function \(f\) is Borel measurable: for every Borel set \(D \subseteq \mathbb{R}\), the set \(f^{-1}(D)\) is one of \(\varnothing\), \(A\), \(\mathbb{R} \setminus A\), \(\mathbb{R}\), each of which is Borel.

Step 4: \(f\) has the required property.

First note the key density property: for every nonempty open interval \(I\),

\begin{equation*} |A \cap I| > 0 \quad \text{and} \quad |I \setminus A| > 0 . \end{equation*}

Indeed, \(I\) contains some interval \(I_n\) with rational endpoints, and then \(K_n \subseteq A \cap I\) gives \(|A \cap I| \geq |K_n| > 0\), while \(L_n \subseteq I \setminus A\) gives \(|I \setminus A| \geq |L_n| > 0\) (monotonicity of outer measure).

Suppose now that \(B \subseteq \mathbb{R}\) satisfies \(|\mathbb{R} \setminus B| = 0\); we show \(f|_B\) is not continuous.

Since \(|A| \geq |K_1| > 0\) and \(A \subseteq (A \cap B) \cup (\mathbb{R} \setminus B)\), subadditivity gives

\begin{equation*} 0 < |A| \leq |A \cap B| + |\mathbb{R} \setminus B| = |A \cap B| , \end{equation*}

so \(A \cap B \neq \varnothing\). Choose \(x \in A \cap B\), so \(f(x) = 1\).

Suppose, for contradiction, that \(f|_B\) is continuous at \(x\). Taking \(\varepsilon = \frac{1}{2}\), there exists \(\delta > 0\) such that \(|f(y) - f(x)| < \frac{1}{2}\) for all \(y \in B\) with \(|y - x| < \delta\). Since \(f\) takes only the values \(0\) and \(1\), this forces \(f(y) = 1\), i.e. \(y \in A\), for all such \(y\). In other words, with \(I = (x - \delta, x + \delta)\),

\begin{equation*} B \cap I \subseteq A . \end{equation*}

Consequently

\begin{equation*} I \setminus A \subseteq I \setminus (B \cap I) \subseteq \mathbb{R} \setminus B , \end{equation*}

so \(|I \setminus A| \leq |\mathbb{R} \setminus B| = 0\) by monotonicity. This contradicts \(|I \setminus A| > 0\) from the density property.

Hence \(f|_B\) is not continuous at \(x\), and in particular \(f|_B\) is not a continuous function on \(B\). Since \(B\) was an arbitrary subset of \(\mathbb{R}\) with \(|\mathbb{R} \setminus B| = 0\), the Borel measurable function \(f = \chi_A\) is the required example.

The statement is false. The corresponding result for a countable family is true (2.53: the pointwise supremum of a sequence of \(\mathcal{S}\)-measurable functions is \(\mathcal{S}\)-measurable), but its proof uses the identity

\begin{equation*} h^{-1}\bigl((a, \infty]\bigr) = \bigcup_{k=1}^{\infty} f_k^{-1}\bigl((a, \infty]\bigr) \quad\text{for } h = \sup_k f_k , \end{equation*}

and a \(\sigma\)-algebra is closed only under countable unions. With an uncountable index set the corresponding union is uncountable, and that is exactly what goes wrong.

Counterexample. By 2.67, there exists a set \(A \subseteq \mathbb{R}\) that is not a Borel set. For each \(t \in \mathbb{R}\) define \(f_t : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f_t = \begin{cases} \chi_{\{t\}} & \text{if } t \in A,\\[2pt] 0 & \text{if } t \notin A .\end{cases} \end{equation*}

Each \(f_t\) is Borel measurable. Indeed, if \(t \notin A\) then \(f_t\) is the constant function \(0\), and constant functions are Borel measurable. If \(t \in A\) then \(f_t = \chi_{\{t\}}\), and for every Borel set \(D \subseteq \mathbb{R}\) the inverse image \(f_t^{-1}(D)\) is one of \(\varnothing\), \(\{t\}\), \(\mathbb{R} \setminus \{t\}\), \(\mathbb{R}\), each of which is a Borel set (the singleton \(\{t\}\) is closed, hence Borel).

Now compute \(f(x) = \sup\{f_t(x) : t \in \mathbb{R}\}\) for \(x \in \mathbb{R}\).

If \(x \in A\), then taking \(t = x\) gives \(f_x(x) = \chi_{\{x\}}(x) = 1\), and since every \(f_t\) takes values in \(\{0,1\}\) we get \(f(x) = 1\).

If \(x \notin A\), fix any \(t \in \mathbb{R}\). If \(t \notin A\) then \(f_t(x) = 0\). If \(t \in A\), then \(t \neq x\) (because \(x \notin A\)), so \(f_t(x) = \chi_{\{t\}}(x) = 0\). Hence \(f_t(x) = 0\) for every \(t\), and \(f(x) = 0\).

Therefore \(f = \chi_A\). Now \((\frac{1}{2}, \infty]\) is a Borel subset of \([-\infty, \infty]\), because by 2.50 this only requires \((\frac{1}{2}, \infty] \cap \mathbb{R} = (\frac{1}{2}, \infty)\) to be a Borel subset of \(\mathbb{R}\), which it is, being open. Since \(f\) takes only the values \(0\) and \(1\),

\begin{equation*} f^{-1}\bigl( (\tfrac{1}{2}, \infty] \bigr) = \{x \in \mathbb{R} : f(x) > \tfrac{1}{2}\} = A , \end{equation*}

which is not a Borel set. By the definition 2.51 of measurability for functions with values in \([-\infty, \infty]\), applied with \(\mathcal{S} = \mathcal{B}\) the \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}\), this shows that \(f\) is not Borel measurable.

Note that all the functions \(f_t\) here are bounded by \(1\), so \(f\) is even real-valued and bounded; the failure has nothing to do with the value \(\infty\) being allowed, only with the uncountability of the index set.

Here \(|\cdot|\) denotes outer measure, so no measurability of \(f\) is needed: everything follows from monotonicity and countable subadditivity of outer measure (2.8) together with the formula for the outer measure of an interval (2.14).

Write

\begin{equation*} E = \{x \in \mathbb{R} : f(x) < 1\}, \qquad N = \mathbb{R} \setminus E = \{x \in \mathbb{R} : f(x) \geq 1\}, \end{equation*}

where the value \(f(x) = \infty\) is counted as \(\geq 1\). The strategy: the exceptional set \(N\) is covered by intervals whose lengths form a convergent series, so \(N\) has finite outer measure; a set whose complement has finite outer measure must have infinite outer measure.

Step 1: covering \(N\) by short intervals.

For each \(k \in \mathbb{Z}^+\) let

\begin{equation*} A_k = \Bigl[ b_k - \frac{1}{2^{k-1}}, \ b_k + \frac{1}{2^{k-1}} \Bigr] = \{x \in \mathbb{R} : |x - b_k| \leq 2^{1-k}\}. \end{equation*}

By 2.14,

\begin{equation*} |A_k| = \frac{2}{2^{k-1}} = \frac{4}{2^{k}} . \end{equation*}

We claim

\begin{equation*} N \subseteq \bigcup_{k=1}^{\infty} A_k . \end{equation*}

Equivalently, we show that \(x \notin \bigcup_{k=1}^{\infty} A_k\) implies \(f(x) < 1\). So suppose \(x \notin A_k\) for every \(k\), which means

\begin{equation*} |x - b_k| > \frac{1}{2^{k-1}} \quad \text{for every } k \in \mathbb{Z}^+ . \end{equation*}

In particular \(|x - b_k| > 0\) for every \(k\), so \(x \notin \{b_1, b_2, \ldots\}\) and hence \(f(x)\) is given by the series. For each \(k\), using \(4^k = 2^{2k}\),

\begin{equation*} \frac{1}{4^k |x - b_k|} < \frac{1}{4^k \cdot 2^{1-k}} = \frac{1}{2^{2k} \cdot 2^{1-k}} = \frac{1}{2^{k+1}} . \end{equation*}

Summing the geometric series,

\begin{equation*} f(x) = \sum_{k=1}^{\infty} \frac{1}{4^k |x - b_k|} \leq \sum_{k=1}^{\infty} \frac{1}{2^{k+1}} = \frac{1}{2} < 1 . \end{equation*}

Thus \(x \in E\), which proves the claimed inclusion. (Note that this argument also shows \(E \neq \varnothing\), and in fact that \(f\) is finite off a set of finite measure, no matter how the \(b_k\) are chosen; that answers the concern raised in the bracketed remark.)

Step 2: \(N\) has finite outer measure.

By monotonicity and countable subadditivity of outer measure (2.8),

\begin{equation*} |N| \leq \Bigl| \bigcup_{k=1}^{\infty} A_k \Bigr| \leq \sum_{k=1}^{\infty} |A_k| = \sum_{k=1}^{\infty} \frac{4}{2^k} = 4 . \end{equation*}

Step 3: conclusion.

Let \(n \in \mathbb{Z}^+\). Since \(E \cup N = \mathbb{R}\), we have \([-n, n] \subseteq (E \cap [-n,n]) \cup N\), so subadditivity and 2.14 give

\begin{equation*} 2n = |[-n,n]| \leq |E \cap [-n,n]| + |N| \leq |E| + 4 , \end{equation*}

using monotonicity in the last step. Hence

\begin{equation*} |E| \geq 2n - 4 \quad \text{for every } n \in \mathbb{Z}^+ , \end{equation*}

and letting \(n \to \infty\) forces \(|E| = \infty\). That is,

\begin{equation*} |\{x \in \mathbb{R} : f(x) < 1\}| = \infty , \end{equation*}

as desired.

Two comments. First, the sequence \(b_1, b_2, \ldots\) was completely arbitrary: it may be dense in \(\mathbb{R}\), may repeat values, and may be unbounded, yet the set where \(f < 1\) always has infinite measure while the set where \(f \geq 1\) always has outer measure at most \(4\). The rapidly decaying weights \(4^{-k}\) do all the work, since the set where the \(k\)-th term alone is at least \(2^{-(k+1)}\) is an interval of length \(4 \cdot 2^{-k}\). Second, the same argument with \(A_k\) replaced by \(\{x : |x - b_k| \le \varepsilon 2^{1-k}\}\) shows \(|\{x : f(x) \geq \varepsilon^{-1}\}| \le 4\varepsilon\) for every \(\varepsilon>0\), hence \(|\{x \in \mathbb{R} : f(x) = \infty\}| = 0\): the series converges for almost every \(x \in \mathbb{R}\).

The strategy is to extend \(f\) to all of \(\mathbb{R}\), apply 2.95 (every Lebesgue measurable function is almost Borel measurable) to the extension, and then restrict back to \(B\). The hypothesis that \(B\) is a Borel set is exactly what makes the last step work.

Step 1: extend \(f\) to \(\mathbb{R}\).

Define \(\tilde{f} : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} \tilde{f}(x) = \begin{cases} f(x) & \text{if } x \in B, \\ 0 & \text{if } x \in \mathbb{R} \setminus B. \end{cases} \end{equation*}

We claim \(\tilde{f}\) is a Lebesgue measurable function in the sense of 2.94, i.e. \(\tilde{f}^{-1}(A)\) is a Lebesgue measurable subset of \(\mathbb{R}\) for every Borel set \(A \subseteq \mathbb{R}\).

So let \(A \subseteq \mathbb{R}\) be a Borel set. Splitting \(\mathbb{R}\) into \(B\) and \(\mathbb{R} \setminus B\), on which \(\tilde{f}\) is respectively \(f\) and the constant \(0\), gives

\begin{equation*} \tilde{f}^{-1}(A) = \begin{cases} f^{-1}(A) & \text{if } 0 \notin A, \\ f^{-1}(A) \cup (\mathbb{R} \setminus B) & \text{if } 0 \in A. \end{cases} \end{equation*}

Here \(f^{-1}(A)\) means \(\{x \in B : f(x) \in A\}\), which is a Lebesgue measurable subset of \(\mathbb{R}\) because \(f\) is a Lebesgue measurable function and \(A\) is a Borel set. Also \(\mathbb{R} \setminus B\) is a Borel set (the Borel sets form a \(\sigma\)-algebra and \(B\) is Borel), hence is Lebesgue measurable, because every Borel set is Lebesgue measurable (the \(\sigma\)-algebra of Lebesgue measurable sets contains all open sets, hence contains the smallest \(\sigma\)-algebra containing the open sets). In either case \(\tilde{f}^{-1}(A)\) is a union of at most two Lebesgue measurable sets and so is Lebesgue measurable. Thus \(\tilde{f}\) is a Lebesgue measurable function from \(\mathbb{R}\) to \(\mathbb{R}\).

Step 2: apply 2.95.

By 2.95, there exists a Borel measurable function \(\tilde{g} : \mathbb{R} \to \mathbb{R}\) such that

\begin{equation*} |\{x \in \mathbb{R} : \tilde{g}(x) \neq \tilde{f}(x)\}| = 0. \end{equation*}

Step 3: restrict to \(B\).

Define \(g : B \to \mathbb{R}\) by \(g = \tilde{g}|_B\), that is, \(g(x) = \tilde{g}(x)\) for \(x \in B\).

First, \(g\) is Borel measurable. Indeed, let \(A \subseteq \mathbb{R}\) be a Borel set. Then

\begin{equation*} g^{-1}(A) = \{x \in B : \tilde{g}(x) \in A\} = B \cap \tilde{g}^{-1}(A). \end{equation*}

Now \(\tilde{g}^{-1}(A)\) is a Borel subset of \(\mathbb{R}\) because \(\tilde{g}\) is Borel measurable, and \(B\) is a Borel set by hypothesis; hence \(g^{-1}(A)\) is a Borel set, being the intersection of two Borel sets. Thus \(g : B \to \mathbb{R}\) is a Borel measurable function.

Second, the exceptional set is small. If \(x \in B\) and \(g(x) \neq f(x)\), then \(\tilde{g}(x) = g(x) \neq f(x) = \tilde{f}(x)\) (the last equality because \(x \in B\)). Hence

\begin{equation*} \{x \in B : g(x) \neq f(x)\} \subseteq \{x \in \mathbb{R} : \tilde{g}(x) \neq \tilde{f}(x)\}. \end{equation*}

Outer measure preserves order (2.5), so

\begin{equation*} |\{x \in B : g(x) \neq f(x)\}| \le |\{x \in \mathbb{R} : \tilde{g}(x) \neq \tilde{f}(x)\}| = 0. \end{equation*}

Therefore \(|\{x \in B : g(x) \neq f(x)\}| = 0\), as desired.

Remark on the hypothesis that \(B\) is Borel.

The assumption that \(B\) is a Borel set is not merely a convenience: it is necessary. By the definition in 2.94, a function \(g : B \to \mathbb{R}\) is Borel measurable when \(g^{-1}(A)\) is a Borel set for every Borel set \(A \subseteq \mathbb{R}\); taking \(A = \mathbb{R}\) forces \(B = g^{-1}(\mathbb{R})\) to be a Borel set. So if \(B\) were Lebesgue measurable but not Borel, no Borel measurable function on \(B\) would exist at all, and the assertion of the exercise would fail for such a \(B\), because its hypothesis can be met: the identity map \(f(x) = x\) on \(B\) is Lebesgue measurable, since \(f^{-1}(A) = B \cap A\) is Lebesgue measurable for every Borel set \(A\). This is also where the argument above would break down: in Step 3 the set \(g^{-1}(A) = B \cap \tilde{g}^{-1}(A)\) is Borel precisely because \(B\) is.

Remark: a self-contained version of Step 2.

For completeness, here is the argument of 2.95 adapted directly to \(B\), avoiding the extension. Let \(\mathcal{L}_B\) denote the \(\sigma\)-algebra of Lebesgue measurable subsets of \(B\) and \(\mathcal{B}_B\) the \(\sigma\)-algebra of Borel subsets of \(B\); because \(B\) is a Borel set, a subset of \(B\) lies in \(\mathcal{B}_B\) exactly when it is a Borel subset of \(\mathbb{R}\), and a subset of \(B\) lies in \(\mathcal{L}_B\) exactly when it is a Lebesgue measurable subset of \(\mathbb{R}\). As noted after 2.94, \(f\) is an \(\mathcal{L}_B\)-measurable function on \(B\). So by 2.89 applied to the measurable space \((B, \mathcal{L}_B)\) there is a sequence \(f_1, f_2, \ldots\) of simple \(\mathcal{L}_B\)-measurable functions from \(B\) to \(\mathbb{R}\) converging pointwise on \(B\) to \(f\). Fix \(k \in \mathbb{Z}^+\) and write

\begin{equation*} f_k = c_1 \chi_{A_1} + \cdots + c_n \chi_{A_n} \end{equation*}

where \(c_1, \ldots, c_n\) are the distinct values taken by \(f_k\) and \(A_j = f_k^{-1}(\{c_j\})\), so that \(A_1, \ldots, A_n\) are disjoint sets in \(\mathcal{L}_B\) whose union is \(B\). Each \(A_j\) is therefore a Lebesgue measurable subset of \(\mathbb{R}\), so by the equivalence of (a) and (d) in 2.71 there is a Borel set \(B_j \subseteq A_j\) with \(|A_j \setminus B_j| = 0\). Put \(g_k = c_1 \chi_{B_1} + \cdots + c_n \chi_{B_n}\), viewed as a function on \(B\); each \(B_j\) is a Borel subset of \(B\), and \(g_k\) takes only the values \(c_1, \ldots, c_n, 0\), each on a set that is a finite union of the Borel sets \(B_1, \ldots, B_n, B \setminus (B_1 \cup \cdots \cup B_n)\), so \(g_k : B \to \mathbb{R}\) is Borel measurable. Moreover

\begin{equation*} \{x \in B : g_k(x) \neq f_k(x)\} \subseteq (A_1 \setminus B_1) \cup \cdots \cup (A_n \setminus B_n) \end{equation*}

has outer measure \(0\) by subadditivity (2.8). Let

\begin{equation*} E = \Bigl\{x \in B : \lim_{k \to \infty} g_k(x) \text{ exists in } \mathbb{R}\Bigr\}, \end{equation*}

which lies in \(\mathcal{B}_B\) by Exercise 14(b) in Section 2B applied to the measurable space \((B, \mathcal{B}_B)\) and the \(\mathcal{B}_B\)-measurable functions \(g_1, g_2, \ldots\); since \(B\) is Borel, \(E\) is thus a Borel subset of \(\mathbb{R}\). If \(x \in B\) lies outside \(\bigcup_{k=1}^{\infty} \{x \in B : g_k(x) \neq f_k(x)\}\), then \(g_k(x) = f_k(x)\) for every \(k\), so \(\lim_{k \to \infty} g_k(x) = f(x)\) exists and \(x \in E\). Hence

\begin{equation*} B \setminus E \subseteq \bigcup_{k=1}^{\infty} \{x \in B : g_k(x) \neq f_k(x)\}, \end{equation*}

so \(|B \setminus E| = 0\) by countable subadditivity (2.8). Define \(g(x) = \lim_{k \to \infty} (\chi_E g_k)(x)\) for \(x \in B\); the limit exists for every \(x \in B\) (by definition of \(E\) when \(x \in E\), and because \((\chi_E g_k)(x) = 0\) for all \(k\) when \(x \in B \setminus E\)). Each \(\chi_E g_k\) is \(\mathcal{B}_B\)-measurable, being the product of the \(\mathcal{B}_B\)-measurable functions \(\chi_E\) (measurable because \(E \in \mathcal{B}_B\)) and \(g_k\), by 2.46(a); so \(g\) is \(\mathcal{B}_B\)-measurable by 2.48, and hence \(g : B \to \mathbb{R}\) is Borel measurable. Finally

\begin{equation*} \{x \in B : g(x) \neq f(x)\} \subseteq (B \setminus E) \cup \bigcup_{k=1}^{\infty} \{x \in B : g_k(x) \neq f_k(x)\}, \end{equation*}

because for \(x \in E\) with \(g_k(x) = f_k(x)\) for all \(k\) we get \(g(x) = \lim_{k \to \infty} g_k(x) = f(x)\). The right side has outer measure \(0\), so \(|\{x \in B : g(x) \neq f(x)\}| = 0\).

Suppose \(E \in \mathcal{S}\) satisfies \(\mu(E) = \infty\), and suppose toward a contradiction that

\begin{equation*} c := \inf_E f > 0 . \end{equation*}

Note first that \(E \neq \varnothing\), since \(\mu(\varnothing) = 0 \neq \infty\); so the infimum above is an infimum of a nonempty subset of \([0, \infty]\) and \(c \in (0, \infty]\).

Let \(\mathcal{P}\) be the \(\mathcal{S}\)-partition of \(X\) consisting of the two sets \(E\) and \(X \setminus E\). (These are disjoint sets in \(\mathcal{S}\) whose union is \(X\), so \(\mathcal{P}\) is indeed an \(\mathcal{S}\)-partition in the sense of 3.1; if \(E = X\) then the second set is \(\varnothing\), which is harmless.) By the definition 3.2 of the lower Lebesgue sum,

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \mu(E) \inf_E f + \mu(X \setminus E) \inf_{X \setminus E} f \geq \mu(E) \inf_E f = \infty \cdot c = \infty, \end{equation*}

where the inequality holds because the discarded term is an element of \([0, \infty]\) and hence is nonnegative, and where the last equality holds because \(c > 0\).

By the definition 3.3 of the integral, \(\int f \, d\mu = \sup \{ \mathcal{L}(f, \mathcal{P}^{\prime}) : \mathcal{P}^{\prime} \text{ is an } \mathcal{S}\text{-partition of } X \} \geq \mathcal{L}(f, \mathcal{P}) = \infty\). Thus \(\int f \, d\mu = \infty\), contradicting the hypothesis that \(\int f \, d\mu < \infty\).

This contradiction shows that the assumption \(\inf_E f > 0\) is untenable. Since \(f\) takes values in \([0, \infty]\) we always have \(\inf_E f \geq 0\), so \(\inf_E f = 0\), which is what we wanted to prove.

In words: a set of infinite measure on which \(f\) has a positive lower bound would already contribute an infinite lower Lebesgue sum, so a function with finite integral must come arbitrarily close to \(0\) on every set of infinite measure.

Throughout, \(\mathcal{L}(f, \mathcal{P})\) denotes the lower Lebesgue sum of 3.2 taken with respect to the measure \(\delta_c\).

Step 1: we show \(\int f \, d\delta_c \leq f( c)\).

Let \(\mathcal{P}\) be an \(\mathcal{S}\)-partition \(A_1, \ldots, A_m\) of \(X\). Because \(A_1, \ldots, A_m\) are disjoint sets whose union is \(X\), the point \(c\) lies in exactly one of them; say \(c \in A_{j_0}\). Then \(\delta_c(A_{j_0}) = 1\) and \(\delta_c(A_j) = 0\) for every \(j \neq j_0\). Hence, using the convention that \(0 \cdot \infty = 0\),

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \sum_{j=1}^m \delta_c(A_j) \inf_{A_j} f = \inf_{A_{j_0}} f \leq f( c), \end{equation*}

the last inequality because \(c \in A_{j_0}\). Taking the supremum over all \(\mathcal{S}\)-partitions \(\mathcal{P}\) of \(X\) and using the definition 3.3 of the integral gives \(\int f \, d\delta_c \leq f( c)\).

Note that this half of the argument used nothing about \(f\) beyond its being a function into \([0, \infty]\).

Step 2: we show \(\int f \, d\delta_c \geq f( c)\).

Here is where measurability of \(f\) enters, and here is the point of the warning in the exercise: we cannot simply use the partition \(\{c\}, X \setminus \{c\}\), because \(\{c\}\) need not belong to \(\mathcal{S}\). Instead we use level sets of \(f\), which do belong to \(\mathcal{S}\).

If \(f( c) = 0\) the desired inequality is trivial, since integrals of nonnegative functions are nonnegative. So suppose \(f( c) > 0\) and let \(t \in [0, f( c))\) be arbitrary. Set

\begin{equation*} A = f^{-1}\bigl((t, \infty]\bigr) = \{ x \in X : f(x) > t \}. \end{equation*}

The set \((t, \infty]\) is a Borel subset of \([-\infty, \infty]\), because its intersection with \(\mathbb{R}\) is the open interval \((t, \infty)\) (see 2.50). Since \(f\) is \(\mathcal{S}\)-measurable, the definition 2.51 of measurability gives \(A \in \mathcal{S}\). Also \(c \in A\), since \(f( c) > t\).

Let \(\mathcal{P}\) be the \(\mathcal{S}\)-partition of \(X\) consisting of \(A\) and \(X \setminus A\). Since \(c \in A\), we have \(\delta_c(A) = 1\) and \(\delta_c(X \setminus A) = 0\). Therefore, again using \(0 \cdot \infty = 0\),

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \delta_c(A) \inf_A f + \delta_c(X \setminus A) \inf_{X \setminus A} f = \inf_A f \geq t, \end{equation*}

because \(f(x) > t\) for every \(x \in A\). Hence \(\int f \, d\delta_c \geq t\).

Since \(t \in [0, f( c))\) was arbitrary, taking the supremum over such \(t\) gives \(\int f \, d\delta_c \geq f( c)\). (When \(f( c) = \infty\) this says \(\int f \, d\delta_c \geq t\) for every \(t \in [0, \infty)\), so the integral equals \(\infty = f( c)\).)

Combining Steps 1 and 2 gives \(\int f \, d\delta_c = f( c)\).

Write

\begin{equation*} E = \{ x \in X : f(x) > 0 \} = f^{-1}\bigl((0, \infty]\bigr) \quad\text{and}\quad E_k = \{ x \in X : f(x) > \tfrac{1}{k} \} = f^{-1}\bigl((\tfrac{1}{k}, \infty]\bigr) \end{equation*}

for \(k \in \mathbb{Z}^+\). Each of \((0, \infty]\) and \((\frac{1}{k}, \infty]\) is a Borel subset of \([-\infty, \infty]\), since its intersection with \(\mathbb{R}\) is an open interval (see 2.50). Because \(f\) is \(\mathcal{S}\)-measurable, the definition 2.51 gives \(E \in \mathcal{S}\) and \(E_k \in \mathcal{S}\) for each \(k\).

First suppose \(\mu(E) > 0\); we show \(\int f \, d\mu > 0\).

Because \(\frac{1}{1} \geq \frac{1}{2} \geq \cdots\), we have \(E_1 \subseteq E_2 \subseteq \cdots\). Moreover \(\bigcup_{k=1}^{\infty} E_k = E\): the inclusion \(\subseteq\) is clear since \(\frac{1}{k} > 0\); conversely if \(f(x) > 0\) then \(\frac{1}{k} < f(x)\) for some \(k \in \mathbb{Z}^+\), so \(x \in E_k\). Hence 2.59 (measure of an increasing union) gives

\begin{equation*} \lim_{k \to \infty} \mu(E_k) = \mu(E) > 0 . \end{equation*}

Therefore there exists \(k \in \mathbb{Z}^+\) with \(\mu(E_k) > 0\). Fix such a \(k\).

Now \(\frac{1}{k} \chi_{E_k}(x) \leq f(x)\) for every \(x \in X\): if \(x \in E_k\) then \(f(x) > \frac{1}{k}\), and if \(x \notin E_k\) then the left side is \(0 \leq f(x)\). Consequently, by 3.8 (integration is order preserving) and then 3.7 (integral of a simple function, applied with the single set \(E_k\) and the coefficient \(\frac{1}{k}\)),

\begin{equation*} \int f \, d\mu \geq \int \tfrac{1}{k} \chi_{E_k} \, d\mu = \tfrac{1}{k} \mu(E_k) > 0 , \end{equation*}

as desired.

Conversely suppose \(\mu(E) = 0\); we show \(\int f \, d\mu = 0\), which proves the remaining implication in contrapositive form.

Let \(\mathcal{P}\) be an arbitrary \(\mathcal{S}\)-partition \(A_1, \ldots, A_m\) of \(X\). Fix \(j \in \{1, \ldots, m\}\) and consider the term \(\mu(A_j) \inf_{A_j} f\) of the lower Lebesgue sum. There are two cases.

If \(A_j \subseteq E\), then \(\mu(A_j) \leq \mu(E) = 0\) by 2.57(a), so \(\mu(A_j) = 0\) and the term equals \(0 \cdot \inf_{A_j} f = 0\) (using the convention \(0 \cdot \infty = 0\)).

If \(A_j \not\subseteq E\), then there exists \(x \in A_j\) with \(x \notin E\), which means \(f(x) = 0\) (recall \(f\) takes values in \([0, \infty]\)). Hence \(\inf_{A_j} f = 0\) and the term equals \(\mu(A_j) \cdot 0 = 0\) (using the convention \(\infty \cdot 0 = 0\)).

In either case the term is \(0\), so \(\mathcal{L}(f, \mathcal{P}) = 0\). Since \(\mathcal{P}\) was an arbitrary \(\mathcal{S}\)-partition of \(X\), the definition 3.3 gives

\begin{equation*} \int f \, d\mu = \sup \{ \mathcal{L}(f, \mathcal{P}) : \mathcal{P} \text{ is an } \mathcal{S}\text{-partition of } X \} = 0 . \end{equation*}

Thus \(\int f \, d\mu > 0\) if and only if \(\mu(E) > 0\).

Let \(r_1, r_2, \ldots\) be an enumeration (a bijective listing) of the rational numbers in \([0, 1]\); such a listing exists because \(\mathbb{Q} \cap [0, 1]\) is countably infinite. Define \(f : [0, 1] \to (0, \infty)\) by

\begin{equation*} f(x) = \begin{cases} \dfrac{1}{k} & \text{if } x = r_k \text{ for some } k \in \mathbb{Z}^+, \\[2mm] 1 & \text{if } x \text{ is irrational.} \end{cases} \end{equation*}

Because the listing is bijective, each rational \(x \in [0, 1]\) equals \(r_k\) for exactly one \(k\), so \(f\) is well defined. Every value of \(f\) lies in \((0, 1] \subseteq (0, \infty)\), so \(f\) indeed maps into \((0, \infty)\); note also that \(f\) is bounded, so its lower Riemann integral on \([0, 1]\) is defined.

The function \(f\) is Borel measurable. Indeed, let \(B \subseteq \mathbb{R}\) be a Borel set. Then

\begin{equation*} f^{-1}(B) = \{ r_k : k \in \mathbb{Z}^+ \text{ and } \tfrac{1}{k} \in B \} \; \cup \; \begin{cases} [0,1] \setminus \mathbb{Q} & \text{if } 1 \in B, \\ \varnothing & \text{if } 1 \notin B. \end{cases} \end{equation*}

The first set is countable, hence Borel (a countable set is a countable union of singletons, and singletons are closed, hence Borel). The second set is either \(\varnothing\) or \([0,1] \setminus \mathbb{Q}\), and \([0,1] \setminus \mathbb{Q}\) is the intersection of the Borel set \([0,1]\) with the complement of the countable, hence Borel, set \(\mathbb{Q}\). So \(f^{-1}(B)\) is a union of two Borel sets and is therefore Borel. Thus \(f\) is Borel measurable.

Now we compute the lower Riemann integral. Let \(P\) be an arbitrary partition \(0 = x_0 < x_1 < \cdots < x_n = 1\) of \([0, 1]\), and fix \(j \in \{1, \ldots, n\}\). The interval \([x_{j-1}, x_j]\) is a nondegenerate closed subinterval of \([0, 1]\), so it contains infinitely many rational numbers. Since the listing \(r_1, r_2, \ldots\) is injective, the index set

\begin{equation*} K_j = \{ k \in \mathbb{Z}^+ : r_k \in [x_{j-1}, x_j] \} \end{equation*}

is infinite, and therefore contains arbitrarily large integers. Hence

\begin{equation*} \inf_{[x_{j-1}, x_j]} f \; \leq \; \inf \Bigl\{ \tfrac{1}{k} : k \in K_j \Bigr\} = 0 , \end{equation*}

and since \(f > 0\) this infimum equals \(0\). Therefore, by the definition 1.3 of the lower Riemann sum,

\begin{equation*} L(f, P, [0,1]) = \sum_{j=1}^n (x_j - x_{j-1}) \inf_{[x_{j-1}, x_j]} f = \sum_{j=1}^n (x_j - x_{j-1}) \cdot 0 = 0 . \end{equation*}

Because \(P\) was an arbitrary partition of \([0, 1]\), taking the supremum over all partitions and using the definition 1.7 of the lower Riemann integral gives

\begin{equation*} L(f, [0,1]) = 0 , \end{equation*}

as required.

For contrast, let \(\lambda\) denote Lebesgue measure on the Borel subsets of \([0, 1]\). Since \(f(x) > 0\) for every \(x \in [0,1]\), we have \(\lambda(\{x \in [0,1] : f(x) > 0\}) = \lambda([0,1]) = 1 > 0\), so the previous exercise (3A.3) gives \(\int f \, d\lambda > 0\). In fact the value can be computed exactly: \(\chi_{[0,1] \setminus \mathbb{Q}} \leq f \leq \chi_{[0,1]}\) pointwise on \([0,1]\), so 3.8 and 3.4 give

\begin{equation*} 1 = \lambda\bigl([0,1] \setminus \mathbb{Q}\bigr) \leq \int f \, d\lambda \leq \lambda([0,1]) = 1 , \end{equation*}

so \(\int f \, d\lambda = 1\). The lower Riemann integral fails here because Riemann partitions are forced to use intervals, and every nondegenerate interval contains points where \(f\) is nearly \(0\); Lebesgue’s definition may use arbitrary Borel sets in a partition, and the set \([0,1] \setminus \mathbb{Q}\) (on which \(f\) equals \(1\)) is allowed as a partition piece.

Example 3.6 asserts the following. Suppose \(\mu\) is counting measure on \(\mathbb{Z}^+\) and \(b_1, b_2, \ldots\) is a sequence of nonnegative numbers; think of \(b\) as the function from \(\mathbb{Z}^+\) to \([0, \infty)\) defined by \(b(k) = b_k\). Then

\begin{equation*} \int b \, d\mu = \sum_{k=1}^{\infty} b_k . \end{equation*}

Here \(\mathcal{S}\) is the \(\sigma\)-algebra of all subsets of \(\mathbb{Z}^+\) and \(\mu\) is counting measure (2.55): \(\mu(E)\) is the number of elements of \(E\) when \(E\) is finite, and \(\mu(E) = \infty\) when \(E\) is infinite. Every function on \(\mathbb{Z}^+\) is \(\mathcal{S}\)-measurable, since every subset of \(\mathbb{Z}^+\) is in \(\mathcal{S}\). Note that \(\sum_{k=1}^{\infty} b_k\) is a series of nonnegative terms and so converges in \([0, \infty]\), its value being \(\lim_{n \to \infty} \sum_{k=1}^n b_k\); equivalently it is \(\sup \{ \sum_{k \in F} b_k : F \subseteq \mathbb{Z}^+ \text{ finite} \}\).

Both inequalities are proved directly from the definitions 3.2 and 3.3.

First we show \(\int b \, d\mu \geq \sum_{k=1}^{\infty} b_k\).

Fix \(n \in \mathbb{Z}^+\) and let \(\mathcal{P}_n\) be the \(\mathcal{S}\)-partition of \(\mathbb{Z}^+\) consisting of the \(n+1\) sets

\begin{equation*} \{1\}, \{2\}, \ldots, \{n\}, \quad \{n+1, n+2, \ldots\} . \end{equation*}

These are disjoint and their union is \(\mathbb{Z}^+\). Since \(\mu(\{k\}) = 1\) and \(\inf_{\{k\}} b = b_k\), and since the last term of the lower Lebesgue sum is an element of \([0, \infty]\) and hence nonnegative,

\begin{equation*} \mathcal{L}(b, \mathcal{P}_n) = \sum_{k=1}^n b_k + \mu(\{n+1, n+2, \ldots\}) \inf_{k > n} b_k \geq \sum_{k=1}^n b_k . \end{equation*}

Hence \(\int b \, d\mu \geq \sum_{k=1}^n b_k\) for every \(n \in \mathbb{Z}^+\). Letting \(n \to \infty\) gives \(\int b \, d\mu \geq \sum_{k=1}^{\infty} b_k\).

Next we show \(\int b \, d\mu \leq \sum_{k=1}^{\infty} b_k\).

Let \(\mathcal{P}\) be an arbitrary \(\mathcal{S}\)-partition \(A_1, \ldots, A_m\) of \(\mathbb{Z}^+\). We claim that for each \(j\),

\begin{equation*} \mu(A_j) \inf_{A_j} b \leq \sum_{k \in A_j} b_k , \end{equation*}

where the right side denotes \(\sup \{ \sum_{k \in F} b_k : F \subseteq A_j \text{ finite} \} \in [0, \infty]\). To see this, put \(c_j = \inf_{A_j} b\) and consider three cases.

If \(A_j = \varnothing\), then the left side is \(0 \cdot \infty = 0\) and the right side is \(0\).

If \(A_j\) is finite and nonempty, say with \(n_j\) elements, then \(\mu(A_j) = n_j\) and each of the \(n_j\) numbers \(b_k\) with \(k \in A_j\) is at least \(c_j\), so \(\mu(A_j) c_j = n_j c_j \leq \sum_{k \in A_j} b_k\).

If \(A_j\) is infinite, then \(\mu(A_j) = \infty\). If \(c_j = 0\) the left side is \(\infty \cdot 0 = 0\) and the inequality is clear. If \(c_j > 0\), then for every \(N \in \mathbb{Z}^+\) we may choose a subset \(F \subseteq A_j\) with \(N\) elements, giving \(\sum_{k \in A_j} b_k \geq \sum_{k \in F} b_k \geq N c_j\); letting \(N \to \infty\) shows \(\sum_{k \in A_j} b_k = \infty\), so the inequality holds.

This proves the claim. Summing the claim over \(j = 1, \ldots, m\) gives

\begin{equation*} \mathcal{L}(b, \mathcal{P}) = \sum_{j=1}^m \mu(A_j) \inf_{A_j} b \leq \sum_{j=1}^m \sum_{k \in A_j} b_k \leq \sum_{k=1}^{\infty} b_k . \end{equation*}

To justify the last inequality, write \(s_j = \sum_{k \in A_j} b_k \in [0, \infty]\) and let \(t_1, \ldots, t_m \in [0, \infty)\) be any real numbers with \(t_j < s_j\) whenever \(s_j > 0\) and \(t_j = 0\) whenever \(s_j = 0\). By the definition of \(s_j\) as a supremum, for each \(j\) there is a finite set \(F_j \subseteq A_j\) with \(\sum_{k \in F_j} b_k \geq t_j\). The sets \(F_1, \ldots, F_m\) are disjoint (being subsets of the disjoint sets \(A_1, \ldots, A_m\)) and their union is a finite subset of \(\mathbb{Z}^+\), so

\begin{equation*} \sum_{j=1}^m t_j \; \leq \; \sum_{j=1}^m \sum_{k \in F_j} b_k \; = \sum_{k \in F_1 \cup \cdots \cup F_m} b_k \; \leq \; \sum_{k=1}^{\infty} b_k . \end{equation*}

Now take the supremum over all such tuples \((t_1, \ldots, t_m)\). That supremum equals \(\sum_{j=1}^m s_j\): if every \(s_j\) is finite the \(t_j\) may be taken arbitrarily close to \(s_j\), while if some \(s_j = \infty\) the corresponding \(t_j\) may be taken arbitrarily large, making both sides \(\infty\). Hence \(\sum_{j=1}^m \sum_{k \in A_j} b_k \leq \sum_{k=1}^{\infty} b_k\).

Since \(\mathcal{P}\) was arbitrary, the definition 3.3 gives \(\int b \, d\mu \leq \sum_{k=1}^{\infty} b_k\).

Combining the two inequalities gives \(\int b \, d\mu = \sum_{k=1}^{\infty} b_k\), verifying Example 3.6.

Alternatively, once the Monotone Convergence Theorem 3.11 is available there is a two-line proof. For \(n \in \mathbb{Z}^+\) define \(b^{(n)} = \sum_{k=1}^n b_k \chi_{\{k\}}\). Then \(0 \leq b^{(1)} \leq b^{(2)} \leq \cdots\) (because each \(b_k \geq 0\)) and \(\lim_{n \to \infty} b^{(n)}(k) = b_k\) for every \(k \in \mathbb{Z}^+\) (indeed \(b^{(n)}(k) = b_k\) once \(n \geq k\)). Hence 3.11 and 3.7 give

\begin{equation*} \int b \, d\mu = \lim_{n \to \infty} \int b^{(n)} \, d\mu = \lim_{n \to \infty} \sum_{k=1}^n b_k \mu(\{k\}) = \lim_{n \to \infty} \sum_{k=1}^n b_k = \sum_{k=1}^{\infty} b_k . \end{equation*}

This is the analogue for lower Lebesgue sums of 1.5, which says that refining a partition increases the lower Riemann sum.

Write \(\mathcal{P}\) as the list \(A_1, \ldots, A_m\) and \(\mathcal{P}^{\prime}\) as the list \(B_1, \ldots, B_n\). By hypothesis, for each \(k \in \{1, \ldots, n\}\) there exists an index \(j(k) \in \{1, \ldots, m\}\) with

\begin{equation*} B_k \subseteq A_{j(k)} . \end{equation*}

Fix such a choice of \(j(1), \ldots, j(n)\) once and for all (if some \(B_k\) is contained in more than one \(A_j\), which can happen only when \(B_k = \varnothing\), choose either).

We first record a set-theoretic fact.

Claim: for each \(j \in \{1, \ldots, m\}\),

\begin{equation*} A_j = \bigcup_{\{k \, : \, j(k) = j\}} B_k , \end{equation*}

and the sets in this union are disjoint.

To prove the claim, note first that disjointness is inherited from \(\mathcal{P}^{\prime}\). The inclusion \(\supseteq\) is immediate, since \(j(k) = j\) gives \(B_k \subseteq A_j\). For \(\subseteq\), let \(x \in A_j\). Because \(\mathcal{P}^{\prime}\) is a partition of \(X\), there is a (unique) \(k\) with \(x \in B_k\). Then \(x \in B_k \subseteq A_{j(k)}\), so \(x \in A_j \cap A_{j(k)}\). Since \(A_1, \ldots, A_m\) are disjoint and this intersection is nonempty, \(j(k) = j\). Thus \(x\) lies in the union on the right, proving the claim.

By finite additivity of \(\mu\) (a consequence of the definition 2.54 of a measure), the claim gives

\begin{equation*} \mu(A_j) = \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \qquad \text{for each } j \in \{1, \ldots, m\}. \end{equation*}

We also note the elementary distributive law in \([0, \infty]\) that will be used: if \(c \in [0, \infty]\) and \(t_1, \ldots, t_p \in [0, \infty]\), then \(\sum_{i=1}^p c \, t_i = c \sum_{i=1}^p t_i\), where \(0 \cdot \infty = \infty \cdot 0 = 0\). Indeed, for \(c \in (0, \infty)\) this is the usual distributive law (both sides being \(\infty\) exactly when some \(t_i = \infty\)); for \(c = 0\) both sides equal \(0\); and for \(c = \infty\) both sides equal \(0\) if every \(t_i = 0\) and equal \(\infty\) otherwise.

Now we compute. The index sets \(\{k : j(k) = j\}\), for \(j = 1, \ldots, m\), are disjoint and their union is \(\{1, \ldots, n\}\), so we may group the terms of \(\mathcal{L}(f, \mathcal{P}^{\prime})\) according to \(j\):

\begin{equation*} \begin{aligned} \mathcal{L}(f, \mathcal{P}^{\prime}) &= \sum_{k=1}^n \mu(B_k) \inf_{B_k} f \\ &= \sum_{j=1}^m \; \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \inf_{B_k} f \\ &\geq \sum_{j=1}^m \; \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \inf_{A_j} f \\ &= \sum_{j=1}^m \Bigl( \inf_{A_j} f \Bigr) \sum_{\{k \, : \, j(k) = j\}} \mu(B_k) \\ &= \sum_{j=1}^m \mu(A_j) \inf_{A_j} f \\ &= \mathcal{L}(f, \mathcal{P}) . \end{aligned} \end{equation*}

The inequality in the third line holds because \(B_k \subseteq A_j\) whenever \(j(k) = j\), and the infimum of \(f\) over a subset is at least the infimum over the larger set, so \(\inf_{B_k} f \geq \inf_{A_j} f\); multiplying by \(\mu(B_k) \in [0, \infty]\) preserves this inequality (with the convention \(0 \cdot \infty = 0\), since if \(\mu(B_k) = 0\) both sides are \(0\)). The fourth line uses the distributive law recorded above, and the fifth line uses finite additivity as recorded above.

Hence \(\mathcal{L}(f, \mathcal{P}) \leq \mathcal{L}(f, \mathcal{P}^{\prime})\), as desired.

Note the degenerate cases cause no trouble: if \(B_k = \varnothing\) then \(\inf_{B_k} f = \infty\) but \(\mu(B_k) = 0\), so that term contributes \(0\) on both sides; similarly for empty \(A_j\).

Notation. For a function \(g : X \to [0, \infty]\) and a set \(A \subseteq X\), write

\begin{equation*} \sum_{x \in A} g(x) = \sup \Bigl\{ \sum_{x \in F} g(x) : F \subseteq A \text{ and } F \text{ is finite} \Bigr\} \in [0, \infty] , \end{equation*}

with the convention that the empty sum is \(0\) (so the supremum is over a nonempty collection and is well defined, and equals \(0\) when \(A = \varnothing\)). Throughout, \(0 \cdot \infty = \infty \cdot 0 = 0\).

Note that every function \(f : X \to [0, \infty]\) is \(\mathcal{S}\)-measurable, because \(\mathcal{S}\) consists of all subsets of \(X\); so the integral \(\int f \, d\mu\) is defined.

Preliminary 0 (a supremum lemma, used repeatedly). Suppose \(A_1, \ldots, A_m\) are disjoint subsets of \(X\) with union contained in a set \(A \subseteq X\), and \(g : X \to [0, \infty]\). Then

\begin{equation*} \sum_{j=1}^m \; \sum_{x \in A_j} g(x) \; \leq \; \sum_{x \in A} g(x) . \end{equation*}

To see this, write \(s_j = \sum_{x \in A_j} g(x)\) and let \(t_1, \ldots, t_m \in [0, \infty)\) satisfy \(t_j < s_j\) when \(s_j > 0\) and \(t_j = 0\) when \(s_j = 0\). By definition of \(s_j\) as a supremum, choose a finite \(F_j \subseteq A_j\) with \(\sum_{x \in F_j} g(x) \geq t_j\). The sets \(F_1, \ldots, F_m\) are disjoint and \(F_1 \cup \cdots \cup F_m\) is a finite subset of \(A\), so

\begin{equation*} \sum_{j=1}^m t_j \leq \sum_{j=1}^m \sum_{x \in F_j} g(x) = \sum_{x \in F_1 \cup \cdots \cup F_m} g(x) \leq \sum_{x \in A} g(x) . \end{equation*}

Taking the supremum over all such tuples \((t_1, \ldots, t_m)\) gives \(\sum_{j=1}^m s_j\) on the left: if every \(s_j\) is finite the \(t_j\) may be taken arbitrarily close to \(s_j\), and if some \(s_j = \infty\) the corresponding \(t_j\) may be taken arbitrarily large, forcing both sides to be \(\infty\). This proves the lemma.

Preliminary 1: \(\mu\) is a measure. (This is asserted without proof in 2.55; here is the verification.) Clearly \(\mu(\varnothing) = 0\), since the only finite subset of \(\varnothing\) is \(\varnothing\) and the empty sum is \(0\). Let \(E_1, E_2, \ldots\) be disjoint subsets of \(X\) with union \(E\). If \(F \subseteq E\) is finite, then \(F\) meets only finitely many \(E_i\), so for some \(n\) we have \(\sum_{x \in F} w(x) = \sum_{i=1}^n \sum_{x \in F \cap E_i} w(x) \leq \sum_{i=1}^n \mu(E_i) \leq \sum_{i=1}^{\infty} \mu(E_i)\), the first inequality because \(F \cap E_i\) is a finite subset of \(E_i\); taking the supremum over such \(F\) gives \(\mu(E) \leq \sum_{i=1}^{\infty} \mu(E_i)\). Conversely, for each \(n\), Preliminary 0 applied to the disjoint sets \(E_1, \ldots, E_n \subseteq E\) and to \(g = w\) gives \(\sum_{i=1}^n \mu(E_i) \leq \mu(E)\); letting \(n \to \infty\) gives \(\sum_{i=1}^{\infty} \mu(E_i) \leq \mu(E)\). Hence \(\mu\) is countably additive and is a measure.

Preliminary 2: \(\mu(\{x\}) = w(x)\) for each \(x \in X\), since the only finite subsets of \(\{x\}\) are \(\varnothing\) and \(\{x\}\), with sums \(0\) and \(w(x)\).

Preliminary 3: suppose \(c \in [0, \infty]\), \(A \subseteq X\), and \(g : X \to [0, \infty]\) satisfies \(g(x) \geq c \, w(x)\) for all \(x \in A\). Then for every finite \(F \subseteq A\) we have \(\sum_{x \in F} g(x) \geq \sum_{x \in F} c \, w(x) = c \sum_{x \in F} w(x)\), the last equality being the distributive law for a finite sum in \([0, \infty]\) (for \(c \in (0, \infty)\) it is standard; for \(c = 0\) both sides are \(0\); for \(c = \infty\) both sides are \(0\) if every \(w(x)\) with \(x \in F\) is \(0\) and are \(\infty\) otherwise). Hence

\begin{equation*} \sum_{x \in A} g(x) \geq \sup_{F} \; c \sum_{x \in F} w(x) = c \sum_{x \in A} w(x) . \end{equation*}

The last equality is the identity \(\sup_F (c \, t_F) = c \sup_F t_F\) for \(c \in [0, \infty]\) and \(t_F \in [0, \infty]\): for \(c \in (0, \infty)\) this is standard; for \(c = 0\) both sides are \(0\); and for \(c = \infty\) both sides are \(0\) if every \(t_F = 0\) and both are \(\infty\) otherwise.

Now we prove the two inequalities. Write \(S = \sum_{x \in X} w(x) f(x)\).

Step 1: \(\int f \, d\mu \geq S\).

Let \(F = \{x_1, \ldots, x_p\}\) be an arbitrary finite subset of \(X\), with \(x_1, \ldots, x_p\) distinct. Let \(\mathcal{P}\) be the \(\mathcal{S}\)-partition of \(X\) consisting of the \(p + 1\) sets

\begin{equation*} \{x_1\}, \ldots, \{x_p\}, \quad X \setminus F , \end{equation*}

which are disjoint subsets of \(X\) (hence in \(\mathcal{S}\)) whose union is \(X\). Using Preliminary 2 and \(\inf_{\{x_i\}} f = f(x_i)\), and discarding the final nonnegative term,

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \sum_{i=1}^p w(x_i) f(x_i) + \mu(X \setminus F) \inf_{X \setminus F} f \; \geq \; \sum_{x \in F} w(x) f(x) . \end{equation*}

Therefore \(\int f \, d\mu \geq \sum_{x \in F} w(x) f(x)\) by 3.3. Taking the supremum over all finite \(F \subseteq X\) gives \(\int f \, d\mu \geq S\).

Step 2: \(\int f \, d\mu \leq S\).

Let \(\mathcal{P}\) be an arbitrary \(\mathcal{S}\)-partition \(A_1, \ldots, A_m\) of \(X\), and fix \(j\). Put \(c_j = \inf_{A_j} f\). Then \(f(x) \geq c_j\) for every \(x \in A_j\), so \(w(x) f(x) \geq c_j w(x)\) for every \(x \in A_j\) (multiplication by \(w(x) \in [0, \infty]\) preserves the inequality under our conventions: if \(w(x) = 0\) both sides are \(0\)). Applying Preliminary 3 with \(g(x) = w(x) f(x)\), \(A = A_j\), \(c = c_j\) gives

\begin{equation*} \sum_{x \in A_j} w(x) f(x) \; \geq \; c_j \sum_{x \in A_j} w(x) \; = \; c_j \, \mu(A_j) = \mu(A_j) \inf_{A_j} f . \end{equation*}

(If \(A_j = \varnothing\), then \(\mu(A_j) = 0\) and both sides are \(0\).) Summing over \(j = 1, \ldots, m\),

\begin{equation*} \mathcal{L}(f, \mathcal{P}) = \sum_{j=1}^m \mu(A_j) \inf_{A_j} f \; \leq \; \sum_{j=1}^m \; \sum_{x \in A_j} w(x) f(x) \; \leq \; S . \end{equation*}

The last inequality is Preliminary 0 applied to the disjoint sets \(A_1, \ldots, A_m\) (whose union is \(X\)), with \(A = X\) and \(g(x) = w(x) f(x)\).

Since \(\mathcal{P}\) was arbitrary, 3.3 gives \(\int f \, d\mu \leq S\).

Combining Steps 1 and 2,

\begin{equation*} \int f \, d\mu = \sum_{x \in X} w(x) f(x) . \end{equation*}

Taking \(w(x) = 1\) for all \(x\) makes \(\mu\) counting measure on \(X\), and the result then contains Exercise 3A.5 (with \(X = \mathbb{Z}^+\)) as a special case; taking \(w = \chi_{\{c\}}\) makes \(\mu\) the Dirac measure \(\delta_c\) on the \(\sigma\)-algebra of all subsets of \(X\), recovering Exercise 3A.2 in that case.

For each \(k \in \mathbb{Z}^+\) define

\begin{equation*} f_k = \chi_{[k,\, k+1]}. \end{equation*}

Each \(f_k\) takes only the values \(0\) and \(1\), so each \(f_k\) is a simple function from \(\mathbb{R}\) to \([0, \infty)\), and each \(f_k\) is Borel measurable because \([k, k+1]\) is a Borel set.

Pointwise convergence to \(0\): fix \(x \in \mathbb{R}\) and choose \(N \in \mathbb{Z}^+\) with \(N > x\). If \(k \geq N\) then \(k > x\), so \(x \notin [k, k+1]\) and hence \(f_k(x) = 0\). Thus \(f_k(x) = 0\) for all \(k \geq N\), and in particular \(\lim_{k \to \infty} f_k(x) = 0\). Because \(x\) was arbitrary, the sequence converges pointwise to \(0\) on all of \(\mathbb{R}\).

Integrals: by 3.4 (the integral of a characteristic function is the measure of the set),

\begin{equation*} \int f_k \, d\lambda = \lambda\bigl([k, k+1]\bigr) = 1 \end{equation*}

for every \(k \in \mathbb{Z}^+\). Hence \(\lim_{k \to \infty} \int f_k \, d\lambda = 1\), while \(\int \bigl(\lim_{k \to \infty} f_k\bigr) d\lambda = \int 0 \, d\lambda = 0\).

This example shows that the hypothesis in the Monotone Convergence Theorem (3.11) that the sequence be increasing cannot simply be dropped: here the sequence is not monotone, and the limit of the integrals is not the integral of the limit.

Another example with the same behavior is \(f_k = \frac{1}{k}\chi_{[0,\,k]}\); then \(0 \leq f_k(x) \leq \frac{1}{k}\) for every \(x\), so \(f_k \to 0\) pointwise, while \(\int f_k \, d\lambda = \frac{1}{k} \cdot \lambda([0,k]) = 1\) for every \(k\). Note that in this second example the convergence to \(0\) is even uniform on \(\mathbb{R}\), which shows that uniform convergence alone does not allow the interchange of limits and integrals over a set of infinite measure.

Throughout, \(\chi_A f\) denotes the function on \(X\) whose value at \(x\) is \(f(x)\) if \(x \in A\) and \(0\) if \(x \notin A\) (so no product of the form \(0 \cdot \infty\) arises; with the usual convention \(0 \cdot \infty = 0\) this is the pointwise product).

Step 1: \(\nu\) is well defined. Fix \(A \in \mathcal{S}\). The function \(\chi_A f\) maps \(X\) to \([0, \infty]\), and for \(a \in \mathbb{R}\) we have

\begin{equation*} (\chi_A f)^{-1}\bigl((a, \infty]\bigr) = \begin{cases} A \cap f^{-1}\bigl((a, \infty]\bigr) & \text{if } a \geq 0,\\ X & \text{if } a < 0, \end{cases} \end{equation*}

because \(\chi_A f\) vanishes off \(A\) and agrees with \(f\) on \(A\). Both of these sets are in \(\mathcal{S}\) (using that \(A \in \mathcal{S}\) and that \(f\) is \(\mathcal{S}\)-measurable), so \(\chi_A f\) is \(\mathcal{S}\)-measurable by 2.52. Hence \(\int \chi_A f \, d\mu\) is defined and lies in \([0, \infty]\), and \(\nu(A) \in [0, \infty]\).

Step 2: \(\nu(\emptyset) = 0\). We have \(\chi_\emptyset f = 0\), the function that is identically \(0\). For every \(\mathcal{S}\)-partition \(A_1, \ldots, A_m\) of \(X\) we get \(\mathcal{L}(0, P) = \sum_{j=1}^m \mu(A_j) \cdot 0 = 0\), so \(\int 0 \, d\mu = 0\). Thus \(\nu(\emptyset) = 0\).

Step 3: countable additivity. Suppose \(A_1, A_2, \ldots\) is a disjoint sequence of sets in \(\mathcal{S}\) and let \(A = \bigcup_{k=1}^\infty A_k\); note \(A \in \mathcal{S}\). For \(n \in \mathbb{Z}^+\) define

\begin{equation*} g_n = \sum_{k=1}^{n} \chi_{A_k} f = \chi_{A_1 \cup \cdots \cup A_n} f, \end{equation*}

where the second equality holds because the \(A_k\) are disjoint: at a point \(x\) lying in exactly one \(A_k\) with \(k \leq n\) the sum equals \(f(x)\), and at a point lying in none of \(A_1, \ldots, A_n\) every term is \(0\). By Step 1 (applied to the set \(A_1 \cup \cdots \cup A_n \in \mathcal{S}\)), each \(g_n\) is \(\mathcal{S}\)-measurable.

The sequence \(g_1, g_2, \ldots\) is increasing, since \(g_{n+1} - g_n = \chi_{A_{n+1}} f \geq 0\) pointwise (more precisely, \(g_{n+1}(x) = g_n(x)\) unless \(x \in A_{n+1}\), in which case \(g_n(x) = 0 \le f(x) = g_{n+1}(x)\)). Moreover

\begin{equation*} \lim_{n \to \infty} g_n(x) = (\chi_A f)(x) \qquad \text{for every } x \in X: \end{equation*}

if \(x \in A\) then \(x \in A_k\) for exactly one \(k\), and \(g_n(x) = f(x)\) for all \(n \geq k\); if \(x \notin A\) then \(g_n(x) = 0\) for all \(n\), and \((\chi_A f)(x) = 0\) as well.

The Monotone Convergence Theorem (3.11) now gives

\begin{equation*} \lim_{n \to \infty} \int g_n \, d\mu = \int \chi_A f \, d\mu = \nu(A). \end{equation*}

On the other hand, additivity of integration for nonnegative functions (3.16), applied \(n - 1\) times, gives

\begin{equation*} \int g_n \, d\mu = \int \Bigl( \sum_{k=1}^n \chi_{A_k} f \Bigr) d\mu = \sum_{k=1}^{n} \int \chi_{A_k} f \, d\mu = \sum_{k=1}^{n} \nu(A_k). \end{equation*}

Combining the two displays,

\begin{equation*} \nu(A) = \lim_{n \to \infty} \sum_{k=1}^{n} \nu(A_k) = \sum_{k=1}^{\infty} \nu(A_k), \end{equation*}

the last equality being the definition of a sum of a series in \([0, \infty]\) (the partial sums are increasing, so the limit exists in \([0, \infty]\)).

Steps 1 through 3 show that \(\nu\) is a function from \(\mathcal{S}\) to \([0, \infty]\) with \(\nu(\emptyset) = 0\) that is countably additive. Hence \(\nu\) is a measure on \((X, \mathcal{S})\), as required by 2.54.

For \(n \in \mathbb{Z}^+\) define the partial sum

\begin{equation*} g_n = \sum_{k=1}^{n} f_k . \end{equation*}

Measurability of \(g_n\). It suffices to check that the sum of two \(\mathcal{S}\)-measurable functions \(u, v : X \to [0, \infty]\) is \(\mathcal{S}\)-measurable, and then induct. For \(a < 0\) we have \((u + v)^{-1}((a, \infty]) = X \in \mathcal{S}\). For \(a \geq 0\) we claim

\begin{equation*} (u + v)^{-1}\bigl((a, \infty]\bigr) = \bigcup_{q \in \mathbb{Q}} \Bigl( u^{-1}\bigl((q, \infty]\bigr) \cap v^{-1}\bigl((a - q, \infty]\bigr) \Bigr). \end{equation*}

Indeed, if \(u(x) > q\) and \(v(x) > a - q\) then \(u(x) + v(x) > a\). Conversely suppose \(u(x) + v(x) > a\). If \(u(x)\) and \(v(x)\) are both finite, pick a rational \(q\) with \(a - v(x) < q < u(x)\); then \(u(x) > q\) and \(v(x) > a - q\). If \(u(x) = \infty\), pick a rational \(q > a\); then \(u(x) > q\) and \(v(x) \geq 0 > a - q\). If \(v(x) = \infty\), pick a rational \(q < 0\); then \(u(x) \geq 0 > q\) and \(v(x) = \infty > a - q\). The claimed identity holds, the right side is a countable union of sets in \(\mathcal{S}\), and so \(u + v\) is \(\mathcal{S}\)-measurable by 2.52. By induction each \(g_n\) is \(\mathcal{S}\)-measurable.

Monotonicity and limit. Because each \(f_k\) is nonnegative, \(g_{n+1}(x) - g_n(x) = f_{n+1}(x) \geq 0\), so \(0 \leq g_1 \leq g_2 \leq \cdots\). Moreover, by the definition of the sum of a series of nonnegative terms as the limit of its partial sums (a limit that always exists in \([0, \infty]\) because the partial sums increase),

\begin{equation*} \lim_{n \to \infty} g_n(x) = \sum_{k=1}^{\infty} f_k(x) = f(x) \qquad \text{for every } x \in X . \end{equation*}

In particular \(f\) is \(\mathcal{S}\)-measurable by 2.53 (it is the pointwise supremum of the \(g_n\)).

Applying the theorems. Additivity of integration for nonnegative functions (3.16), used \(n - 1\) times, gives

\begin{equation*} \int g_n \, d\mu = \sum_{k=1}^{n} \int f_k \, d\mu \end{equation*}

for each \(n \in \mathbb{Z}^+\). The Monotone Convergence Theorem (3.11) applies to the increasing sequence \(g_1 \le g_2 \le \cdots\) of nonnegative \(\mathcal{S}\)-measurable functions with pointwise limit \(f\), giving

\begin{equation*} \int f \, d\mu = \lim_{n \to \infty} \int g_n \, d\mu = \lim_{n \to \infty} \sum_{k=1}^{n} \int f_k \, d\mu = \sum_{k=1}^{\infty} \int f_k \, d\mu , \end{equation*}

where the last equality is again the definition of the sum of a series of terms in \([0, \infty]\). This is the desired equation; note that both sides may equal \(\infty\), and the identity holds in \([0, \infty]\) with no integrability hypothesis.

Each \(|f_k|\) is \(\mathcal{S}\)-measurable: for \(a < 0\) we have \(|f_k|^{-1}((a, \infty]) = X\), and for \(a \geq 0\)

\begin{equation*} |f_k|^{-1}\bigl((a, \infty]\bigr) = f_k^{-1}\bigl((a, \infty)\bigr) \cup f_k^{-1}\bigl((-\infty, -a)\bigr) \in \mathcal{S}, \end{equation*}

so 2.52 applies. (Alternatively, \(|f_k| = f_k^+ + f_k^-\) in the notation of 3.17.)

Define \(g : X \to [0, \infty]\) by

\begin{equation*} g(x) = \sum_{k=1}^{\infty} |f_k(x)| . \end{equation*}

The function \(g\) is \(\mathcal{S}\)-measurable: it is the pointwise supremum of the increasing sequence of partial sums \(\sum_{k=1}^{n} |f_k|\), each of which is \(\mathcal{S}\)-measurable (finite sums of \([0,\infty]\)-valued \(\mathcal{S}\)-measurable functions are \(\mathcal{S}\)-measurable, as shown in the solution to 3A.10), so 2.53 applies. Now the previous exercise (3A.10), applied to the nonnegative \(\mathcal{S}\)-measurable functions \(|f_1|, |f_2|, \ldots\), gives

\begin{equation*} \int g \, d\mu = \sum_{k=1}^{\infty} \int |f_k| \, d\mu < \infty , \end{equation*}

the finiteness being the hypothesis.

Now set

\begin{equation*} E = g^{-1}\bigl([0, \infty)\bigr) = \{ x \in X : g(x) < \infty \} . \end{equation*}

Since \(g\) is \(\mathcal{S}\)-measurable, \(E \in \mathcal{S}\), and \(X \setminus E = g^{-1}(\{\infty\}) \in \mathcal{S}\).

Claim: \(\mu(X \setminus E) = 0\). Fix \(t \in (0, \infty)\). For every \(x \in X\) we have

\begin{equation*} g(x) \geq t \, \chi_{X \setminus E}(x), \end{equation*}

because \(g(x) = \infty \geq t\) when \(x \in X \setminus E\), while the right side is \(0\) when \(x \in E\). Hence by 3.8 (integration is order preserving), 3.20 (homogeneity, or directly from the definition) and 3.4 (the integral of a characteristic function),

\begin{equation*} \int g \, d\mu \geq \int t \, \chi_{X \setminus E} \, d\mu = t \, \mu(X \setminus E). \end{equation*}

Therefore

\begin{equation*} \mu(X \setminus E) \leq \frac{1}{t} \int g \, d\mu \end{equation*}

for every \(t \in (0, \infty)\). Because \(\int g \, d\mu < \infty\), letting \(t \to \infty\) gives \(\mu(X \setminus E) = 0\).

Finally, suppose \(x \in E\). Then \(\sum_{k=1}^{\infty} |f_k(x)| = g(x) < \infty\), so this is a convergent series of nonnegative real numbers. Writing \(s_n = \sum_{k=1}^{n} |f_k(x)|\), the sequence \(s_1, s_2, \ldots\) converges to the real number \(g(x)\), so

\begin{equation*} |f_n(x)| = s_n - s_{n-1} \to g(x) - g(x) = 0 \end{equation*}

as \(n \to \infty\) (with \(s_0 = 0\)). Hence \(\lim_{k \to \infty} f_k(x) = 0\) for every \(x \in E\).

Thus \(E \in \mathcal{S}\) satisfies \(\mu(X \setminus E) = 0\) and \(f_k(x) \to 0\) for every \(x \in E\), completing the proof. In the language of Chapter 3, the conclusion says that \(f_k \to 0\) almost everywhere.

The function must take values in \((0, \infty)\), so it must be finite at every point, yet fail to be integrable over every interval. We build it as a series of poles placed at the rationals, with coefficients decaying fast enough that the series converges off a set of measure \(0\); on that exceptional set we simply redefine the function to be \(1\).

Construction. Let \(r_1, r_2, \ldots\) be an enumeration of \(\mathbb{Q}\). For \(k \in \mathbb{Z}^+\) define \(h_k : \mathbb{R} \to (0, \infty]\) by

\begin{equation*} h_k(x) = \frac{1}{|x - r_k|} \quad \text{if } x \neq r_k, \qquad h_k(r_k) = \infty . \end{equation*}

Each \(h_k\) is Borel measurable: if \(a > 0\) then \(h_k^{-1}((a, \infty]) = (r_k - \frac{1}{a},\, r_k + \frac{1}{a})\), which is open, and if \(a \leq 0\) then \(h_k^{-1}((a, \infty]) = \mathbb{R}\); now apply 2.52. Define \(g : \mathbb{R} \to (0, \infty]\) by

\begin{equation*} g(x) = \sum_{k=1}^{\infty} 8^{-k} h_k(x). \end{equation*}

Each partial sum is Borel measurable and \(g\) is their pointwise supremum, so \(g\) is Borel measurable by 2.53. Every term is strictly positive, so \(g(x) > 0\) for all \(x\).

The set where \(g\) is infinite is null. Let

\begin{equation*} B = \bigcap_{n=1}^{\infty} \; \bigcup_{k=n}^{\infty} \bigl( r_k - 4^{-k},\, r_k + 4^{-k} \bigr), \end{equation*}

a Borel set, consisting of those \(x\) with \(|x - r_k| < 4^{-k}\) for infinitely many \(k\). For each \(n\), monotonicity and countable subadditivity of \(\lambda\) (2.57 and 2.58) give

\begin{equation*} \lambda(B) \leq \sum_{k=n}^{\infty} 2 \cdot 4^{-k} = \frac{8}{3} \cdot 4^{-n} . \end{equation*}

Letting \(n \to \infty\) yields \(\lambda(B) = 0\).

Now suppose \(x \notin B \cup \mathbb{Q}\). Since \(x \notin B\) there is \(N \in \mathbb{Z}^+\) with \(|x - r_k| \geq 4^{-k}\) for all \(k \geq N\), whence

\begin{equation*} 8^{-k} h_k(x) = \frac{8^{-k}}{|x - r_k|} \leq 8^{-k} \cdot 4^{k} = 2^{-k} \qquad (k \geq N). \end{equation*}

Since \(x \notin \mathbb{Q}\), each of the finitely many terms with \(k < N\) is a finite real number. Therefore

\begin{equation*} g(x) \leq \sum_{k=1}^{N-1} 8^{-k} h_k(x) + \sum_{k=N}^{\infty} 2^{-k} < \infty . \end{equation*}

Consequently the Borel set \(S = g^{-1}(\{\infty\})\) satisfies \(S \subseteq B \cup \mathbb{Q}\), and hence

\begin{equation*} \lambda(S) \leq \lambda(B) + \lambda(\mathbb{Q}) = 0 . \end{equation*}

Definition of \(f\). Define \(f : \mathbb{R} \to (0, \infty)\) by

\begin{equation*} f(x) = \begin{cases} g(x) & \text{if } x \notin S, \\ 1 & \text{if } x \in S. \end{cases} \end{equation*}

For \(x \notin S\) we have \(0 < g(x) < \infty\), so indeed \(f\) takes values in \((0, \infty)\). For \(a \in \mathbb{R}\),

\begin{equation*} f^{-1}\bigl((a, \infty)\bigr) = \Bigl( g^{-1}\bigl((a, \infty]\bigr) \setminus S \Bigr) \cup \begin{cases} S & \text{if } a < 1, \\ \emptyset & \text{if } a \geq 1, \end{cases} \end{equation*}

which is a Borel set; so \(f\) is Borel measurable by 2.52.

Verification. Fix a nonempty open interval \(I \subseteq \mathbb{R}\).

Step 1: \(\int \chi_I h_j \, d\lambda = \infty\) whenever \(r_j \in I\). Choose \(\delta > 0\) with \((r_j - \delta, r_j + \delta) \subseteq I\). For \(m = 0, 1, 2, \ldots\) put

\begin{equation*} A_m = \bigl( r_j + \delta 2^{-m-2},\; r_j + \delta 2^{-m-1} \bigr] . \end{equation*}

These are disjoint Borel subsets of \((r_j, r_j + \delta/2] \subseteq I\), with \(\lambda(A_m) = \delta 2^{-m-2}\). For \(x \in A_m\) we have \(0 < x - r_j \leq \delta 2^{-m-1}\), so \(h_j(x) \geq 2^{m+1}/\delta\). Hence for each \(M\),

\begin{equation*} \sum_{m=0}^{M} \frac{2^{m+1}}{\delta} \, \chi_{A_m}(x) \leq \chi_I(x) h_j(x) \qquad \text{for all } x \in \mathbb{R}, \end{equation*}

so 3.9 gives

\begin{equation*} \int \chi_I h_j \, d\lambda \geq \sum_{m=0}^{M} \frac{2^{m+1}}{\delta} \cdot \delta 2^{-m-2} = \frac{M+1}{2}. \end{equation*}

Letting \(M \to \infty\) proves Step 1.

Step 2: \(\int \chi_I g \, d\lambda = \infty\). Since \(\mathbb{Q}\) is dense and \(I\) is a nonempty open interval, there is \(j\) with \(r_j \in I\). The nonnegative Borel functions \(8^{-k}\chi_I h_k\) sum pointwise to \(\chi_I g\), so by 3A.10 and homogeneity of integration (3.20),

\begin{equation*} \int \chi_I g \, d\lambda = \sum_{k=1}^{\infty} 8^{-k} \int \chi_I h_k \, d\lambda \geq 8^{-j} \int \chi_I h_j \, d\lambda = \infty . \end{equation*}

Step 3: \(\int \chi_{I \cap S} \, g \, d\lambda = 0\). The function \(\chi_{I \cap S} \, g\) equals \(\infty\) on \(I \cap S\) and \(0\) elsewhere. Suppose \(A_1, \ldots, A_m\) are disjoint Borel sets and \(c_1, \ldots, c_m \in [0, \infty)\) satisfy \(\sum_{j=1}^m c_j \chi_{A_j} \leq \chi_{I \cap S} \, g\) pointwise. If \(c_j > 0\) and \(x \in A_j\), then the left side is positive at \(x\), forcing \(x \in I \cap S\); thus \(A_j \subseteq S\) and \(\lambda(A_j) = 0\). Therefore \(\sum_{j=1}^m c_j \lambda(A_j) = 0\) for every such choice, and 3.9 gives \(\int \chi_{I \cap S} \, g \, d\lambda = 0\).

Step 4: conclusion. Pointwise \(\chi_I g = \chi_{I \cap S} \, g + \chi_{I \setminus S} \, g\), so additivity of integration (3.16) together with Steps 2 and 3 gives

\begin{equation*} \infty = \int \chi_I g \, d\lambda = 0 + \int \chi_{I \setminus S} \, g \, d\lambda , \end{equation*}

so \(\int \chi_{I \setminus S} \, g \, d\lambda = \infty\). Finally \(\chi_I f \geq \chi_{I \setminus S} \, g\) pointwise, since the two agree on \(I \setminus S\) and the right side is \(0\) off \(I \setminus S\). Hence by 3.8,

\begin{equation*} \int \chi_I f \, d\lambda = \infty . \end{equation*}

Because \(I\) was an arbitrary nonempty open interval, the Borel measurable function \(f : \mathbb{R} \to (0, \infty)\) has the required property.

Let \(\lambda\) denote Lebesgue measure on \(\mathbb{R}\), with \(\mathcal{S}\) the \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}\). For \(k \in \mathbb{Z}^+\) define \(f_k : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f_k = -\chi_{[k, \infty)}, \qquad \text{that is,} \qquad f_k(x) = \begin{cases} -1 & \text{if } x \geq k, \\ 0 & \text{if } x < k. \end{cases} \end{equation*}

Each \(f_k\) is a Borel measurable simple function.

The sequence is increasing. If \(x < k\) then \(f_k(x) = 0 = f_{k+1}(x)\). If \(k \leq x < k+1\) then \(f_k(x) = -1 < 0 = f_{k+1}(x)\). If \(x \geq k+1\) then \(f_k(x) = -1 = f_{k+1}(x)\). In every case \(f_k(x) \leq f_{k+1}(x)\), so \(f_1 \leq f_2 \leq \cdots\). What fails is only the hypothesis \(f_k \geq 0\).

The pointwise limit is \(0\). Fix \(x \in \mathbb{R}\) and take \(N \in \mathbb{Z}^+\) with \(N > x\). Then \(f_k(x) = 0\) for all \(k \geq N\), so

\begin{equation*} f(x) := \lim_{k \to \infty} f_k(x) = 0 \qquad \text{for every } x \in \mathbb{R}. \end{equation*}

Hence \(\int f \, d\lambda = \int 0 \, d\lambda = 0\).

The integrals. In the notation of 3.17, \(f_k^{\,+} = 0\) and \(f_k^{\,-} = \chi_{[k,\infty)}\). Thus \(\int f_k^{\,+} d\lambda = 0\) is finite, so \(\int f_k \, d\lambda\) is defined by 3.18, and by 3.4

\begin{equation*} \int f_k \, d\lambda = \int f_k^{\,+} \, d\lambda - \int f_k^{\,-} \, d\lambda = 0 - \lambda\bigl([k, \infty)\bigr) = -\infty . \end{equation*}

Therefore

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\lambda = -\infty \neq 0 = \int f \, d\lambda , \end{equation*}

and the conclusion of the Monotone Convergence Theorem fails for this increasing sequence of measurable functions that are not nonnegative.

Remark. The failure is caused entirely by the infinite negative mass at the bottom of the sequence, and no example can avoid that. Indeed, suppose \(f_1 \leq f_2 \leq \cdots\) are \(\mathcal{S}\)-measurable real-valued functions with \(\int |f_1| \, d\mu < \infty\). Then \(f_k - f_1\) is a nonnegative increasing sequence converging pointwise to \(f - f_1\), so 3.11 gives \(\int (f_k - f_1) \, d\mu \to \int (f - f_1) \, d\mu\); adding the finite number \(\int f_1 \, d\mu\) to both sides recovers \(\int f_k \, d\mu \to \int f \, d\mu\). (This is the increasing half of Exercise 20 in this section.) In particular, in any counterexample \(\int |f_1| \, d\mu\) must be infinite, as it is above: \(\int |f_1| \, d\lambda = \lambda([1, \infty)) = \infty\). So the nonnegativity hypothesis in 3.11 cannot simply be deleted, but it can be weakened to \(\int |f_1| \, d\mu < \infty\).

Let \(\lambda\) denote Lebesgue measure on \(\mathbb{R}\), with the Borel \(\sigma\)-algebra. For \(k \in \mathbb{Z}^+\) define

\begin{equation*} f_k = \chi_{[k, \infty)}, \qquad \text{that is,} \qquad f_k(x) = \begin{cases} 1 & \text{if } x \geq k, \\ 0 & \text{if } x < k. \end{cases} \end{equation*}

Each \(f_k\) is a nonnegative Borel measurable simple function, so every hypothesis of 3.11 holds except that the sequence decreases rather than increases.

The sequence is decreasing. Since \([k+1, \infty) \subseteq [k, \infty)\), we have \(\chi_{[k+1,\infty)} \leq \chi_{[k,\infty)}\) pointwise, that is \(f_1 \geq f_2 \geq \cdots \geq 0\).

The pointwise limit is \(0\). Fix \(x \in \mathbb{R}\) and choose \(N \in \mathbb{Z}^+\) with \(N > x\). Then \(x \notin [k, \infty)\) for every \(k \geq N\), so \(f_k(x) = 0\) for all \(k \geq N\). Hence

\begin{equation*} f(x) := \lim_{k \to \infty} f_k(x) = 0 \qquad \text{for every } x \in \mathbb{R}, \end{equation*}

and therefore \(\int f \, d\lambda = 0\).

The integrals. By 3.4,

\begin{equation*} \int f_k \, d\lambda = \lambda\bigl([k, \infty)\bigr) = \infty \end{equation*}

for every \(k \in \mathbb{Z}^+\). Thus

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\lambda = \infty \neq 0 = \int f \, d\lambda , \end{equation*}

so the conclusion of the Monotone Convergence Theorem fails for this decreasing sequence.

Remark. The failure is driven entirely by the infinite mass of \(f_1\), and it must be. If \(f_1 \geq f_2 \geq \cdots \geq 0\) are \(\mathcal{S}\)-measurable with \(f_1\) real-valued and \(\int f_1 \, d\mu < \infty\), then \(f_1 - f_1, f_1 - f_2, \ldots\) is an increasing sequence of nonnegative \(\mathcal{S}\)-measurable functions with pointwise limit \(f_1 - f\), so 3.11 gives \(\int (f_1 - f_k) \, d\mu \to \int (f_1 - f) \, d\mu\); subtracting this from the finite number \(\int f_1 \, d\mu\) yields \(\int f_k \, d\mu \to \int f \, d\mu\). This is the decreasing half of Exercise 20 in this section, and its hypothesis \(\int |f_1| \, d\mu < \infty\) fails in the example above, where \(\int f_1 \, d\lambda = \infty\).

In particular, no counterexample can have \(f_1\) bounded on a space of finite measure: if \(\mu(X) < \infty\) and \(c = \sup_X f_1 < \infty\), then \(f_1 \leq c \, \chi_X\) pointwise, so 3.8 and 3.4 give \(\int f_1 \, d\mu \leq c \, \mu(X) < \infty\). But finite measure alone is not enough. Let \(X = (0, 1]\) with the Borel \(\sigma\)-algebra and Lebesgue measure, and define

\begin{equation*} f_k(x) = \begin{cases} 1/x & \text{if } 0 < x \leq 1/k, \\ 0 & \text{if } 1/k < x \leq 1. \end{cases} \end{equation*}

Each \(f_k\) is Borel measurable, and \(f_1 \geq f_2 \geq \cdots \geq 0\) because the sets \((0, 1/k]\) decrease while the formula \(1/x\) does not depend on \(k\). For each \(x \in (0,1]\) we have \(f_k(x) = 0\) once \(1/k < x\), so \(f_k \to 0\) pointwise. Yet \(\int f_k \, d\lambda = \infty\) for every \(k\): the sets \(A_m = (2^{-m-1}/k, \; 2^{-m}/k]\) for \(m = 0, 1, 2, \ldots\) are disjoint subsets of \((0, 1/k]\) with \(\lambda(A_m) = 2^{-m-1}/k\), and \(f_k \geq k \, 2^{m}\) on \(A_m\), so \(\sum_{m=0}^{M} k \, 2^{m} \chi_{A_m} \leq f_k\) pointwise and 3.9 gives \(\int f_k \, d\lambda \geq \sum_{m=0}^{M} k \, 2^{m} \cdot 2^{-m-1}/k = (M+1)/2\) for every \(M\). Here \(\lambda\bigl((0,1]\bigr) = 1 < \infty\), but \(f_1\) is unbounded and \(\int f_1 \, d\lambda = \infty\), so Exercise 20 again does not apply.

Throughout, \(\mathcal{B}\) denotes the \(\sigma\)-algebra of Borel subsets of \(\mathbb{R}\), and \(\lambda\) is Lebesgue measure on \(\mathcal{B}\); thus \(\lambda(A) = |A|\) (outer measure) for every \(A \in \mathcal{B}\), by 2.68 and 2.69.

Preliminaries on the two changes of variable.

Fix \(t \in \mathbb{R}\) and let \(\tau : \mathbb{R} \to \mathbb{R}\) be \(\tau(x) = x - t\). For \(A \subseteq \mathbb{R}\) we have \(\tau^{-1}(A) = t + A\). Because \(\tau\) is continuous it is Borel measurable (2.41), so

\begin{equation*} A \in \mathcal{B} \implies t + A \in \mathcal{B}. \end{equation*}

Moreover \(A \mapsto t + A\) is a bijection of \(\mathcal{B}\) onto \(\mathcal{B}\) (its inverse is \(A \mapsto -t + A\)), it preserves disjointness, and it satisfies \((t + A_1) \cup \cdots \cup (t + A_m) = t + (A_1 \cup \cdots \cup A_m)\). Finally \(\lambda(t + A) = \lambda(A)\) for all \(A \in \mathcal{B}\), by 2.7.

Now fix \(t \in \mathbb{R} \setminus \{0\}\) and let \(\sigma(x) = tx\), so that \(\sigma^{-1}(A) = \frac{1}{t} A\), where \(cA := \{ca : a \in A\}\). Again \(\sigma\) is continuous, hence Borel measurable, so \(A \in \mathcal{B}\) implies \(\frac{1}{t}A \in \mathcal{B}\). The map \(A \mapsto \frac{1}{t}A\) is a bijection of \(\mathcal{B}\) onto \(\mathcal{B}\) (inverse \(A \mapsto tA\)), preserves disjointness and unions, and by Exercise 2 in Section 2A we have \(|cA| = |c| \, |A|\) for all \(c \in \mathbb{R}\) and \(A \subseteq \mathbb{R}\); hence

\begin{equation*} \lambda\Bigl(\tfrac{1}{t}A\Bigr) = \frac{1}{|t|}\,\lambda(A) \qquad \text{for all } A \in \mathcal{B}. \end{equation*}

Measurability of \(f_t\). In case (a), for every Borel subset \(B\) of \([-\infty,\infty]\),

\begin{equation*} f_t^{-1}(B) = \{x : f(x - t) \in B\} = t + f^{-1}(B), \end{equation*}

which is a Borel subset of \(\mathbb{R}\) by the above. Thus \(f_t\) is Borel measurable. In case (b), \(f_t^{-1}(B) = \{x : f(tx) \in B\} = \frac{1}{t} f^{-1}(B)\), again Borel. So in both cases \(f_t\) is Borel measurable.

(a) The nonnegative case. Assume first \(f \ge 0\). Let \(P\) be a Borel partition \(A_1, \dots, A_m\) of \(\mathbb{R}\), and let \(P_t\) denote \(t + A_1, \dots, t + A_m\), which by the preliminaries is again a Borel partition of \(\mathbb{R}\). Because \(x \in t + A_j\) if and only if \(x - t \in A_j\),

\begin{equation*} \inf_{t + A_j} f_t = \inf\{f(x - t) : x \in t + A_j\} = \inf\{f(y) : y \in A_j\} = \inf_{A_j} f . \end{equation*}

Combining this with \(\lambda(t + A_j) = \lambda(A_j)\) gives

\begin{equation*} L(f_t, P_t) = \sum_{j=1}^{m} \lambda(t + A_j) \inf_{t+A_j} f_t = \sum_{j=1}^{m} \lambda(A_j) \inf_{A_j} f = L(f, P). \end{equation*}

Since \(P \mapsto P_t\) is a bijection from the set of Borel partitions of \(\mathbb{R}\) onto itself, the two sets of lower Lebesgue sums \(\{L(f,P) : P\}\) and \(\{L(f_t, Q) : Q\}\) coincide. Taking suprema and using the definition 3.3 of the integral,

\begin{equation*} \int f_t \, d\lambda = \int f \, d\lambda . \end{equation*}

(a) The general case. For arbitrary \(f : \mathbb{R} \to [-\infty,\infty]\) note the pointwise identities

\begin{equation*} (f_t)^+ (x) = \max\{f(x-t), 0\} = (f^+)_t(x), \qquad (f_t)^-(x) = (f^-)_t(x). \end{equation*}

By the nonnegative case applied to \(f^+\) and to \(f^-\),

\begin{equation*} \int (f_t)^+ \, d\lambda = \int f^+ \, d\lambda \qquad \text{and} \qquad \int (f_t)^- \, d\lambda = \int f^- \, d\lambda . \end{equation*}

Because \(\int f \, d\lambda\) is defined, at least one of \(\int f^+ \, d\lambda\), \(\int f^- \, d\lambda\) is finite; hence at least one of \(\int (f_t)^+ \, d\lambda\), \(\int (f_t)^- \, d\lambda\) is finite, so \(\int f_t \, d\lambda\) is defined, and by 3.18

\begin{equation*} \int f_t \, d\lambda = \int (f_t)^+ d\lambda - \int (f_t)^- d\lambda = \int f^+ d\lambda - \int f^- d\lambda = \int f \, d\lambda . \end{equation*}

(b) The nonnegative case. Fix \(t \ne 0\) and assume \(f \ge 0\). Let \(P\) be a Borel partition \(A_1, \dots, A_m\) of \(\mathbb{R}\) and let \(P^{(t)}\) denote \(\frac{1}{t}A_1, \dots, \frac{1}{t}A_m\), a Borel partition of \(\mathbb{R}\) by the preliminaries. Since \(x \in \frac{1}{t}A_j\) if and only if \(tx \in A_j\),

\begin{equation*} \inf_{\frac{1}{t}A_j} f_t = \inf\{f(tx) : x \in \tfrac{1}{t}A_j\} = \inf\{f(y) : y \in A_j\} = \inf_{A_j} f . \end{equation*}

Therefore

\begin{equation*} L\bigl(f_t, P^{(t)}\bigr) = \sum_{j=1}^{m} \lambda\Bigl(\tfrac{1}{t}A_j\Bigr) \inf_{A_j} f = \frac{1}{|t|} \sum_{j=1}^{m} \lambda(A_j) \inf_{A_j} f = \frac{1}{|t|} L(f, P). \end{equation*}

As \(P \mapsto P^{(t)}\) is a bijection of the set of Borel partitions of \(\mathbb{R}\) onto itself, the set of lower Lebesgue sums for \(f_t\) is exactly \(\frac{1}{|t|}\) times the set of lower Lebesgue sums for \(f\). Since multiplication by the positive constant \(\frac{1}{|t|}\) commutes with taking suprema in \([0,\infty]\), definition 3.3 gives

\begin{equation*} \int f_t \, d\lambda = \frac{1}{|t|} \int f \, d\lambda . \end{equation*}

(b) The general case. Again \((f_t)^{\pm} = (f^{\pm})_t\) pointwise, so by the nonnegative case

\begin{equation*} \int (f_t)^{\pm} \, d\lambda = \frac{1}{|t|} \int f^{\pm} \, d\lambda . \end{equation*}

Since \(\frac{1}{|t|} \in (0,\infty)\), one of these is finite exactly when the corresponding integral of \(f^{\pm}\) is finite; hence \(\int f_t \, d\lambda\) is defined and

\begin{equation*} \int f_t \, d\lambda = \frac{1}{|t|} \int f^+ d\lambda - \frac{1}{|t|} \int f^- d\lambda = \frac{1}{|t|} \int f \, d\lambda , \end{equation*}

the last step being valid in \([-\infty,\infty]\) because at most one of the two terms is infinite.

First note that every \(\mathcal{S}\)-measurable function is \(\mathcal{T}\)-measurable: if \(g\) is \(\mathcal{S}\)-measurable and \(B\) is a Borel subset of \([-\infty,\infty]\), then \(g^{-1}(B) \in \mathcal{S} \subseteq \mathcal{T}\). In particular \(f\) is \(\mathcal{T}\)-measurable, so \(\int f \, d\mu_2\) makes sense.

One inequality is immediate. Every \(\mathcal{S}\)-partition \(A_1, \dots, A_m\) of \(X\) is also a \(\mathcal{T}\)-partition of \(X\), and since \(\mu_1(A_j) = \mu_2(A_j)\) for each \(j\) the two lower Lebesgue sums agree:

\begin{equation*} \sum_{j=1}^{m} \mu_1(A_j) \inf_{A_j} f = \sum_{j=1}^{m} \mu_2(A_j) \inf_{A_j} f . \end{equation*}

Hence the supremum defining \(\int f \, d\mu_2\) is taken over a collection of numbers containing all the lower Lebesgue sums that define \(\int f \, d\mu_1\), so by 3.3

\begin{equation*} \int f \, d\mu_1 \le \int f \, d\mu_2 . \end{equation*}

The reverse inequality is the real content, and it is where the \(\mathcal{S}\)-measurability of \(f\) (as opposed to mere \(\mathcal{T}\)-measurability) is used. We prove the equality outright by approximation.

Step 1: the result for simple \(\mathcal{S}\)-measurable functions. Suppose \(h : X \to [0,\infty)\) is a simple \(\mathcal{S}\)-measurable function. Write \(h\) in its standard representation

\begin{equation*} h = \sum_{k=1}^{n} c_k \chi_{E_k}, \end{equation*}

where \(c_1, \dots, c_n \in [0,\infty)\) are the distinct values of \(h\) and \(E_k = h^{-1}(\{c_k\})\); thus \(E_1, \dots, E_n\) are disjoint sets in \(\mathcal{S}\) whose union is \(X\). Because \(\mathcal{S} \subseteq \mathcal{T}\), these sets also lie in \(\mathcal{T}\), so 3.7 applies in each of the two measure spaces and gives

\begin{equation*} \int h \, d\mu_1 = \sum_{k=1}^{n} c_k \, \mu_1(E_k) = \sum_{k=1}^{n} c_k \, \mu_2(E_k) = \int h \, d\mu_2 , \end{equation*}

the middle equality holding because each \(E_k \in \mathcal{S}\), where \(\mu_1\) and \(\mu_2\) agree.

Step 2: an increasing sequence of nonnegative simple \(\mathcal{S}\)-measurable functions converging to \(f\). By 2.89, applied to the measurable space \((X, \mathcal{S})\) and to \(f\), there is a sequence \(u_1, u_2, \dots\) of simple \(\mathcal{S}\)-measurable functions from \(X\) to \(\mathbb{R}\) such that \(|u_k(x)| \le |u_{k+1}(x)| \le |f(x)|\) for all \(k \in \mathbb{Z}^+\) and all \(x \in X\), and \(\lim_{k \to \infty} u_k(x) = f(x)\) for every \(x \in X\). Define

\begin{equation*} f_k = \max\{u_1^+, u_2^+, \dots, u_k^+\} \qquad (k \in \mathbb{Z}^+), \end{equation*}

where \(u^+ = \max\{u, 0\}\). Each \(f_k\) takes only finitely many values (each value is a value of some \(u_j^+\)), and for \(a \in \mathbb{R}\) we have \(f_k^{-1}\bigl((a,\infty]\bigr) = \bigcup_{j=1}^{k} (u_j^+)^{-1}\bigl((a,\infty]\bigr) \in \mathcal{S}\), so by 2.52 each \(f_k\) is a simple nonnegative \(\mathcal{S}\)-measurable function. By construction \(f_1 \le f_2 \le \cdots\). Since \(f \ge 0\) we have \(|f| = f\), so \(u_j^+ \le |u_j| \le f\) for every \(j\), whence \(f_k \le f\); and \(f_k \ge u_k^+ \to f^+ = f\) pointwise. Therefore

\begin{equation*} 0 \le f_1 \le f_2 \le \cdots \quad \text{and} \quad \lim_{k \to \infty} f_k(x) = f(x) \ \text{ for every } x \in X . \end{equation*}

(This is the same use of 2.89 that the book makes in its proof of 3.16.)

Each \(f_k\) is also a simple \(\mathcal{T}\)-measurable function, and \(f\) is \(\mathcal{T}\)-measurable. Hence the Monotone Convergence Theorem 3.11 applies both in the measure space \((X, \mathcal{S}, \mu_1)\) and in the measure space \((X, \mathcal{T}, \mu_2)\). Using Step 1 in the middle,

\begin{equation*} \int f \, d\mu_1 = \lim_{k \to \infty} \int f_k \, d\mu_1 = \lim_{k \to \infty} \int f_k \, d\mu_2 = \int f \, d\mu_2 , \end{equation*}

which completes the proof.

For each \(k \in \mathbb{Z}^+\) define \(g_k : X \to [0, \infty]\) by

\begin{equation*} g_k(x) = \inf\{f_j(x) : j \ge k\} = \inf\{f_k(x), f_{k+1}(x), \dots\}. \end{equation*}

(a) Each \(g_k\) is the pointwise infimum of the sequence of \(\mathcal{S}\)-measurable functions \(f_k, f_{k+1}, \dots\), hence \(g_k\) is \(\mathcal{S}\)-measurable by 2.53. For each \(x \in X\) the sequence \(g_1(x) \le g_2(x) \le \cdots\) is increasing (the infimum is taken over a smaller set as \(k\) grows), so

\begin{equation*} f(x) = \liminf_{k \to \infty} f_k(x) = \lim_{k \to \infty} g_k(x) = \sup\{g_k(x) : k \in \mathbb{Z}^+\} . \end{equation*}

Thus \(f\) is the pointwise supremum of the sequence \(g_1, g_2, \dots\) of \(\mathcal{S}\)-measurable functions, so \(f\) is \(\mathcal{S}\)-measurable by 2.53.

(b) By part (a), \(0 \le g_1 \le g_2 \le \cdots\) is an increasing sequence of nonnegative \(\mathcal{S}\)-measurable functions with \(\lim_{k \to \infty} g_k(x) = f(x)\) for every \(x \in X\). Hence the Monotone Convergence Theorem 3.11 gives

\begin{equation*} \int f \, d\mu = \lim_{k \to \infty} \int g_k \, d\mu . \end{equation*}

Fix \(k \in \mathbb{Z}^+\). For every \(j \ge k\) we have \(g_k(x) \le f_j(x)\) for all \(x \in X\), so 3.8 gives \(\int g_k \, d\mu \le \int f_j \, d\mu\) for every \(j \ge k\). Taking the infimum over \(j \ge k\),

\begin{equation*} \int g_k \, d\mu \le \inf\Bigl\{ \int f_j \, d\mu : j \ge k \Bigr\} . \end{equation*}

Now let \(k \to \infty\). The left side tends to \(\int f \, d\mu\), and by the definition of \(\liminf\) the right side tends to \(\liminf_{k \to \infty} \int f_k \, d\mu\). Since the inequality is preserved in the limit,

\begin{equation*} \int f \, d\mu \le \liminf_{k \to \infty} \int f_k \, d\mu , \end{equation*}

as desired.

(c) Let \(X = [0,1]\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of Borel subsets of \([0,1]\), and let \(\mu = \lambda\) be Lebesgue measure on \(\mathcal{S}\), so \(\mu(X) = 1 < \infty\). Define

\begin{equation*} f_k = \begin{cases} \chi_{[0, 1/2]} & \text{if } k \text{ is odd,} \\ \chi_{(1/2, \, 1]} & \text{if } k \text{ is even.} \end{cases} \end{equation*}

Each \(f_k\) is a nonnegative Borel measurable function with \(0 \le f_k \le 1\), so the family \(\{f_k\}_{k \in \mathbb{Z}^+}\) is uniformly bounded.

For each \(x \in [0,1]\), the sequence \(f_1(x), f_2(x), \dots\) takes the value \(0\) for infinitely many \(k\) (namely for all even \(k\) if \(x \in [0, 1/2]\), and for all odd \(k\) if \(x \in (1/2, 1]\)). Hence \(\inf\{f_j(x) : j \ge k\} = 0\) for every \(k\), and therefore

\begin{equation*} f(x) = \liminf_{k \to \infty} f_k(x) = 0 \qquad \text{for every } x \in [0,1]. \end{equation*}

Consequently \(\int f \, d\lambda = 0\). On the other hand, by 3.4,

\begin{equation*} \int f_k \, d\lambda = \lambda\bigl([0, \tfrac{1}{2}]\bigr) = \tfrac{1}{2} \quad \text{for odd } k, \qquad \int f_k \, d\lambda = \lambda\bigl((\tfrac{1}{2}, 1]\bigr) = \tfrac{1}{2} \quad \text{for even } k . \end{equation*}

Thus \(\liminf_{k \to \infty} \int f_k \, d\lambda = \frac{1}{2}\), and

\begin{equation*} 0 = \int f \, d\lambda < \liminf_{k \to \infty} \int f_k \, d\lambda = \tfrac{1}{2}, \end{equation*}

so the inequality in (b) is strict here.

Take

\begin{equation*} x_k = \frac{(-1)^k}{k} \qquad (k \in \mathbb{Z}^+), \end{equation*}

so that \(x_1 = -1\), \(x_2 = \frac{1}{2}\), \(x_3 = -\frac{1}{3}\), and so on.

The partial sums converge. The sequence \(\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \dots\) decreases to \(0\), so the alternating series test shows that \(\lim_{n \to \infty} \sum_{k=1}^{n} x_k\) exists in \(\mathbb{R}\); in fact

\begin{equation*} \lim_{n \to \infty} \sum_{k=1}^{n} \frac{(-1)^k}{k} = -\ln 2 . \end{equation*}

(For an explicit argument that the limit exists, write \(s_n = \sum_{k=1}^{n} x_k\). The odd partial sums increase, since \(s_{2n+1} - s_{2n-1} = \frac{1}{2n} - \frac{1}{2n+1} > 0\); the even partial sums decrease, since \(s_{2n+2} - s_{2n} = -\frac{1}{2n+1} + \frac{1}{2n+2} < 0\); and \(s_{2n} - s_{2n+1} = \frac{1}{2n+1} > 0\), so every even partial sum exceeds every odd one. Thus \(s_1 \le s_3 \le \cdots\) is increasing and bounded above by \(s_2\), and \(s_2 \ge s_4 \ge \cdots\) is decreasing and bounded below by \(s_1\); both subsequences converge in \(\mathbb{R}\), and \(s_{2n} - s_{2n+1} = \frac{1}{2n+1} \to 0\) forces the two limits to agree. Hence \(\lim_{n \to \infty} s_n\) exists in \(\mathbb{R}\).)

The integral is not defined. Recall from 3.18 that \(\int x \, d\mu\) is defined only when at least one of \(\int x^+ \, d\mu\) and \(\int x^- \, d\mu\) is finite. Here

\begin{equation*} x^+(k) = \begin{cases} \dfrac{1}{k} & \text{if } k \text{ is even,} \\[4pt] 0 & \text{if } k \text{ is odd,} \end{cases} \qquad x^-(k) = \begin{cases} 0 & \text{if } k \text{ is even,} \\[4pt] \dfrac{1}{k} & \text{if } k \text{ is odd.} \end{cases} \end{equation*}

Both \(x^+\) and \(x^-\) are nonnegative functions on \(\mathbb{Z}^+\), and every subset of \(\mathbb{Z}^+\) belongs to the \(\sigma\)-algebra on which counting measure is defined, so both are measurable. By Example 3.6, integration with respect to counting measure on \(\mathbb{Z}^+\) is summation, so

\begin{equation*} \int x^+ \, d\mu = \sum_{k=1}^{\infty} x^+(k) = \sum_{m=1}^{\infty} \frac{1}{2m} = \frac{1}{2} \sum_{m=1}^{\infty} \frac{1}{m} = \infty \end{equation*}

and

\begin{equation*} \int x^- \, d\mu = \sum_{k=1}^{\infty} x^-(k) = \sum_{m=1}^{\infty} \frac{1}{2m - 1} \ge \sum_{m=1}^{\infty} \frac{1}{2m} = \infty , \end{equation*}

both because the harmonic series diverges. Since \(\int x^+ \, d\mu = \int x^- \, d\mu = \infty\), the expression \(\int x^+ \, d\mu - \int x^- \, d\mu\) is of the form \(\infty - \infty\), and hence \(\int x \, d\mu\) is not defined.

The point of the example is that \(\lim_{n \to \infty} \sum_{k=1}^{n} x_k\) takes the terms in a particular order and relies on cancellation, whereas integration with respect to counting measure is an order-free notion: it is defined only for series that converge absolutely (equivalently, only when \(\int |x| \, d\mu = \sum_{k=1}^{\infty} |x_k| < \infty\)) or for series in which only one of the two sign parts is infinite. Here \(\sum_{k=1}^{\infty} \frac{1}{k} = \infty\), so the convergence is merely conditional.

Throughout we use the book’s convention, stated alongside 3.2, that \(0 \cdot \infty\) and \(\infty \cdot 0\) are interpreted to be \(0\). The quantities \(\inf_X f\) and \(\sup_X f\) presuppose \(X \ne \varnothing\), so assume that; then \(\inf_X f \in [0, \infty)\) and \(\sup_X f \in [0, \infty]\). (If \(X = \varnothing\) the assertion is vacuous: \(\mu(X) = 0\) and \(\int f \, d\mu = 0\).)

The left inequality. Let \(P\) be the \(\mathcal{S}\)-partition of \(X\) consisting of the single set \(A_1 = X\). (This is a legitimate \(\mathcal{S}\)-partition: it is a finite collection of pairwise disjoint sets in \(\mathcal{S}\) whose union is \(X\).) By definition 3.2 its lower Lebesgue sum is

\begin{equation*} L(f, P) = \mu(A_1) \inf_{A_1} f = \mu(X) \inf_X f . \end{equation*}

Since \(\int f \, d\mu\) is by definition 3.3 the supremum of \(L(f, Q)\) over all \(\mathcal{S}\)-partitions \(Q\) of \(X\), and \(P\) is one such partition,

\begin{equation*} \mu(X) \inf_X f = L(f, P) \le \int f \, d\mu . \end{equation*}

The right inequality. Let \(c = \sup_X f \in [0, \infty]\). Then \(f(x) \le c\) for every \(x \in X\), that is,

\begin{equation*} f(x) \le c \, \chi_X(x) \qquad \text{for every } x \in X . \end{equation*}

Both \(f\) and \(c \chi_X\) are \(\mathcal{S}\)-measurable functions from \(X\) to \([0, \infty]\), so 3.8 (integration is order preserving) gives

\begin{equation*} \int f \, d\mu \le \int c \, \chi_X \, d\mu . \end{equation*}

Applying 3.15 with \(n = 1\), \(E_1 = X \in \mathcal{S}\), and \(c_1 = c \in [0, \infty]\) yields

\begin{equation*} \int c \, \chi_X \, d\mu = c \, \mu(X) = \mu(X) \sup_X f . \end{equation*}

Combining the last two displays gives \(\int f \, d\mu \le \mu(X) \sup_X f\), which together with the left inequality completes the proof.

Two remarks on the degenerate cases. If \(\mu(X) = \infty\) and \(\inf_X f = 0\), the left inequality reads \(0 \le \int f \, d\mu\), which is true (and by Exercise 1 in this section, \(\inf_E f = 0\) is forced on every \(E\) of infinite measure whenever \(\int f \, d\mu < \infty\)). If \(\mu(X) = 0\), then for every \(\mathcal{S}\)-partition \(A_1, \dots, A_m\) of \(X\) we have \(\mu(A_j) \le \mu(X) = 0\) by 2.57, so each term \(\mu(A_j) \inf_{A_j} f\) is \(0 \cdot t\) with \(t \in [0, \infty]\) and hence equals \(0\) by the convention above; thus every lower Lebesgue sum vanishes, \(\int f \, d\mu = 0\), and both inequalities hold.

The functions \(f_k\) take values in \([-\infty, \infty]\); because the sequence is monotone, \(\lim_{k \to \infty} f_k(x)\) exists in \([-\infty, \infty]\) for each \(x\), so \(f\) is well defined, and \(f\) is \(\mathcal{S}\)-measurable by 2.53 (it is the pointwise supremum of the \(f_k\) in the increasing case and the pointwise infimum in the decreasing case).

We first record two lemmas, then treat the increasing case, then deduce the decreasing case.

Lemma A. If \(N \in \mathcal{S}\) with \(\mu(N) = 0\) and \(h : X \to [0, \infty]\) is \(\mathcal{S}\)-measurable, then \(h \chi_N\) is \(\mathcal{S}\)-measurable and \(\int h \chi_N \, d\mu = 0\).

Proof of Lemma A. For \(a \ge 0\) we have \((h \chi_N)^{-1}\bigl((a, \infty]\bigr) = h^{-1}\bigl((a,\infty]\bigr) \cap N \in \mathcal{S}\), and for \(a < 0\) we have \((h \chi_N)^{-1}\bigl((a, \infty]\bigr) = X \in \mathcal{S}\); hence \(h \chi_N\) is \(\mathcal{S}\)-measurable by 2.52. Now let \(P\) be any \(\mathcal{S}\)-partition \(A_1, \dots, A_m\) of \(X\) and fix \(j\). If \(A_j \not\subseteq N\), choose \(y \in A_j \setminus N\); then \((h\chi_N)(y) = 0\), so \(\inf_{A_j} h \chi_N = 0\) and therefore \(\mu(A_j) \inf_{A_j} h \chi_N = 0\) (using the convention \(\infty \cdot 0 = 0\)). If \(A_j \subseteq N\), then \(\mu(A_j) \le \mu(N) = 0\) by 2.57, so again \(\mu(A_j) \inf_{A_j} h \chi_N = 0\) (using \(0 \cdot \infty = 0\)). Hence \(L(h\chi_N, P) = 0\) for every \(\mathcal{S}\)-partition \(P\), and definition 3.3 gives \(\int h \chi_N \, d\mu = 0\).

Lemma B. Suppose \(N \in \mathcal{S}\) with \(\mu(N) = 0\), put \(M = X \setminus N\), and suppose \(g : X \to [-\infty,\infty]\) is \(\mathcal{S}\)-measurable and \(\int g \, d\mu\) is defined. Then \(g \chi_M\) is \(\mathcal{S}\)-measurable, \(\int g \chi_M \, d\mu\) is defined, and \(\int g \chi_M \, d\mu = \int g \, d\mu\).

Proof of Lemma B. Measurability: for \(a \ge 0\), \((g\chi_M)^{-1}\bigl((a,\infty]\bigr) = g^{-1}\bigl((a,\infty]\bigr) \cap M \in \mathcal{S}\), and for \(a < 0\), \((g \chi_M)^{-1}\bigl((a,\infty]\bigr) = \bigl(g^{-1}((a,\infty]) \cap M\bigr) \cup N \in \mathcal{S}\); apply 2.52. Next, \((g\chi_M)^+ = g^+ \chi_M\) and \((g \chi_M)^- = g^- \chi_M\) pointwise. Since \(g^{\pm} = g^{\pm} \chi_M + g^{\pm} \chi_N\) pointwise, additivity 3.16 and Lemma A give

\begin{equation*} \int g^{\pm} \, d\mu = \int g^{\pm} \chi_M \, d\mu + \int g^{\pm} \chi_N \, d\mu = \int (g \chi_M)^{\pm} \, d\mu . \end{equation*}

Thus \(\int (g\chi_M)^+ d\mu\) and \(\int (g \chi_M)^- d\mu\) equal \(\int g^+ d\mu\) and \(\int g^- d\mu\) respectively; in particular at least one is finite, so \(\int g\chi_M \, d\mu\) is defined and equals \(\int g \, d\mu\).

The increasing case. Assume \(f_1 \le f_2 \le \cdots\).

Step 0: all the integrals are defined. For each \(k\) we have \(f_k \ge f_1\), hence \(f_k^- \le f_1^- \le |f_1|\) pointwise, so 3.8 gives \(\int f_k^- \, d\mu \le \int |f_1| \, d\mu < \infty\). Thus \(\int f_k \, d\mu\) is defined (and lies in \((-\infty, \infty]\)). The same argument with \(f\) in place of \(f_k\), using \(f \ge f_1\), shows \(\int f \, d\mu\) is defined.

Step 1: the set where \(f_1\) is infinite is null. Let \(N = f_1^{-1}\bigl(\{-\infty, \infty\}\bigr)\), which lies in \(\mathcal{S}\) because \(f_1\) is \(\mathcal{S}\)-measurable and \(\{-\infty, \infty\}\) is a Borel subset of \([-\infty, \infty]\) (its intersection with \(\mathbb{R}\) is \(\varnothing\); see 2.50). For every \(n \in \mathbb{Z}^+\) we have \(|f_1| \ge n \chi_N\) pointwise, so 3.8 together with 3.15 (with \(n = 1\) summand, \(E_1 = N\), \(c_1 = n\)) gives \(\int |f_1| \, d\mu \ge n \mu(N)\). Since \(\int |f_1| \, d\mu < \infty\) and \(n\) is arbitrary, \(\mu(N) = 0\).

Step 2: reduction to the case where \(f_1\) is real valued. Let \(M = X \setminus N\) and set \(\widetilde{f_k} = f_k \chi_M\) and \(\widetilde{f} = f \chi_M\). By Lemma B these are \(\mathcal{S}\)-measurable and

\begin{equation*} \int \widetilde{f_k} \, d\mu = \int f_k \, d\mu \quad (k \in \mathbb{Z}^+), \qquad \int \widetilde{f} \, d\mu = \int f \, d\mu . \end{equation*}

On \(N\) every \(\widetilde{f_k}\) and \(\widetilde{f}\) equals \(0\); on \(M\) they equal \(f_k\) and \(f\). Hence \(\widetilde{f_1} \le \widetilde{f_2} \le \cdots\) is still increasing, \(\lim_{k \to \infty} \widetilde{f_k}(x) = \widetilde{f}(x)\) for every \(x \in X\), and \(|\widetilde{f_1}| \le |f_1|\) gives \(\int |\widetilde{f_1}| \, d\mu < \infty\) by 3.8. Moreover \(\widetilde{f_1}\) is real valued. So it suffices to prove the result under the additional hypothesis that \(f_1(x) \in \mathbb{R}\) for every \(x \in X\), which we now assume.

Step 3: the auxiliary nonnegative sequence. Because \(f_k \ge f_1 > -\infty\), each \(f_k\) takes values in \((-\infty, \infty]\), and likewise \(f\) does. Since \(f_1\) is real valued, the differences

\begin{equation*} g_k := f_k - f_1 \quad \text{and} \quad g := f - f_1 \end{equation*}

are well defined functions from \(X\) to \([0, \infty]\). They are \(\mathcal{S}\)-measurable: for \(a \in \mathbb{R}\),

\begin{equation*} g_k^{-1}\bigl((a, \infty]\bigr) = \{x : f_k(x) > a + f_1(x)\} = \bigcup_{r \in \mathbb{Q}} \Bigl( f_k^{-1}\bigl((r, \infty]\bigr) \cap f_1^{-1}\bigl([-\infty, r - a)\bigr) \Bigr) , \end{equation*}

because \(f_k(x) > a + f_1(x)\) (with \(a + f_1(x) \in \mathbb{R}\)) holds if and only if there is a rational \(r\) with \(a + f_1(x) < r < f_k(x)\), i.e. \(f_k(x) > r\) and \(f_1(x) < r - a\). The right side is a countable union of sets in \(\mathcal{S}\), so \(g_k\) is \(\mathcal{S}\)-measurable by 2.52; the same computation applies to \(g\). Clearly \(0 \le g_1 \le g_2 \le \cdots\) and \(\lim_{k \to \infty} g_k(x) = g(x)\) for every \(x \in X\). Hence the Monotone Convergence Theorem 3.11 gives

\begin{equation*} \lim_{k \to \infty} \int g_k \, d\mu = \int g \, d\mu . \end{equation*}

Step 4: splitting off \(f_1\). We claim that if \(h : X \to (-\infty, \infty]\) is \(\mathcal{S}\)-measurable with \(h \ge f_1\), then

\begin{equation*} \int h \, d\mu = \int (h - f_1) \, d\mu + \int f_1 \, d\mu . \end{equation*}

First, \(h - f_1\) is a well-defined \(\mathcal{S}\)-measurable function from \(X\) to \([0, \infty]\), by the rational-splitting computation of Step 3 with \(h\) in place of \(f_k\) (that computation used only that \(f_1\) is real valued, that \(h\) is \(\mathcal{S}\)-measurable, and that \(h \ge f_1\)). Next, note the pointwise identity between functions with values in \([0, \infty]\)

\begin{equation*} h^+ + f_1^- = (h - f_1) + f_1^+ + h^- , \end{equation*}

which is verified by substituting \(h - f_1 = h^+ - h^- - f_1^+ + f_1^-\) (legitimate at every \(x\), since \(f_1(x)\) is real and at most one of \(h^+(x), h^-(x)\) is nonzero) and cancelling; when \(h(x) = \infty\) both sides equal \(\infty\). Integrating both sides and using additivity 3.16 twice,

\begin{equation*} \int h^+ \, d\mu + \int f_1^- \, d\mu = \int (h - f_1) \, d\mu + \int f_1^+ \, d\mu + \int h^- \, d\mu . \end{equation*}

Now \(\int f_1^+ \, d\mu\) and \(\int f_1^- \, d\mu\) are finite because \(f_1^{\pm} \le |f_1|\) and \(\int |f_1| \, d\mu < \infty\), and \(\int h^- \, d\mu \le \int f_1^- \, d\mu < \infty\) because \(h \ge f_1\) forces \(h^- \le f_1^-\). Subtracting the finite numbers \(\int f_1^- \, d\mu\) and \(\int h^- \, d\mu\) from both sides gives

\begin{equation*} \int h^+ \, d\mu - \int h^- \, d\mu = \int (h - f_1) \, d\mu + \int f_1^+ \, d\mu - \int f_1^- \, d\mu , \end{equation*}

which by 3.18 is exactly the claim (both sides possibly equal to \(+\infty\)).

Step 5: conclusion in the increasing case. Applying Step 4 with \(h = f_k\) and with \(h = f\),

\begin{equation*} \int f_k \, d\mu = \int g_k \, d\mu + \int f_1 \, d\mu \quad \text{and} \quad \int f \, d\mu = \int g \, d\mu + \int f_1 \, d\mu . \end{equation*}

Since \(\int f_1 \, d\mu\) is a real number, adding it commutes with taking limits in \((-\infty, \infty]\), so Step 3 gives

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\mu = \lim_{k \to \infty} \int g_k \, d\mu + \int f_1 \, d\mu = \int g \, d\mu + \int f_1 \, d\mu = \int f \, d\mu . \end{equation*}

The decreasing case. Now assume \(f_1 \ge f_2 \ge \cdots\). Put \(h_k = -f_k\). Each \(h_k\) is \(\mathcal{S}\)-measurable, since \(h_k^{-1}\bigl((a, \infty]\bigr) = f_k^{-1}\bigl([-\infty, -a)\bigr) \in \mathcal{S}\) for every \(a \in \mathbb{R}\) (use 2.52). Then \(h_1 \le h_2 \le \cdots\) is increasing, \(\lim_{k \to \infty} h_k(x) = -f(x)\) for every \(x\), and \(\int |h_1| \, d\mu = \int |f_1| \, d\mu < \infty\). Also \(\int f_k \, d\mu\) is defined for each \(k\): from \(f_k \le f_1\) we get \(f_k^+ \le f_1^+ \le |f_1|\), so \(\int f_k^+ \, d\mu < \infty\) by 3.8; similarly \(\int f \, d\mu\) is defined. By the increasing case already proved,

\begin{equation*} \lim_{k \to \infty} \int (-f_k) \, d\mu = \int (-f) \, d\mu . \end{equation*}

By homogeneity 3.20 with \(c = -1\) (applicable because \(\int f_k \, d\mu\) and \(\int f \, d\mu\) are defined), \(\int (-f_k) \, d\mu = -\int f_k \, d\mu\) and \(\int(-f) \, d\mu = -\int f \, d\mu\). Substituting and multiplying by \(-1\) yields

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\mu = \int f \, d\mu , \end{equation*}

completing the proof.

Remark. Exercise 14 shows that the conclusion can fail for a decreasing sequence without some integrability hypothesis (for instance \(f_k = \chi_{[k, \infty)}\) with Lebesgue measure on \(\mathbb{R}\), where \(\int f_k \, d\lambda = \infty\) for every \(k\) but \(f = 0\)). The hypothesis \(\int |f_1| \, d\mu < \infty\) is exactly what rules this out: it pins the whole decreasing sequence below an integrable function. This is why the note after Exercise 14 points to the present exercise.

Recall the statement of 3.15: if \((X, \mathcal{S}, \mu)\) is a measure space, \(E_1, \dots, E_n \in \mathcal{S}\), and \(c_1, \dots, c_n \in [0, \infty]\), then

\begin{equation*} \int \sum_{k=1}^{n} c_k \chi_{E_k} \, d\mu = \sum_{k=1}^{n} c_k \, \mu(E_k). \end{equation*}

Everything in Lebesgue’s parable is contained in this formula.

The pocket, literally. Model the money as follows. Let \(X\) be the (finite) set of bills and coins in the pocket, let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(X\), and let \(\mu\) be counting measure on \(\mathcal{S}\), so that \(\mu(E)\) is the number of items in \(E\). Let \(v : X \to [0, \infty)\) assign to each item its face value. The total to be paid is \(\int v \, d\mu\): indeed, writing \(v = \sum_{x \in X} v(x) \chi_{\{x\}}\) (a finite sum, since \(X\) is finite) and applying 3.15 with \(\mu(\{x\}) = 1\) gives \(\int v \, d\mu = \sum_{x \in X} v(x)\), literally the sum of the values of all the items in the pocket. (This is the finite case of Example 3.6, where integration against counting measure is summation.)

Lebesgue’s method is 3.15 read from left to right. Let \(c_1, \dots, c_n\) be the distinct denominations present, and let

\begin{equation*} E_k = v^{-1}(\{c_k\}) \end{equation*}

be the heap of all items of value \(c_k\). These heaps are exactly what Lebesgue means by ordering the money “according to identical values”: each \(E_k\) is a piece of the domain cut out by a single value in the range. Since \(v = \sum_{k=1}^{n} c_k \chi_{E_k}\), formula 3.15 gives

\begin{equation*} \int v \, d\mu = \sum_{k=1}^{n} c_k \, \mu(E_k) = \sum_{k=1}^{n} (\text{denomination } c_k) \times (\text{number of items in the heap}) . \end{equation*}

That is precisely “paying the several heaps one after the other”: you hand over the pile of tens, then the pile of fives, and so on, and the sum is a sum over the \(n\) values in the range, not over the many items in the domain. Riemann’s method, by contrast, is “in the order I find them”: you fix an enumeration of the items in the pocket, i.e. an ordering of the domain, and accumulate the values one item at a time, running along the domain.

Why 3.15 is a range partition. Note what 3.15 asks of the sets \(E_1, \dots, E_n\): only that they belong to \(\mathcal{S}\), so that the numbers \(\mu(E_k)\) make sense. They need not be intervals, and 3.15 does not even require them to be disjoint. The numbers \(c_1, \dots, c_n\), on the other hand, are values of the function. So the data of the sum \(\sum_k c_k \mu(E_k)\) is: a list of values, and the measure of the set on which each value is taken.

This is exactly how the integral of an arbitrary nonnegative measurable function is computed. Let \(f : X \to [0, \infty)\) be \(\mathcal{S}\)-measurable. For each \(j \in \mathbb{Z}^+\) chop the part \([0, j)\) of the range into the \(j 2^{j}\) half-open intervals \([k 2^{-j}, (k+1) 2^{-j})\) with \(0 \le k < j 2^{j}\), lump the rest of the range together, and let

\begin{equation*} E_{j,k} = f^{-1}\bigl([k 2^{-j}, (k+1) 2^{-j})\bigr), \qquad E_{j,\infty} = f^{-1}\bigl([j, \infty)\bigr) . \end{equation*}

These sets lie in \(\mathcal{S}\) precisely because \(f\) is \(\mathcal{S}\)-measurable, and for fixed \(j\) they are disjoint with union \(X\). Put

\begin{equation*} s_j = \sum_{k=0}^{j 2^{j} - 1} k 2^{-j} \, \chi_{E_{j,k}} + j \, \chi_{E_{j, \infty}} , \end{equation*}

a finite linear combination of characteristic functions, so that 3.15 applies verbatim and gives

\begin{equation*} \int s_j \, d\mu = \sum_{k=0}^{j 2^{j} - 1} k 2^{-j} \, \mu(E_{j,k}) + j \, \mu(E_{j,\infty}) . \end{equation*}

Pointwise \(s_j = \min\{ 2^{-j} \lfloor 2^{j} f \rfloor, \, j \}\); since \(2^{-j} \lfloor 2^{j} t \rfloor\) increases with \(j\) for each fixed \(t \ge 0\), we get \(0 \le s_1 \le s_2 \le \cdots\), and \(s_j(x) \to f(x)\) for every \(x\) (once \(j > f(x)\) we have \(0 \le f(x) - s_j(x) \le 2^{-j}\)). This is the construction behind 2.89 in the case of a nonnegative function, and the Monotone Convergence Theorem 3.11 now gives \(\int f \, d\mu\) as the limit of the sums displayed above. The partition being refined at each stage is a partition of the range; the domain pieces \(E_{j,k}\) are whatever the function’s level sets happen to be, and the only thing demanded of them is that they be measurable.

Contrast with Riemann integration. A Riemann sum for \(f\) on \([a, b]\) starts from a partition \(a = t_0 < t_1 < \cdots < t_n = b\) of the domain and forms \(\sum_j f(s_j)(t_j - t_{j-1})\) with \(s_j \in [t_{j-1}, t_j]\); the lower Riemann sums \(L(f, P, [a,b])\) of 1.3 replace \(f(s_j)\) by \(\inf_{[t_{j-1}, t_j]} f\). The accuracy of such a sum is governed by the oscillation of \(f\) on each subinterval, so the method works only when the function is nearly constant on small intervals of the domain. Nothing about the range is used to choose the partition.

The two methods therefore fail and succeed in different places. For \(f = \chi_{\mathbb{Q} \cap [0,1]}\) every subinterval of \([0,1]\) contains both rationals and irrationals, so \(\inf f = 0\) and \(\sup f = 1\) on every piece of every domain partition; hence \(L(f, [0,1]) = 0\) while \(U(f, [0,1]) = 1\), and \(f\) is not Riemann integrable. But the range of \(f\) has only the two points \(0\) and \(1\), so Lebesgue’s sorting produces just two heaps, \(\mathbb{Q} \cap [0,1]\) and \([0,1] \setminus \mathbb{Q}\), and 3.15 gives at once

\begin{equation*} \int \chi_{\mathbb{Q} \cap [0,1]} \, d\lambda = 1 \cdot \lambda\bigl(\mathbb{Q} \cap [0,1]\bigr) + 0 \cdot \lambda\bigl([0,1] \setminus \mathbb{Q}\bigr) = 1 \cdot 0 + 0 \cdot 1 = 0 , \end{equation*}

in agreement with 3.5. A wildly oscillating function still has a small range, and grouping the domain by value tames it.

The price of Lebesgue’s method is visible in the same formula: to pay the heaps one must be able to count each heap, that is, one must know \(\mu(E_k)\) for level sets \(E_k\) that may be very complicated sets. This is exactly why a theory of measure (Chapter 2) has to be built before the integral can be defined, and it is why the definition 3.3 of \(\int f \, d\mu\) via lower Lebesgue sums allows arbitrary sets of \(\mathcal{S}\) in its partitions rather than only intervals, as the lower Riemann sums do. Once arbitrary measurable sets are allowed, a domain partition can be chosen to follow the level sets of \(f\), and the domain-partition definition 3.3 becomes equivalent to the range-partition picture described above.

Take

\begin{equation*} f_k = \chi_{\{k\}}, \end{equation*}

that is, \(f_k : \mathbb{Z}^+ \to [0, \infty)\) is defined by \(f_k(m) = 1\) if \(m = k\) and \(f_k(m) = 0\) if \(m \neq k\). Here the \(\sigma\)-algebra is the collection of all subsets of \(\mathbb{Z}^+\), so every function on \(\mathbb{Z}^+\) is measurable.

Pointwise convergence to \(0\): fix \(m \in \mathbb{Z}^+\). If \(k > m\) then \(k \neq m\) and hence \(f_k(m) = 0\). Thus \(f_k(m) = 0\) for all \(k > m\), and in particular

\begin{equation*} \lim_{k \to \infty} f_k(m) = 0 . \end{equation*}

The integrals: because \(f_k\) is the characteristic function of the set \(\{k\}\), the formula for the integral of a characteristic function (3.4) gives

\begin{equation*} \int f_k \, d\mu = \mu(\{k\}) = 1 \end{equation*}

for every \(k \in \mathbb{Z}^+\), since counting measure assigns to each singleton the value \(1\). Hence

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\mu = 1 \neq 0 = \int \Bigl( \lim_{k \to \infty} f_k \Bigr) d\mu . \end{equation*}

Thus the sequence \(f_1, f_2, \ldots\) has the required properties. Note that this example is consistent with the Dominated Convergence Theorem (3.31): the smallest function dominating all the \(f_k\) is \(g \equiv 1\), and \(\int g \, d\mu = \mu(\mathbb{Z}^+) = \infty\), so no dominating function with finite integral exists. The example is also consistent with the Monotone Convergence Theorem (3.11), since the sequence \(f_1, f_2, \ldots\) is not increasing. The mass \(1\) carried by each \(f_k\) simply escapes to infinity.

For \(k \in \mathbb{Z}^+\) define \(f_k : \mathbb{R} \to [0,1]\) by

\begin{equation*} f_k(x) = \max\Bigl\{ 0, \; 1 - \frac{|x - k^2|}{k} \Bigr\} . \end{equation*}

Thus \(f_k\) is the tent function that equals \(1\) at \(x = k^2\), decreases linearly to \(0\) at the two points \(k^2 - k\) and \(k^2 + k\), and equals \(0\) outside the interval \([k^2 - k, \, k^2 + k]\).

Each \(f_k\) is continuous, since \(x \mapsto 1 - |x - k^2| / k\) is continuous and the maximum of two continuous functions is continuous. Clearly \(0 \le f_k(x) \le 1\) for all \(x\), so \(f_k\) maps \(\mathbb{R}\) into \([0,1]\). Being continuous, each \(f_k\) is Borel measurable (2.41).

Pointwise convergence to \(0\): fix \(x \in \mathbb{R}\). Because \(k^2 - k \to \infty\) as \(k \to \infty\), there exists \(K \in \mathbb{Z}^+\) such that \(k^2 - k > |x|\) for all \(k \ge K\). For such \(k\) we have \(x < k^2 - k\), hence \(x \notin [k^2 - k, \, k^2 + k]\) and therefore \(f_k(x) = 0\). Thus

\begin{equation*} \lim_{k \to \infty} f_k(x) = 0 \end{equation*}

for every \(x \in \mathbb{R}\).

The integrals: \(f_k\) vanishes outside \([k^2 - k, \, k^2 + k]\) and is continuous, hence Riemann integrable on that interval; by 3.34 its Lebesgue integral over the interval equals its Riemann integral there, which is the area of a triangle with base of length \(2k\) and height \(1\). Thus

\begin{equation*} \int f_k \, d\lambda = \int_{[k^2 - k, \, k^2 + k]} f_k \, d\lambda = \frac{1}{2} \cdot 2k \cdot 1 = k . \end{equation*}

(The integral over the complement of \([k^2-k,\,k^2+k]\) is \(0\) because \(f_k\) is \(0\) there.) Hence

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\lambda = \lim_{k \to \infty} k = \infty , \end{equation*}

while \(\int (\lim_{k \to \infty} f_k) \, d\lambda = \int 0 \, d\lambda = 0\).

As in the previous exercise, the point is that the bumps run off to infinity while growing wider. Note that the Bounded Convergence Theorem (3.26) does not apply because \(\lambda(\mathbb{R}) = \infty\), and the Dominated Convergence Theorem (3.31) does not apply because any \(g\) with \(g \ge f_k\) for all \(k\) satisfies \(\int g \, d\lambda \ge \int f_k \, d\lambda = k\) for every \(k\), so \(\int g \, d\lambda = \infty\).

First we check that \(g\) is well defined and real valued. Fix \(x \in \mathbb{R}\). By 3.24 the expression \(\int_{(-\infty, x)} f \, d\lambda\) means \(\int \chi_{(-\infty, x)} f \, d\lambda\). Since \(|\chi_{(-\infty, x)} f| \le |f|\) pointwise, 3.8 gives

\begin{equation*} \int \bigl| \chi_{(-\infty, x)} f \bigr| \, d\lambda \le \int |f| \, d\lambda < \infty , \end{equation*}

so both \((\chi_{(-\infty,x)} f)^+\) and \((\chi_{(-\infty,x)} f)^-\) have finite integrals; hence the integral defining \(g(x)\) is defined (3.18) and is a real number. Also \(\chi_{(-\infty,x)} f\) is Borel measurable, being the product of a Borel measurable function and the characteristic function of a Borel set.

Now let \(\varepsilon > 0\). Apply 3.28 (integrals on small sets are small) to the nonnegative Borel measurable function \(|f|\), whose integral is finite: there exists \(\delta > 0\) such that

\begin{equation*} \int_B |f| \, d\lambda < \varepsilon \quad \text{whenever } B \text{ is a Borel set with } \lambda(B) < \delta . \end{equation*}

Suppose now that \(x, y \in \mathbb{R}\) with \(|x - y| < \delta\); without loss of generality \(x \le y\). Then

\begin{equation*} \chi_{(-\infty, \, y)} f = \chi_{(-\infty, \, x)} f + \chi_{[x, \, y)} f , \end{equation*}

because the intervals \((-\infty, x)\) and \([x,y)\) are disjoint with union \((-\infty, y)\). Both functions on the right have absolute value bounded by \(|f|\) and hence have finite integrals of their absolute values, so additivity of integration (3.21) applies and gives

\begin{equation*} g(y) - g(x) = \int \chi_{[x, \, y)} f \, d\lambda = \int_{[x, \, y)} f \, d\lambda . \end{equation*}

Therefore, using 3.23 and then 3.8,

\begin{equation*} |g(y) - g(x)| = \left| \int_{[x, \, y)} f \, d\lambda \right| \le \int_{[x, \, y)} |f| \, d\lambda . \end{equation*}

Since \(\lambda([x,y)) = y - x = |x - y| < \delta\), the choice of \(\delta\) gives

\begin{equation*} \int_{[x, \, y)} |f| \, d\lambda < \varepsilon . \end{equation*}

Hence \(|g(y) - g(x)| < \varepsilon\).

We have shown: for every \(\varepsilon > 0\) there exists \(\delta > 0\) such that \(|g(x) - g(y)| < \varepsilon\) for all \(x, y \in \mathbb{R}\) with \(|x - y| < \delta\). Because \(\delta\) depends only on \(\varepsilon\) and not on the points \(x, y\), this is exactly the statement that \(g\) is uniformly continuous on \(\mathbb{R}\).

Throughout, for an \(\mathcal{S}\)-partition \(P\) of \(X\) given by \(A_1, \ldots, A_m\) we write, in parallel with the lower Lebesgue sum \(L(f, P) = \sum_{j=1}^m \mu(A_j) \inf_{A_j} f\) of 3.2, the upper Lebesgue sum

\begin{equation*} U(f, P) = \sum_{j=1}^{m} \mu(A_j) \sup_{A_j} f , \end{equation*}

and we set \(U(f) = \inf \{ U(f, P) : P \text{ is an } \mathcal{S}\text{-partition of } X \}\). The assertion of (a) is that \(\int f \, d\mu = U(f)\).

Part (a).

If \(X = \varnothing\) then \(\mu(X) = 0\) and both sides equal \(0\), so assume \(X \neq \varnothing\). The definition 3.1 of an \(\mathcal{S}\)-partition permits empty sets, but a term of \(L(f, P)\) or of \(U(f, P)\) coming from an empty \(A_j\) has the factor \(\mu(\varnothing) = 0\), hence is \(0\) whatever value is assigned to \(\inf_{\varnothing} f\) or \(\sup_{\varnothing} f\), by the convention (stated just before 3.2) that \(0 \cdot \infty = \infty \cdot 0 = 0\). Deleting the empty sets from a partition of the nonempty set \(X\) therefore leaves an \(\mathcal{S}\)-partition of \(X\) with the same lower and upper sums. So we may assume throughout that all sets appearing in a partition are nonempty; then \(\inf_{A_j} f\) and \(\sup_{A_j} f\) are genuine real numbers, because \(f\) is bounded. Let \(c \in [0, \infty)\) be such that \(0 \le f(x) \le c\) for all \(x \in X\). Since \(\mu(X) < \infty\), every \(L(f, P)\) and every \(U(f, P)\) is a real number in \([0, c\,\mu(X)]\).

Step 1: refining a partition increases \(L\) and decreases \(U\). Let \(P\) be the partition \(A_1, \ldots, A_m\) and let \(P^{\prime}\) be the partition \(B_1, \ldots, B_n\). Let \(Q\) be the \(\mathcal{S}\)-partition consisting of the nonempty sets among \(A_i \cap B_j\) for \(i \in \{1, \ldots, m\}\) and \(j \in \{1, \ldots, n\}\); these sets are disjoint, lie in \(\mathcal{S}\), and their union is \(X\). Fix \(i\). The sets \(A_i \cap B_1, \ldots, A_i \cap B_n\) are disjoint with union \(A_i\), so additivity of the measure \(\mu\) gives \(\mu(A_i) = \sum_{j=1}^n \mu(A_i \cap B_j)\). Moreover \(\inf_{A_i} f \le \inf_{A_i \cap B_j} f\) whenever \(A_i \cap B_j \neq \varnothing\), because the infimum is taken over a smaller set. Hence

\begin{equation*} \mu(A_i) \inf_{A_i} f = \sum_{j=1}^{n} \mu(A_i \cap B_j) \inf_{A_i} f \le \sum_{j=1}^{n} \mu(A_i \cap B_j) \inf_{A_i \cap B_j} f , \end{equation*}

where terms with \(A_i \cap B_j = \varnothing\) contribute \(0\) on both sides. Summing over \(i \in \{1, \ldots, m\}\) yields \(L(f, P) \le L(f, Q)\). The same computation with \(B_j\) in place of \(A_i\) and suprema in place of infima (now using \(\sup_{A_i \cap B_j} f \le \sup_{B_j} f\)) yields \(U(f, Q) \le U(f, P^{\prime})\).

Step 2: \(\int f \, d\mu \le U(f)\). Since \(\inf_{A} f \le \sup_{A} f\) for every nonempty \(A\), we have \(L(f, Q) \le U(f, Q)\) for every partition \(Q\). Combining this with Step 1, for any two \(\mathcal{S}\)-partitions \(P\) and \(P^{\prime}\) we get

\begin{equation*} L(f, P) \le L(f, Q) \le U(f, Q) \le U(f, P^{\prime}) , \end{equation*}

where \(Q\) is the common refinement constructed above. Fix \(P^{\prime}\) and take the supremum over all \(P\): by the definition 3.3 of the integral this gives \(\int f \, d\mu \le U(f, P^{\prime})\). Now take the infimum over all \(P^{\prime}\) to obtain \(\int f \, d\mu \le U(f)\).

Step 3: \(U(f) \le \int f \, d\mu\). If \(c = 0\) then \(f \equiv 0\) and both quantities are \(0\); so assume \(c > 0\). Let \(\varepsilon > 0\) and choose \(n \in \mathbb{Z}^+\) with \(\frac{c}{n} \mu(X) < \varepsilon\) (possible because \(\mu(X) < \infty\)). Define

\begin{equation*} A_j = f^{-1}\Bigl( \Bigl[ \tfrac{(j-1)c}{n}, \tfrac{jc}{n} \Bigr) \Bigr) \quad \text{for } j = 1, \ldots, n-1, \qquad A_n = f^{-1}\Bigl( \Bigl[ \tfrac{(n-1)c}{n}, c \Bigr] \Bigr) . \end{equation*}

Each \(A_j\) is in \(\mathcal{S}\) because \(f\) is \(\mathcal{S}\)-measurable and the sets used are Borel subsets of \(\mathbb{R}\); the sets \(A_1, \ldots, A_n\) are disjoint, and their union is \(X\) because \(f(X) \subseteq [0, c]\). So \(P\), the list of nonempty sets among \(A_1, \ldots, A_n\), is an \(\mathcal{S}\)-partition of \(X\). For each nonempty \(A_j\) the values of \(f\) on \(A_j\) lie in an interval of length \(c/n\), so

\begin{equation*} \sup_{A_j} f - \inf_{A_j} f \le \frac{c}{n} . \end{equation*}

Therefore

\begin{equation*} U(f, P) - L(f, P) = \sum_{j} \mu(A_j) \Bigl( \sup_{A_j} f - \inf_{A_j} f \Bigr) \le \frac{c}{n} \sum_{j} \mu(A_j) = \frac{c}{n} \mu(X) < \varepsilon , \end{equation*}

this subtraction being legitimate because all the numbers involved are finite. Hence

\begin{equation*} U(f) \le U(f, P) < L(f, P) + \varepsilon \le \int f \, d\mu + \varepsilon , \end{equation*}

the last inequality holding by 3.3. Since \(\varepsilon > 0\) was arbitrary, \(U(f) \le \int f \, d\mu\).

Steps 2 and 3 together give \(\int f \, d\mu = U(f)\), which is the desired equality.

Part (b).

Let \(X = \mathbb{Z}^+\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(\mathbb{Z}^+\), and define \(\mu\) on \(\mathcal{S}\) by

\begin{equation*} \mu(A) = \sum_{n \in A} 2^{-n} . \end{equation*}

Then \(\mu(\varnothing) = 0\), and \(\mu\) is countably additive because a series of nonnegative terms may be summed over a countable disjoint decomposition of the index set in any order. Note \(\mu(X) = \sum_{n=1}^\infty 2^{-n} = 1 < \infty\). Define \(f : X \to [0, \infty)\) by \(f(n) = n\); every function on \(X\) is \(\mathcal{S}\)-measurable, and \(f\) is unbounded.

First, \(\int f \, d\mu < \infty\). Let \(f_N = \sum_{n=1}^{N} n \chi_{\{n\}}\). Then \(0 \le f_1 \le f_2 \le \cdots\) and \(\lim_{N \to \infty} f_N(m) = f(m)\) for every \(m \in X\) (indeed \(f_N(m) = m\) once \(N \ge m\)). By the Monotone Convergence Theorem (3.11) and the formula 3.7 for the integral of a simple function,

\begin{equation*} \int f \, d\mu = \lim_{N \to \infty} \int f_N \, d\mu = \lim_{N \to \infty} \sum_{n=1}^{N} n \, \mu(\{n\}) = \sum_{n=1}^{\infty} \frac{n}{2^{n}} = 2 < \infty . \end{equation*}

Now let \(A_1, \ldots, A_m\) be any \(\mathcal{S}\)-partition of \(X\). Since \(X = \mathbb{Z}^+\) is infinite and is the union of the finitely many sets \(A_1, \ldots, A_m\), at least one of them, say \(A_k\), is infinite. Then \(A_k\) contains arbitrarily large positive integers, so \(\sup_{A_k} f = \infty\). Also \(A_k \neq \varnothing\), so picking any \(n_0 \in A_k\) gives \(\mu(A_k) \ge 2^{-n_0} > 0\). Hence \(\mu(A_k) \sup_{A_k} f = \infty\), and therefore

\begin{equation*} \sum_{j=1}^{m} \mu(A_j) \sup_{A_j} f = \infty . \end{equation*}

Because this holds for every \(\mathcal{S}\)-partition, the infimum on the right side of the equation in (a) equals \(\infty\), while \(\int f \, d\mu = 2\). Thus the conclusion of (a) fails, even though \(\mu(X) < \infty\) and \(\int f \, d\mu < \infty\); the hypothesis that \(f\) be bounded cannot be dropped.

Part (c).

Let \(X = \mathbb{Z}^+\), let \(\mathcal{S}\) be the \(\sigma\)-algebra of all subsets of \(\mathbb{Z}^+\), and let \(\mu\) be counting measure, so \(\mu(X) = \infty\). Define \(f : X \to [0, \infty)\) by \(f(n) = 2^{-n}\); this \(f\) is \(\mathcal{S}\)-measurable and bounded, since \(0 < f(n) \le \frac12\) for all \(n\).

As in part (b), put \(f_N = \sum_{n=1}^{N} 2^{-n} \chi_{\{n\}}\). These form an increasing sequence of nonnegative simple functions converging pointwise to \(f\), so the Monotone Convergence Theorem (3.11) together with 3.7 gives

\begin{equation*} \int f \, d\mu = \lim_{N \to \infty} \sum_{n=1}^{N} 2^{-n} \mu(\{n\}) = \sum_{n=1}^{\infty} 2^{-n} = 1 . \end{equation*}

(This is the special case of the assertion in Example 3.6 that integration with respect to counting measure is summation.)

Now let \(A_1, \ldots, A_m\) be any \(\mathcal{S}\)-partition of \(X\). As before, at least one \(A_k\) is infinite, so \(\mu(A_k) = \infty\). Since \(A_k \neq \varnothing\) and \(f > 0\) everywhere, \(\sup_{A_k} f \ge f(n_0) > 0\) for any \(n_0 \in A_k\). Hence \(\mu(A_k) \sup_{A_k} f = \infty\) and so

\begin{equation*} \sum_{j=1}^{m} \mu(A_j) \sup_{A_j} f = \infty \end{equation*}

for every \(\mathcal{S}\)-partition. The infimum on the right side of the equation in (a) is therefore \(\infty\), while \(\int f \, d\mu = 1\). Thus the conclusion of (a) fails, even though \(f\) is bounded; the hypothesis \(\mu(X) < \infty\) cannot be deleted.

For \(k \in \mathbb{Z}^+\) define \(f_k : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f_k = \chi_{[-k, \, k]} f , \end{equation*}

so that, by the definition 3.24 of integration over a subset,

\begin{equation*} \int_{[-k, \, k]} f \, d\lambda = \int f_k \, d\lambda . \end{equation*}

Each \(f_k\) is Borel measurable, being the product of the Borel measurable function \(f\) with the characteristic function of the Borel set \([-k, k]\) (see 2.46 for products of measurable functions).

We verify the hypotheses of the Dominated Convergence Theorem (3.31) with dominating function \(g = |f|\).

Pointwise convergence: fix \(x \in \mathbb{R}\). Choose \(K \in \mathbb{Z}^+\) with \(K \ge |x|\). Then for every \(k \ge K\) we have \(x \in [-k, k]\) and hence \(f_k(x) = f(x)\). Thus

\begin{equation*} \lim_{k \to \infty} f_k(x) = f(x) \end{equation*}

for every \(x \in \mathbb{R}\); in fact the sequence is eventually constant at each point.

Domination: for every \(k \in \mathbb{Z}^+\) and every \(x \in \mathbb{R}\),

\begin{equation*} |f_k(x)| = \chi_{[-k, \, k]}(x) \, |f(x)| \le |f(x)| = g(x) , \end{equation*}

since \(\chi_{[-k,k]}(x) \in \{0, 1\}\). The function \(g = |f|\) is Borel measurable and nonnegative, and by hypothesis \(\int g \, d\lambda = \int |f| \, d\lambda < \infty\).

All hypotheses of 3.31 are satisfied, so

\begin{equation*} \lim_{k \to \infty} \int f_k \, d\lambda = \int f \, d\lambda , \end{equation*}

which by the first display is exactly

\begin{equation*} \lim_{k \to \infty} \int_{[-k, \, k]} f \, d\lambda = \int f \, d\lambda , \end{equation*}

as desired. (Note that all the integrals appearing here are defined and finite: \(|f_k| \le |f|\) and \(\int |f| \, d\lambda < \infty\), so by 3.8 each of \(\int f_k^+ \, d\lambda\), \(\int f_k^- \, d\lambda\), \(\int f^+ \, d\lambda\), \(\int f^- \, d\lambda\) is finite, and 3.18 applies.)

Let \(\varphi : \mathbb{R} \to [0, 1]\) be the tent function

\begin{equation*} \varphi(s) = \max\{ 0, \; 1 - |2s - 1| \} , \end{equation*}

so \(\varphi\) is continuous, \(\varphi(s) = 0\) for \(s \notin (0,1)\), \(\varphi(\tfrac12) = 1\), and \(\varphi\) is piecewise linear on \([0,1]\); its graph over \([0,1]\) is a triangle of base \(1\) and height \(1\), so its Riemann integral over \([0,1]\) equals \(\frac12\).

For \(n \in \mathbb{Z}^+\) define \(g_n : [0, \infty) \to \mathbb{R}\) by

\begin{equation*} g_n(x) = \frac{(-1)^{n+1}}{n} \, \varphi(x - n) , \end{equation*}

so \(g_n\) is continuous, \(g_n\) vanishes outside the interval \((n, n+1)\), and \(|g_n| \le \frac1n\) everywhere. Finally define \(f : [0, \infty) \to \mathbb{R}\) by

\begin{equation*} f(x) = \sum_{n=1}^{\infty} g_n(x) . \end{equation*}

For each \(x\) at most one term of this sum is nonzero (the intervals \((n, n+1)\) are disjoint), so \(f\) is well defined and real valued.

\(f\) is continuous. Fix \(N \in \mathbb{Z}^+\). If \(n > N\) then \(g_n\) vanishes on \([0, N+1]\), because \((n, n+1) \cap [0, N+1] = \varnothing\). Hence

\begin{equation*} f(x) = \sum_{n=1}^{N} g_n(x) \quad \text{for all } x \in [0, N+1] , \end{equation*}

and the right side is a finite sum of continuous functions, hence continuous. In particular \(f\) agrees with a continuous function on the set \([0, N+1)\), which is open in \([0, \infty)\). Since continuity is a local property and every point of \([0, \infty)\) lies in \([0, N+1)\) for all large \(N\), the function \(f\) is continuous on \([0, \infty)\). Being continuous, \(f\) is Borel measurable (2.41).

Computing \(\int g_n \, d\lambda\). Since \(g_n\) vanishes outside \([n, n+1]\) and is continuous, hence Riemann integrable on \([n, n+1]\), the identification 3.34 of Riemann and Lebesgue integrals gives

\begin{equation*} \int g_n \, d\lambda = \int_{[n, \, n+1]} g_n \, d\lambda = \frac{(-1)^{n+1}}{n} \int_{0}^{1} \varphi = \frac{(-1)^{n+1}}{2n} . \end{equation*}

In particular \(\int |g_n| \, d\lambda = \frac{1}{2n} < \infty\).

The limit exists. Let \(N \in \mathbb{Z}^+\). On \([0, N]\) we have \(f = \sum_{n=1}^{N-1} g_n\) by the displayed identity above, so additivity of integration (3.21), applicable because each \(\int |g_n| \, d\lambda < \infty\), gives

\begin{equation*} \int_{[0, \, N]} f \, d\lambda = \sum_{n=1}^{N-1} \int_{[0, \, N]} g_n \, d\lambda = \sum_{n=1}^{N-1} \frac{(-1)^{n+1}}{2n} . \end{equation*}

By the alternating series test the series \(\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n}\) converges; its sum is \(\frac{\ln 2}{2}\). Now take any real \(t \ge 1\) and set \(N = \lfloor t \rfloor\). Then \([0, t]\) is the disjoint union of \([0, N]\) and \((N, t]\), and \((N, t] \subseteq (N, N+1]\), where \(|f| \le \frac{1}{N}\). Hence by 3.21 and 3.25,

\begin{equation*} \left| \int_{[0, \, t]} f \, d\lambda - \int_{[0, \, N]} f \, d\lambda \right| = \left| \int_{(N, \, t]} f \, d\lambda \right| \le \lambda\bigl( (N, t] \bigr) \sup_{(N, \, t]} |f| \le 1 \cdot \frac{1}{N} . \end{equation*}

As \(t \to \infty\) we have \(N = \lfloor t \rfloor \to \infty\), so the right side tends to \(0\) and \(\int_{[0, N]} f \, d\lambda\) tends to \(\frac{\ln 2}{2}\). Therefore

\begin{equation*} \lim_{t \to \infty} \int_{[0, \, t]} f \, d\lambda = \frac{\ln 2}{2} \in \mathbb{R} . \end{equation*}

The integral over \([0, \infty)\) is not defined. Note \(g_n \ge 0\) when \(n\) is odd and \(g_n \le 0\) when \(n\) is even, and the \(g_n\) have disjoint supports; hence

\begin{equation*} f^+ = \sum_{n \text{ odd}} g_n \quad \text{and} \quad f^- = - \sum_{n \text{ even}} g_n . \end{equation*}

Fix \(K \in \mathbb{Z}^+\). Then \(f^+ \ge \sum_{k=1}^{K} g_{2k-1} \ge 0\) pointwise, so by 3.8 and additivity of integration for nonnegative functions (3.16),

\begin{equation*} \int f^+ \, d\lambda \ge \sum_{k=1}^{K} \int g_{2k-1} \, d\lambda = \sum_{k=1}^{K} \frac{1}{2(2k-1)} . \end{equation*}

Since \(\sum_{k=1}^{\infty} \frac{1}{2(2k-1)} = \infty\) (its terms are at least \(\frac{1}{4k}\) and the harmonic series diverges), letting \(K \to \infty\) gives \(\int f^+ \, d\lambda = \infty\). The same argument with the even-indexed pieces gives

\begin{equation*} \int f^- \, d\lambda \ge \sum_{k=1}^{K} \frac{1}{2 \cdot 2k} \quad \text{for every } K, \end{equation*}

so \(\int f^- \, d\lambda = \infty\) as well. Because neither \(\int f^+ \, d\lambda\) nor \(\int f^- \, d\lambda\) is finite, the definition 3.18 of the integral of a real-valued function does not apply: \(\int_{[0, \, \infty)} f \, d\lambda\) is not defined.

Thus \(f\) has both required properties. Intuitively, \(f\) is a continuous version of the alternating harmonic series: the truncated integrals converge because of cancellation between consecutive bumps, but the positive and negative parts each have infinite total area.

The idea is the same as in Exercise 6, but now the alternating bumps march toward \(0\) instead of toward \(\infty\); because the intervals shrink, the bumps must grow tall.

Again let \(\varphi : \mathbb{R} \to [0,1]\) be the tent function \(\varphi(s) = \max\{0, \, 1 - |2s-1|\}\), which is continuous and vanishes outside \((0,1)\).

For \(n \in \mathbb{Z}^+\) let

\begin{equation*} I_n = \Bigl( \frac{1}{n+1}, \, \frac{1}{n} \Bigr) , \qquad \text{so} \qquad \lambda(I_n) = \frac{1}{n} - \frac{1}{n+1} = \frac{1}{n(n+1)} , \end{equation*}

and define \(h_n : (0,1) \to \mathbb{R}\) by

\begin{equation*} h_n(x) = (-1)^{n+1} \, 2(n+1) \, \varphi\bigl( n(n+1)x - n \bigr) . \end{equation*}

Note that \(\varphi(n(n+1)x - n) \neq 0\) exactly when \(0 < n(n+1)x - n < 1\), that is, exactly when \(\frac{1}{n+1} < x < \frac{1}{n}\). Thus \(h_n\) is continuous on \((0,1)\), vanishes outside \(I_n\), and \(|h_n| \le 2(n+1)\). The graph of \(|h_n|\) over \(I_n\) is a triangle with base \(\lambda(I_n) = \frac{1}{n(n+1)}\) and height \(2(n+1)\). Since \(h_n\) vanishes outside \(I_n\), we have \(\int h_n \, d\lambda = \int_{[\frac{1}{n+1}, \, \frac{1}{n}]} h_n \, d\lambda\), and \(h_n\) restricted to the closed interval \([\frac{1}{n+1}, \frac{1}{n}]\) is continuous, hence Riemann integrable there; so the identification 3.34 of the Riemann and Lebesgue integrals gives

\begin{equation*} \int h_n \, d\lambda = (-1)^{n+1} \cdot \frac{1}{2} \cdot \frac{1}{n(n+1)} \cdot 2(n+1) = \frac{(-1)^{n+1}}{n} , \qquad \int |h_n| \, d\lambda = \frac{1}{n} . \end{equation*}

Now define \(f : (0,1) \to \mathbb{R}\) by

\begin{equation*} f(x) = \sum_{n=1}^{\infty} h_n(x) . \end{equation*}

The intervals \(I_1, I_2, \ldots\) are disjoint, so for each \(x \in (0,1)\) at most one term is nonzero and \(f\) is a well-defined real-valued function.

\(f\) is continuous. Fix \(N \in \mathbb{Z}^+\). If \(n > N\) then \(I_n \subseteq (0, \frac{1}{N+1})\), so \(h_n\) vanishes on \([\frac{1}{N+1}, 1)\). Hence

\begin{equation*} f(x) = \sum_{n=1}^{N} h_n(x) \quad \text{for all } x \in \Bigl( \frac{1}{N+1}, \, 1 \Bigr) , \end{equation*}

and the right side is a finite sum of continuous functions. Since the open sets \((\frac{1}{N+1}, 1)\) for \(N \in \mathbb{Z}^+\) cover \((0,1)\), the function \(f\) is continuous on \((0,1)\). Being continuous, \(f\) is Borel measurable (2.41). (Note that \(f\) is unbounded near \(0\), which is permitted.)

The limit exists. Fix \(n \ge 2\). By the displayed identity with \(N = n - 1\), we have \(f = \sum_{k=1}^{n-1} h_k\) on \((\frac{1}{n}, 1)\). Each \(h_k\) with \(k \le n-1\) vanishes outside \(I_k\), and \(I_k \subseteq (\frac{1}{n}, 1)\) because \(\frac{1}{k+1} \ge \frac{1}{n}\); hence \(\int_{(\frac{1}{n}, 1)} h_k \, d\lambda = \int h_k \, d\lambda\). Since \(\int |h_k| \, d\lambda = \frac1k < \infty\), additivity of integration (3.21) applies and gives

\begin{equation*} \int_{(\frac{1}{n}, \, 1)} f \, d\lambda = \sum_{k=1}^{n-1} \int h_k \, d\lambda = \sum_{k=1}^{n-1} \frac{(-1)^{k+1}}{k} . \end{equation*}

By the alternating series test the series \(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k}\) converges, with sum \(\ln 2\). Therefore

\begin{equation*} \lim_{n \to \infty} \int_{(\frac{1}{n}, \, 1)} f \, d\lambda = \ln 2 \in \mathbb{R} . \end{equation*}

The integral over \((0,1)\) is not defined. Because \(h_k \ge 0\) for \(k\) odd, \(h_k \le 0\) for \(k\) even, and the \(h_k\) have disjoint supports,

\begin{equation*} f^+ = \sum_{k \text{ odd}} h_k \quad \text{and} \quad f^- = - \sum_{k \text{ even}} h_k . \end{equation*}

Fix \(K \in \mathbb{Z}^+\). Then \(f^+ \ge \sum_{k=1}^{K} h_{2k-1} \ge 0\) pointwise on \((0,1)\), so 3.8 together with additivity for nonnegative functions (3.16) gives

\begin{equation*} \int_{(0, \, 1)} f^+ \, d\lambda \ge \sum_{k=1}^{K} \int h_{2k-1} \, d\lambda = \sum_{k=1}^{K} \frac{1}{2k-1} . \end{equation*}

Letting \(K \to \infty\) and using the divergence of \(\sum_{k=1}^{\infty} \frac{1}{2k-1}\) (its terms are at least \(\frac{1}{2k}\)), we get \(\int_{(0,1)} f^+ \, d\lambda = \infty\). Similarly

\begin{equation*} \int_{(0, \, 1)} f^- \, d\lambda \ge \sum_{k=1}^{K} \frac{1}{2k} \quad \text{for every } K \in \mathbb{Z}^+ , \end{equation*}

so \(\int_{(0,1)} f^- \, d\lambda = \infty\). Since neither of these two integrals is finite, the definition 3.18 does not assign a value: \(\int_{(0, 1)} f \, d\lambda\) is not defined.

Thus \(f\) has both required properties.

Recall the setting of the proof of 3.34. Here \(a < b\) and \(f : [a,b] \to \mathbb{R}\) is bounded. For \(n \in \mathbb{Z}^+\) the partition \(P_n\) cuts \([a,b]\) into \(2^n\) closed subintervals \(I_1, \ldots, I_{2^n}\), each of length \(\delta_n = (b-a)/2^n\), and

\begin{equation*} g_n = \sum_{j=1}^{2^n} \Bigl( \inf_{I_j} f \Bigr) \chi_{I_j}, \qquad h_n = \sum_{j=1}^{2^n} \Bigl( \sup_{I_j} f \Bigr) \chi_{I_j}, \end{equation*}

with the modification described in the proof: at each of the finitely many points \(x\) that lie in two of the intervals \(I_1, \ldots, I_{2^n}\), the value \(g_n(x)\) is redefined to be the infimum of \(f\) over the union of the two intervals containing \(x\), and \(h_n(x)\) is redefined to be the supremum of \(f\) over that union. Finally \(f_L(x) = \lim_{n \to \infty} g_n(x)\) and \(f_U(x) = \lim_{n \to \infty} h_n(x)\), these limits existing because \(g_1 \le g_2 \le \cdots\) and \(h_1 \ge h_2 \ge \cdots\) and \(f\) is bounded.

The assertion 3.38 to be verified is

\begin{equation*} \{ x \in [a,b] : f_U(x) \neq f_L(x) \} = \{ x \in [a,b] : f \text{ is not continuous at } x \}. \end{equation*}

A uniform description of \(g_n\) and \(h_n\). For \(x \in [a,b]\) let

\begin{equation*} J_n(x) = \bigcup \{ I_j : 1 \le j \le 2^n \text{ and } x \in I_j \}. \end{equation*}

Thus \(J_n(x)\) is the union of the one or two intervals of \(P_n\) that contain \(x\); in either case \(J_n(x)\) is a closed interval containing \(x\), of length \(\delta_n\) or \(2\delta_n\). With the modification above in force,

\begin{equation*} g_n(x) = \inf_{J_n(x)} f \qquad \text{and} \qquad h_n(x) = \sup_{J_n(x)} f \end{equation*}

for every \(x \in [a,b]\) and every \(n \in \mathbb{Z}^+\). (This is exactly what the modification buys us: at a point lying in two intervals the unmodified formulas would add two terms rather than give a single infimum or supremum.)

Two consequences will be used repeatedly. First, \(x \in J_n(x)\), so

\begin{equation*} g_n(x) \le f(x) \le h_n(x), \end{equation*}

and letting \(n \to \infty\) gives \(f_L(x) \le f(x) \le f_U(x)\) for all \(x \in [a,b]\); in particular \(f_U(x) = f_L(x)\) forces \(f_L(x) = f(x) = f_U(x)\). Second, because \(g_n(x) \le f_L(x)\) and \(f_U(x) \le h_n(x)\) for every \(n\) (monotonicity of the two sequences), we have

\begin{equation*} 0 \le f_U(x) - f_L(x) \le h_n(x) - g_n(x) = \sup_{J_n(x)} f - \inf_{J_n(x)} f \end{equation*}

for every \(n \in \mathbb{Z}^+\).

Continuity at \(x\) implies \(f_L(x) = f_U(x)\). Suppose \(f\) is continuous at \(x\) and let \(\varepsilon > 0\). Choose \(\eta > 0\) such that \(|f(y) - f(x)| < \varepsilon\) for every \(y \in [a,b]\) with \(|y - x| < \eta\). Choose \(n \in \mathbb{Z}^+\) with \(2\delta_n < \eta\). Since \(J_n(x)\) is an interval containing \(x\) of length at most \(2\delta_n < \eta\), every \(y \in J_n(x)\) satisfies \(|y - x| < \eta\) and hence \(|f(y) - f(x)| < \varepsilon\). Therefore

\begin{equation*} \sup_{J_n(x)} f - \inf_{J_n(x)} f \le 2\varepsilon, \end{equation*}

and so \(0 \le f_U(x) - f_L(x) \le 2\varepsilon\) by the displayed inequality above. As \(\varepsilon > 0\) was arbitrary, \(f_U(x) = f_L(x)\).

\(f_L(x) = f_U(x)\) implies continuity at \(x\). Suppose \(f_L(x) = f_U(x)\); as noted, then \(f_L(x) = f(x) = f_U(x)\). Let \(\varepsilon > 0\). Since \(g_n(x) \to f_L(x)\) and \(h_n(x) \to f_U(x)\) and these two limits are equal, there exists \(n \in \mathbb{Z}^+\) with

\begin{equation*} h_n(x) - g_n(x) < \varepsilon. \end{equation*}

Now we claim that \(x\) is an interior point of \(J_n(x)\) relative to \([a,b]\); that is, there exists \(\eta > 0\) such that

\begin{equation*} \{ y \in [a,b] : |y - x| < \eta \} \subseteq J_n(x). \end{equation*}

Indeed, write \(J_n(x) = [c,d]\). If \(c > a\) then \(c\) is a partition point of \(P_n\) that is not an endpoint of \([a,b]\), so \(c\) lies in two of the intervals \(I_1, \ldots, I_{2^n}\); if \(x\) were equal to \(c\), then \(J_n(x)\) would also contain the interval of \(P_n\) lying immediately to the left of \(c\), contradicting \(\min J_n(x) = c\). Hence \(c > a\) forces \(x > c\). Symmetrically \(d < b\) forces \(x < d\). Consequently either \(x > c\) or \(c = a\), and either \(x < d\) or \(d = b\), which is precisely the statement that \(x\) is interior to \([c,d]\) relative to \([a,b]\). So such an \(\eta > 0\) exists.

Now let \(y \in [a,b]\) with \(|y - x| < \eta\). Then \(y \in J_n(x)\), so

\begin{equation*} g_n(x) = \inf_{J_n(x)} f \le f(y) \le \sup_{J_n(x)} f = h_n(x), \end{equation*}

and the same inequalities hold with \(y\) replaced by \(x\). Subtracting,

\begin{equation*} |f(y) - f(x)| \le h_n(x) - g_n(x) < \varepsilon. \end{equation*}

Thus \(f\) is continuous at \(x\).

The two implications together say that \(f_U(x) = f_L(x)\) if and only if \(f\) is continuous at \(x\), which is 3.38.

The assertion of 3.41 is the following. Suppose \((X, \mathcal{S}, \mu)\) is a measure space, \(E_1, \ldots, E_n\) are disjoint subsets of \(X\), and \(a_1, \ldots, a_n\) are distinct nonzero real numbers. Then

\begin{equation*} a_1 \chi_{E_1} + \cdots + a_n \chi_{E_n} \in L^1(\mu) \end{equation*}

if and only if \(E_k \in \mathcal{S}\) and \(\mu(E_k) < \infty\) for all \(k \in \{1, \ldots, n\}\); and in that case

\begin{equation*} \| a_1 \chi_{E_1} + \cdots + a_n \chi_{E_n} \|_1 = |a_1| \mu(E_1) + \cdots + |a_n| \mu(E_n). \end{equation*}

Note that the sets \(E_k\) are not assumed to be in \(\mathcal{S}\); their measurability is part of what is to be proved in one direction.

Throughout, put

\begin{equation*} f = a_1 \chi_{E_1} + \cdots + a_n \chi_{E_n}, \end{equation*}

a function from \(X\) to \(\mathbb{R}\).

The pointwise description of \(f\). Because \(E_1, \ldots, E_n\) are disjoint, each \(x \in X\) lies in at most one of them, so

\begin{equation*} f(x) = \begin{cases} a_k & \text{if } x \in E_k \text{ for some (necessarily unique) } k, \\ 0 & \text{if } x \notin E_1 \cup \cdots \cup E_n. \end{cases} \end{equation*}

Since \(a_1, \ldots, a_n\) are distinct and nonzero, this shows that for each \(k \in \{1, \ldots, n\}\),

\begin{equation*} E_k = f^{-1}(\{a_k\}). \end{equation*}

Indeed, \(x \in E_k\) gives \(f(x) = a_k\); conversely if \(f(x) = a_k\) then \(f(x) \neq 0\), so \(x\) lies in some \(E_j\) and \(a_j = f(x) = a_k\), which forces \(j = k\) because the numbers \(a_1, \ldots, a_n\) are distinct.

Similarly, by disjointness,

\begin{equation*} |f| = |a_1| \chi_{E_1} + \cdots + |a_n| \chi_{E_n}. \end{equation*}

Suppose \(E_k \in \mathcal{S}\) and \(\mu(E_k) < \infty\) for all \(k\). Each characteristic function \(\chi_{E_k}\) is then \(\mathcal{S}\)-measurable, and a finite linear combination of \(\mathcal{S}\)-measurable real-valued functions is \(\mathcal{S}\)-measurable (by 2.46); hence \(f\) is an \(\mathcal{S}\)-measurable function from \(X\) to \(\mathbb{R}\). The sets \(E_1, \ldots, E_n\) are disjoint sets in \(\mathcal{S}\) and \(|a_1|, \ldots, |a_n| \in [0, \infty)\), so 3.7 applied to the displayed formula for \(|f|\) gives

\begin{equation*} \|f\|_1 = \int |f| \, d\mu = |a_1| \mu(E_1) + \cdots + |a_n| \mu(E_n) < \infty, \end{equation*}

the finiteness holding because each \(\mu(E_k) < \infty\). Thus \(f \in L^1(\mu)\), and the asserted formula for \(\|f\|_1\) holds.

Suppose \(f \in L^1(\mu)\). By the definition of \(L^1(\mu)\) (see 3.40), \(f\) is an \(\mathcal{S}\)-measurable function from \(X\) to \(\mathbb{R}\) with \(\|f\|_1 < \infty\). Fix \(k \in \{1, \ldots, n\}\). The singleton \(\{a_k\}\) is a closed subset of \(\mathbb{R}\) and hence a Borel set, so the definition of \(\mathcal{S}\)-measurability (2.35) gives

\begin{equation*} E_k = f^{-1}(\{a_k\}) \in \mathcal{S}. \end{equation*}

Thus all of \(E_1, \ldots, E_n\) belong to \(\mathcal{S}\), and now 3.7 applies exactly as above and yields

\begin{equation*} |a_1| \mu(E_1) + \cdots + |a_n| \mu(E_n) = \int |f| \, d\mu = \|f\|_1 < \infty. \end{equation*}

Every term on the left is in \([0, \infty]\), so every term is finite; since \(a_k \neq 0\) we have \(|a_k| > 0\), and therefore \(\mu(E_k) < \infty\) for each \(k\).

This proves both implications, and the displayed formula for \(\|f\|_1\) was obtained in each case, completing the verification of 3.41.

A remark on the hypotheses. The assumption that the \(a_k\) are nonzero is needed for the conclusion \(\mu(E_k) < \infty\): if some \(a_k\) were \(0\), then \(E_k\) would contribute nothing to \(f\) and could be a nonmeasurable set or a set of infinite measure. The assumption that the \(a_k\) are distinct is needed for the identification \(E_k = f^{-1}(\{a_k\})\): if \(a_1 = a_2\), then \(f^{-1}(\{a_1\}) = E_1 \cup E_2\), and one could take \(E_1\), \(E_2\) nonmeasurable with measurable union.

(a) First note that \(f^p\) and \(f^r\) are \(\mathcal{S}\)-measurable functions from \(X\) to \([0, \infty)\). Indeed, for \(t > 0\) the map \(s \mapsto s^t\) is continuous on \([0, \infty)\), hence Borel measurable by 2.41, and therefore the composition \(f^t = (s \mapsto s^t) \circ f\) is \(\mathcal{S}\)-measurable by 2.44.

The key pointwise estimate is

\begin{equation*} f(x)^p \le 1 + f(x)^r \qquad \text{for every } x \in X. \end{equation*}

To see this, fix \(x \in X\) and consider two cases. If \(f(x) \le 1\), then \(f(x)^p \le 1 \le 1 + f(x)^r\). If \(f(x) > 1\), then because \(p < r\) and \(f(x) > 1\) we have \(f(x)^p \le f(x)^r \le 1 + f(x)^r\). (The inequality \(s^p \le s^r\) for \(s > 1\) holds since \(s^r / s^p = s^{r-p} \ge 1\) when \(s > 1\) and \(r - p > 0\).) This establishes the estimate in both cases.

Now integrate. The function \(1 + f^r\) is a nonnegative \(\mathcal{S}\)-measurable function, so by the order-preserving property of integration (3.8), additivity of integration for nonnegative functions (3.16), and 3.4 applied to \(\chi_X\),

\begin{equation*} \int f^p \, d\mu \le \int (1 + f^r) \, d\mu = \int 1 \, d\mu + \int f^r \, d\mu = \mu(X) + \int f^r \, d\mu. \end{equation*}

By hypothesis \(\mu(X) < \infty\) and \(\int f^r \, d\mu < \infty\), so the right side is finite. Hence \(\int f^p \, d\mu < \infty\), as desired.

(b) Take \(X = (1, \infty)\), let \(\mathcal{S}\) be the collection of Borel subsets of \((1, \infty)\), and let \(\mu = \lambda\) be Lebesgue measure restricted to \(\mathcal{S}\). Then \(\mu(X) = \infty\). Take

\begin{equation*} p = 1, \qquad r = 2, \qquad f(x) = \frac{1}{x} \quad \text{for } x \in (1, \infty), \end{equation*}

which is a continuous, hence Borel measurable, function from \((1, \infty)\) into \([0, \infty)\).

The functions \(f^2 \chi_{(1,k)}\) for \(k = 2, 3, \ldots\) increase pointwise to \(f^2\) on \((1, \infty)\), so the Monotone Convergence Theorem (3.11), together with 3.34 (which identifies the Lebesgue integral of a continuous function on a bounded closed interval with its Riemann integral), gives

\begin{equation*} \int f^2 \, d\mu = \lim_{k \to \infty} \int_1^k \frac{1}{x^2} \, dx = \lim_{k \to \infty} \Bigl( 1 - \frac{1}{k} \Bigr) = 1 < \infty. \end{equation*}

The same argument applied to \(f\) itself gives

\begin{equation*} \int f^1 \, d\mu = \lim_{k \to \infty} \int_1^k \frac{1}{x} \, dx = \lim_{k \to \infty} \ln k = \infty. \end{equation*}

Thus \(\int f^r \, d\mu < \infty\) while \(\int f^p \, d\mu = \infty\), even though \(p < r\). Hence the conclusion of (a) fails without the hypothesis \(\mu(X) < \infty\).

(An even simpler example, with counting measure: let \(X = \mathbb{Z}^+\), let \(\mathcal{S}\) be all subsets of \(\mathbb{Z}^+\), let \(\mu\) be counting measure, take \(p = 1\), \(r = 2\), and \(f(k) = 1/k\). Then \(\int f^2 \, d\mu = \sum_{k=1}^\infty 1/k^2 < \infty\) but \(\int f \, d\mu = \sum_{k=1}^\infty 1/k = \infty\).)

For each \(k \in \mathbb{Z}^+\) define

\begin{equation*} A_k = \Bigl\{ x \in X : |f(x)| > \tfrac{1}{k} \Bigr\}. \end{equation*}

Because \(f\) is \(\mathcal{S}\)-measurable, so is \(|f|\) (by 2.45), and hence

\begin{equation*} A_k = |f|^{-1}\bigl( (\tfrac{1}{k}, \infty) \bigr) \in \mathcal{S} \end{equation*}

for every \(k \in \mathbb{Z}^+\).

The sets \(A_k\) cover the set in question. Indeed, suppose \(x \in X\) and \(f(x) \ne 0\). Then \(|f(x)| > 0\), so there exists \(k \in \mathbb{Z}^+\) with \(1/k < |f(x)|\), which means \(x \in A_k\). Conversely, if \(x \in A_k\) for some \(k\), then \(|f(x)| > 1/k > 0\) and so \(f(x) \ne 0\). Therefore

\begin{equation*} \{ x \in X : f(x) \ne 0 \} = \bigcup_{k=1}^{\infty} A_k, \end{equation*}

a countable union of sets in \(\mathcal{S}\).

It remains to show that \(\mu(A_k) < \infty\) for each \(k\). Fix \(k \in \mathbb{Z}^+\). From the definition of \(A_k\) we have the pointwise inequality

\begin{equation*} \tfrac{1}{k} \chi_{A_k}(x) \le |f(x)| \qquad \text{for every } x \in X; \end{equation*}

this is clear at points of \(A_k\) (where the left side equals \(1/k < |f(x)|\)) and at points outside \(A_k\) (where the left side equals \(0 \le |f(x)|\)). Both sides are nonnegative \(\mathcal{S}\)-measurable functions, so the order-preserving property of integration (3.8) gives

\begin{equation*} \int \tfrac{1}{k} \chi_{A_k} \, d\mu \le \int |f| \, d\mu = \|f\|_1. \end{equation*}

By 3.7 (with the single set \(A_k \in \mathcal{S}\) and the constant \(1/k\)), the left side equals \(\frac{1}{k} \mu(A_k)\). Hence

\begin{equation*} \mu(A_k) \le k \, \|f\|_1. \end{equation*}

Since \(f \in L^1(\mu)\) we have \(\|f\|_1 < \infty\), and therefore \(\mu(A_k) < \infty\).

Thus \(\{ x \in X : f(x) \ne 0 \}\) is the union of the countably many sets \(A_1, A_2, \ldots\), each of which lies in \(\mathcal{S}\) and has finite \(\mu\)-measure, as required.

Two remarks. First, the sets \(A_k\) increase: \(A_1 \subseteq A_2 \subseteq \cdots\). So the union can also be written as an increasing union, or (replacing \(A_k\) by \(A_k \setminus A_{k-1}\)) as a disjoint countable union of sets of finite measure. Second, the inequality \(\mu(A_k) \le k \|f\|_1\) derived above is the Chebyshev-type estimate \(\mu(\{|f| > t\}) \le \|f\|_1 / t\) for \(t > 0\).

Let \(\lambda\) denote Lebesgue measure on \(\mathbb{R}\). By 3.39, the assertion to be proved is that

\begin{equation*} \lim_{k \to \infty} \int_{(0,1)} f_k \, d\lambda = 0, \end{equation*}

where each \(f_k\) is regarded as a function on \((0,1)\).

Measurability. Each \(f_k\) is continuous on \((0,1)\) (a quotient of continuous functions with nonvanishing denominator), hence Borel measurable by 2.41, hence Lebesgue measurable.

A dominating function. Define \(g : (0,1) \to (0, \infty)\) by

\begin{equation*} g(x) = \frac{1}{\sqrt{x}}. \end{equation*}

For every \(x \in (0,1)\) we have \(0 < 1 - x < 1\), hence \(0 < (1-x)^k \le 1\), and \(|\cos x^k| \le 1\); therefore

\begin{equation*} |f_k(x)| = \frac{(1-x)^k \, |\cos x^k|}{\sqrt{x}} \le \frac{1}{\sqrt{x}} = g(x) \end{equation*}

for every \(k \in \mathbb{Z}^+\) and every \(x \in (0,1)\).

We now check that \(\int_{(0,1)} g \, d\lambda < \infty\). For \(n \in \mathbb{Z}^+\) let \(g_n = g \chi_{(1/n, 1)}\); note \(g_1 = 0\) since \((1,1) = \emptyset\). The intervals \((1/n, 1)\) increase with \(n\) and \(g \ge 0\), so \(g_1 \le g_2 \le \cdots\), and \(g_n(x) \to g(x)\) for every \(x \in (0,1)\), since each \(x \in (0,1)\) satisfies \(x > 1/n\) for all large \(n\). Because \(g\) is continuous on the closed interval \([1/n, 1]\), the Lebesgue and Riemann integrals agree there (3.34), and

\begin{equation*} \int_{(0,1)} g_n \, d\lambda = \int_{1/n}^{1} \frac{dx}{\sqrt{x}} = \Bigl[ 2\sqrt{x} \Bigr]_{1/n}^{1} = 2 - \frac{2}{\sqrt{n}}. \end{equation*}

By the Monotone Convergence Theorem (3.11),

\begin{equation*} \int_{(0,1)} g \, d\lambda = \lim_{n \to \infty} \Bigl( 2 - \frac{2}{\sqrt{n}} \Bigr) = 2 < \infty. \end{equation*}

Pointwise convergence to \(0\). Fix \(x \in (0,1)\). Then \(0 < 1 - x < 1\), so \((1-x)^k \to 0\) as \(k \to \infty\). Since \(|\cos x^k| \le 1\) and \(1/\sqrt{x}\) does not depend on \(k\),

\begin{equation*} |f_k(x)| \le \frac{(1-x)^k}{\sqrt{x}} \longrightarrow 0 \qquad \text{as } k \to \infty. \end{equation*}

Hence \(\lim_{k \to \infty} f_k(x) = 0\) for every \(x \in (0,1)\).

Conclusion. The hypotheses of the Dominated Convergence Theorem (3.31) are satisfied on the measure space \(\bigl( (0,1), \mathcal{B}, \lambda \bigr)\), where \(\mathcal{B}\) denotes the Borel subsets of \((0,1)\): the functions \(f_1, f_2, \ldots\) are measurable, they converge pointwise to the function \(0\), and they are dominated by the measurable function \(g \ge 0\) with \(\int g \, d\lambda = 2 < \infty\). Therefore

\begin{equation*} \lim_{k \to \infty} \int_0^1 f_k = \lim_{k \to \infty} \int_{(0,1)} f_k \, d\lambda = \int_{(0,1)} 0 \, d\lambda = 0, \end{equation*}

as desired.

(Note that the domination also shows \(f_k \in L^1(\lambda)\) for each \(k\), so each of the integrals \(\int_0^1 f_k\) is a well-defined real number.)

The idea is a “typewriter” sequence whose bumps sweep across \([0,1]\) over and over, getting narrower fast enough that the integrals tend to \(0\) but taller without bound, so that at every point infinitely many of the functions are large.

The construction. For \(n \in \mathbb{Z}^+\) let

\begin{equation*} N_n = 1 + 2 + \cdots + n = \frac{n(n+1)}{2}, \end{equation*}

and set \(N_0 = 0\). Since \(N_0 < N_1 < N_2 < \cdots\) and \(N_n \to \infty\), every \(k \in \mathbb{Z}^+\) can be written uniquely as

\begin{equation*} k = N_{n-1} + j \qquad \text{with } n \in \mathbb{Z}^+ \text{ and } j \in \{1, \ldots, n\}. \end{equation*}

For such \(k\) define \(f_k : [0,1] \to [0, \infty)\) by

\begin{equation*} f_k = \sqrt{n} \; \chi_{[\frac{j-1}{n}, \frac{j}{n}]}. \end{equation*}

So the sequence begins \(1 \cdot \chi_{[0,1]}\); then \(\sqrt{2}\,\chi_{[0,1/2]}, \sqrt{2}\,\chi_{[1/2,1]}\); then \(\sqrt{3}\,\chi_{[0,1/3]}, \sqrt{3}\,\chi_{[1/3,2/3]}, \sqrt{3}\,\chi_{[2/3,1]}\); and so on.

Each \(f_k\) is a nonnegative constant times the characteristic function of a closed interval, hence is a nonnegative Borel measurable function on \([0,1]\).

The first condition. Let \(\lambda\) denote Lebesgue measure. If \(k = N_{n-1} + j\) as above, then by 3.7 (or 3.4 together with 3.7),

\begin{equation*} \int_0^1 f_k \, d\lambda = \sqrt{n} \cdot \lambda\Bigl( \Bigl[ \tfrac{j-1}{n}, \tfrac{j}{n} \Bigr] \Bigr) = \sqrt{n} \cdot \frac{1}{n} = \frac{1}{\sqrt{n}}. \end{equation*}

Write \(n(k)\) for the integer \(n\) attached to \(k\) by the decomposition above. For each fixed \(n\), only the finitely many indices \(k \in \{N_{n-1}+1, \ldots, N_n\}\) have \(n(k) = n\); hence only finitely many \(k\) satisfy \(n(k) \le n\), and therefore \(n(k) \to \infty\) as \(k \to \infty\). Consequently

\begin{equation*} \lim_{k \to \infty} \int_0^1 f_k = \lim_{k \to \infty} \frac{1}{\sqrt{n(k)}} = 0. \end{equation*}

The second condition. Fix \(x \in [0,1]\) and \(m \in \mathbb{Z}^+\). Let \(n \in \mathbb{Z}^+\) be arbitrary. The \(n\) intervals

\begin{equation*} \Bigl[ 0, \tfrac{1}{n} \Bigr], \ \Bigl[ \tfrac{1}{n}, \tfrac{2}{n} \Bigr], \ \ldots, \ \Bigl[ \tfrac{n-1}{n}, 1 \Bigr] \end{equation*}

cover \([0,1]\), so there exists \(j \in \{1, \ldots, n\}\) with \(x \in [\frac{j-1}{n}, \frac{j}{n}]\). Put \(k_n = N_{n-1} + j\). Then

\begin{equation*} f_{k_n}(x) = \sqrt{n}. \end{equation*}

Moreover \(k_n > N_{n-1}\), and \(N_{n-1} \to \infty\) as \(n \to \infty\), so \(k_n \ge m\) for all sufficiently large \(n\). Hence for every \(n\) large enough,

\begin{equation*} \sup_{k \ge m} f_k(x) \ge f_{k_n}(x) = \sqrt{n}. \end{equation*}

Letting \(n \to \infty\) gives \(\sup_{k \ge m} f_k(x) = \infty\). Since \(x \in [0,1]\) and \(m \in \mathbb{Z}^+\) were arbitrary, the second condition holds.

Remark. The second condition says \(\limsup_{k \to \infty} f_k(x) = \infty\) for every \(x \in [0,1]\); in particular \(f_1, f_2, \ldots\) converges pointwise at no point of \([0,1]\), even though the integrals tend to \(0\). Note also that \(\int_0^1 |f_k - 0| \, d\lambda = 1/\sqrt{n(k)} \to 0\), so this sequence converges to \(0\) in \(L^1\)-norm while diverging at every point; this shows that convergence in \(L^1\)-norm does not imply pointwise convergence anywhere.

(a) Define \(f : [0,1] \to [0, \infty]\) by \(f(x) = 1/\sqrt{x}\) for \(x \in (0,1]\) and \(f(0) = \infty\). (The value assigned at \(0\) is irrelevant to the integral, since \(\lambda(\{0\}) = 0\); taking \(f(0) = \infty\) makes the monotone approximation below converge at every point of \([0,1]\).)

Measurability. Let \(a \in \mathbb{R}\). If \(a \le 0\) then \(f^{-1}((a, \infty]) = [0,1]\). If \(a > 0\) then, since for \(x \in (0,1]\) we have \(1/\sqrt{x} > a\) if and only if \(x < 1/a^2\),

\begin{equation*} f^{-1}\bigl( (a, \infty] \bigr) = [0,1] \cap \Bigl[ 0, \tfrac{1}{a^2} \Bigr), \end{equation*}

which is \([0, 1/a^2)\) if \(a \ge 1\) and \([0,1]\) if \(0 < a < 1\). In every case the set is a Borel set, so \(f\) is Borel measurable by 2.52.

Monotone approximation. For \(n \in \mathbb{Z}^+\) define \(g_n : [0,1] \to [0, \infty)\) by \(g_n = \min\{f, n\}\); explicitly,

\begin{equation*} g_n(x) = \begin{cases} n & \text{if } 0 \le x \le \dfrac{1}{n^2}, \\[2mm] \dfrac{1}{\sqrt{x}} & \text{if } \dfrac{1}{n^2} \le x \le 1. \end{cases} \end{equation*}

The two formulas agree at \(x = 1/n^2\) (both give \(n\)), so \(g_n\) is continuous on \([0,1]\); in particular \(g_n\) is Borel measurable by 2.41. Clearly \(0 \le g_1 \le g_2 \le \cdots\), and \(g_n(x) \to f(x)\) for every \(x \in [0,1]\): if \(x > 0\) then \(g_n(x) = f(x)\) as soon as \(n \ge 1/\sqrt{x}\), while \(g_n(0) = n \to \infty = f(0)\).

Because \(g_n\) is continuous on \([0,1]\), it is Riemann integrable there, and its Riemann and Lebesgue integrals agree by 3.34. Hence

\begin{equation*} \int_{[0,1]} g_n \, d\lambda = \int_0^1 g_n = n \cdot \frac{1}{n^2} + \int_{1/n^2}^{1} \frac{dx}{\sqrt{x}} = \frac{1}{n} + \Bigl[ 2\sqrt{x} \Bigr]_{1/n^2}^{1} = \frac{1}{n} + \Bigl( 2 - \frac{2}{n} \Bigr) = 2 - \frac{1}{n}. \end{equation*}

By the Monotone Convergence Theorem (3.11),

\begin{equation*} \int_{[0,1]} f \, d\lambda = \lim_{n \to \infty} \int_{[0,1]} g_n \, d\lambda = \lim_{n \to \infty} \Bigl( 2 - \frac{1}{n} \Bigr) = 2. \end{equation*}

(b) Here \(f(x) = 1/(1+x^2)\) is continuous on \(\mathbb{R}\), hence Borel measurable by 2.41, and \(f > 0\).

For \(n \in \mathbb{Z}^+\) let \(g_n = f \chi_{[-n, n]}\). Then \(g_n\) is a nonnegative Borel measurable function, \(g_1 \le g_2 \le \cdots\), and \(g_n(x) \to f(x)\) for every \(x \in \mathbb{R}\) (given \(x\), we have \(g_n(x) = f(x)\) once \(n \ge |x|\)).

Since \(f\) is continuous on \([-n, n]\), it is Riemann integrable there and 3.34 gives

\begin{equation*} \int_{\mathbb{R}} g_n \, d\lambda = \int_{[-n,n]} f \, d\lambda = \int_{-n}^{n} \frac{dx}{1+x^2} = \Bigl[ \arctan x \Bigr]_{-n}^{n} = 2 \arctan n, \end{equation*}

where we used that \(\arctan\) is an antiderivative of \(x \mapsto 1/(1+x^2)\) and that \(\arctan\) is odd.

By the Monotone Convergence Theorem (3.11),

\begin{equation*} \int_{\mathbb{R}} f \, d\lambda = \lim_{n \to \infty} 2 \arctan n = 2 \cdot \frac{\pi}{2} = \pi. \end{equation*}

(c) Let \(f : (0, \infty) \to \mathbb{R}\) be defined by \(f(x) = (\sin x)/x\). This function is continuous on \((0,\infty)\), hence Borel measurable by 2.41. The functions \(s \mapsto \max\{s, 0\}\) and \(s \mapsto \max\{-s, 0\}\) are continuous on \(\mathbb{R}\), hence Borel measurable by 2.41, so \(f^+\) and \(f^-\) are nonnegative Borel measurable functions on \((0,\infty)\) by 2.44.

The integral is not defined. By 3.18, \(\int_{(0,\infty)} f \, d\lambda\) is defined only if at least one of \(\int f^+ \, d\lambda\) and \(\int f^- \, d\lambda\) is finite. We show both are infinite.

For \(k \in \{0, 1, 2, \ldots\}\) set

\begin{equation*} A_k = \bigl( 2k\pi, (2k+1)\pi \bigr) \qquad \text{and} \qquad B_k = \bigl( (2k+1)\pi, (2k+2)\pi \bigr). \end{equation*}

These are open intervals contained in \((0,\infty)\), and \(A_0, B_0, A_1, B_1, \ldots\) are pairwise disjoint (consecutive intervals meet only at the excluded endpoints \(\pi, 2\pi, 3\pi, \ldots\)). On \(A_k\) we have \(\sin x > 0\), so \(f > 0\) there and \(f^+ = f\) on \(A_k\); moreover \(0 < x < (2k+1)\pi\) on \(A_k\), whence

\begin{equation*} f^+(x) = \frac{\sin x}{x} \ge \frac{\sin x}{(2k+1)\pi} \qquad \text{for } x \in A_k. \end{equation*}

Since \(x \mapsto \sin x\) is continuous on the closed interval \([2k\pi, (2k+1)\pi]\), it is Riemann integrable there, and 3.34 together with the Fundamental Theorem of Calculus gives \(\int_{[2k\pi, (2k+1)\pi]} \sin x \, d\lambda = \bigl[ -\cos x \bigr]_{2k\pi}^{(2k+1)\pi} = 1 + 1 = 2\); deleting the two endpoints changes nothing, since a set consisting of two points has \(\lambda\)-measure \(0\) and the integrand is bounded. Hence \(\int \chi_{A_k} \sin x \, d\lambda = 2\), and the order-preserving property 3.8 gives

\begin{equation*} \int f^+ \chi_{A_k} \, d\lambda \ge \frac{2}{(2k+1)\pi}. \end{equation*}

Fix \(n \in \mathbb{Z}^+\). Because \(A_0, \ldots, A_{n-1}\) are disjoint and \(f^+ \ge 0\), we have \(\sum_{k=0}^{n-1} f^+ \chi_{A_k} \le f^+\) pointwise on \((0,\infty)\), so by 3.8 and additivity of integration for nonnegative functions (3.16),

\begin{equation*} \int f^+ \, d\lambda \ \ge \ \sum_{k=0}^{n-1} \int f^+ \chi_{A_k} \, d\lambda \ \ge \ \sum_{k=0}^{n-1} \frac{2}{(2k+1)\pi}. \end{equation*}

The series \(\sum_{k=0}^{\infty} \frac{2}{(2k+1)\pi}\) diverges (its terms are at least \(\frac{1}{(k+1)\pi}\), and the harmonic series diverges). Letting \(n \to \infty\) therefore gives \(\int f^+ \, d\lambda = \infty\).

The same argument with \(B_k\) in place of \(A_k\) handles \(f^-\). On \(B_k\) we have \(\sin x < 0\), so \(f < 0\) and \(f^- = -f\) there, and \(0 < x < (2k+2)\pi\) on \(B_k\), whence \(f^-(x) \ge \frac{-\sin x}{(2k+2)\pi}\) on \(B_k\). Since, exactly as above, \(\int \chi_{B_k} (-\sin x) \, d\lambda = \bigl[ \cos x \bigr]_{(2k+1)\pi}^{(2k+2)\pi} = 1 + 1 = 2\), we get

\begin{equation*} \int f^- \chi_{B_k} \, d\lambda \ge \frac{2}{(2k+2)\pi} = \frac{1}{(k+1)\pi}, \end{equation*}

and summing as before gives \(\int f^- \, d\lambda \ge \sum_{k=0}^{n-1} \frac{1}{(k+1)\pi}\) for every \(n\), so \(\int f^- \, d\lambda = \infty\).

Both \(\int f^+ \, d\lambda\) and \(\int f^- \, d\lambda\) equal \(\infty\), so by 3.18 the integral \(\int_{(0,\infty)} f \, d\lambda\) is not defined.

The limit exists. First observe that \(f\) extends to a continuous function \(\tilde f\) on \([0, \infty)\) by setting \(\tilde f(0) = 1\), because \(\lim_{x \to 0} (\sin x)/x = 1\). Fix \(t > 0\). Then \(\tilde f\) is continuous on \([0,t]\), hence Riemann integrable there. We claim

\begin{equation*} \int_{(0,t)} f \, d\lambda = \int_{[0,t]} \tilde f \, d\lambda = \int_0^t \frac{\sin x}{x} \, dx, \end{equation*}

where the right side denotes a Riemann integral. The second equality is the last assertion of 3.34. For the first: \(\tilde f = f\) on \((0,t)\), and \(\tilde f\) is bounded on \([0,t]\) by some \(c \in (0,\infty)\), so

\begin{equation*} \Bigl| \int_{[0,t]} \tilde f \, d\lambda - \int_{(0,t)} f \, d\lambda \Bigr| = \Bigl| \int \tilde f \chi_{\{0,t\}} \, d\lambda \Bigr| \le \int c \, \chi_{\{0,t\}} \, d\lambda = c \, \lambda(\{0,t\}) = 0 \end{equation*}

by 3.8 and 3.7. (Note also that \(\int_{(0,t)} f \, d\lambda\) is defined, indeed finite, because \(|f| \le c\) on \((0,t)\) and \(\lambda((0,t)) = t < \infty\).)

Now suppose \(t \ge 1\). The functions \(x \mapsto 1/x\) and \(x \mapsto -\cos x\) are continuously differentiable on \([1,t]\), so integration by parts for Riemann integrals gives

\begin{equation*} \int_1^t \frac{\sin x}{x} \, dx = \Bigl[ \frac{-\cos x}{x} \Bigr]_1^t - \int_1^t \frac{\cos x}{x^2} \, dx = \cos 1 - \frac{\cos t}{t} - \int_1^t \frac{\cos x}{x^2} \, dx. \end{equation*}

Consider \(h(x) = (\cos x)/x^2\) on \((1, \infty)\); \(h\) is continuous, hence Borel measurable, and \(|h(x)| \le 1/x^2\). Exactly as in part (a) and (b), the Monotone Convergence Theorem applied to the increasing sequence \(x^{-2} \chi_{(1,n)}\) together with 3.34 gives

\begin{equation*} \int_{(1,\infty)} \frac{1}{x^2} \, d\lambda = \lim_{n \to \infty} \int_1^n \frac{dx}{x^2} = \lim_{n \to \infty} \Bigl( 1 - \frac{1}{n} \Bigr) = 1 < \infty. \end{equation*}

Hence \(h \in L^1\bigl( (1,\infty) \bigr)\).

Let \(t_1, t_2, \ldots\) be any sequence in \((1, \infty)\) with \(t_n \to \infty\). Given \(x \in (1,\infty)\) we have \(t_n > x\) for all large \(n\), so \(h \chi_{(1, t_n)} \to h\) pointwise on \((1,\infty)\); and \(|h \chi_{(1,t_n)}(x)| \le 1/x^2\) for all \(n\), a dominating function with finite integral. Also, \(h\) is continuous on \([1, t_n]\), so 3.34 gives \(\int h \chi_{(1,t_n)} \, d\lambda = \int_1^{t_n} \frac{\cos x}{x^2} \, dx\) with the right side a Riemann integral (again the two endpoints form a \(\lambda\)-null set and \(h\) is bounded there). So the Dominated Convergence Theorem (3.31) gives

\begin{equation*} \lim_{n \to \infty} \int_1^{t_n} \frac{\cos x}{x^2} \, dx = \lim_{n \to \infty} \int h \chi_{(1,t_n)} \, d\lambda = \int_{(1,\infty)} h \, d\lambda. \end{equation*}

Since this holds for every such sequence, the limit

\begin{equation*} L = \lim_{t \to \infty} \int_1^t \frac{\cos x}{x^2} \, dx = \int_{(1,\infty)} \frac{\cos x}{x^2} \, d\lambda \end{equation*}

exists and is a real number.

Also \(|\cos t / t| \le 1/t \to 0\) as \(t \to \infty\). For \(t \ge 1\) additivity of the Riemann integral over adjacent intervals gives \(\int_0^t \frac{\sin x}{x} \, dx = \int_0^1 \frac{\sin x}{x} \, dx + \int_1^t \frac{\sin x}{x} \, dx\) (all three integrands being the continuous function \(\tilde f\)). Therefore, letting \(t \to \infty\) in the integration-by-parts identity,

\begin{equation*} \lim_{t \to \infty} \int_{(0,t)} f \, d\lambda = \int_0^1 \frac{\sin x}{x} \, dx + \lim_{t \to \infty} \int_1^t \frac{\sin x}{x} \, dx = \int_0^1 \frac{\sin x}{x} \, dx + \cos 1 - L, \end{equation*}

which is a real number. Thus \(\lim_{t \to \infty} \int_{(0,t)} f \, d\lambda\) exists in \(\mathbb{R}\), even though \(\int_{(0,\infty)} f \, d\lambda\) is not defined.

(For the record, the value of this limit is \(\pi/2\), but the exercise asks only for existence.)

The statement is false. We produce an open set \(G \subset (0,1)\) for which \(\chi_G\) is not Riemann integrable on \([0,1]\); the point is that \(G\) can be dense in \((0,1)\) and yet have small Lebesgue measure.

Construction of the counterexample. Let \(r_1, r_2, \ldots\) be an enumeration of the rational numbers in \((0,1)\) and set

\begin{equation*} G = (0,1) \cap \bigcup_{k=1}^{\infty} \left( r_k - \frac{1}{2^{k+2}},\ r_k + \frac{1}{2^{k+2}} \right). \end{equation*}

Then \(G\) is an open subset of \((0,1)\), being the intersection of two open sets. Define \(f = \chi_G\), regarded as a function on \([0,1]\); thus \(f \colon [0,1] \to \mathbb{R}\) is bounded (it takes only the values \(0\) and \(1\)), so the lower and upper Riemann integrals \(L(f,[0,1])\) and \(U(f,[0,1])\) of 1.7 make sense.

Two properties of \(G\) drive the argument.

(i) \(G\) is dense in \((0,1)\), because \(G\) contains every rational number in \((0,1)\): indeed \(r_k \in (r_k - 2^{-k-2}, r_k + 2^{-k-2})\) and \(r_k \in (0,1)\).

(ii) \(\lambda(G) \le \tfrac12\), where \(\lambda\) denotes Lebesgue measure; note \(G\) is open, hence Borel, so \(\lambda(G)\) equals the outer measure \(|G|\) (by 2.68 and 2.69). Indeed, an open interval \(I\) is a one-set cover of itself, so the definition 2.2 of outer measure gives \(|I| \le \ell(I)\); combining this with countable subadditivity of outer measure (2.8) and monotonicity (2.5),

\begin{equation*} \lambda(G) \le \sum_{k=1}^{\infty} \left| \left( r_k - \tfrac{1}{2^{k+2}},\ r_k + \tfrac{1}{2^{k+2}} \right) \right| \le \sum_{k=1}^{\infty} \frac{1}{2^{k+1}} = \frac12 . \end{equation*}

The upper Riemann integral equals \(1\). Let \(P\) be the partition \(0 = x_0 < x_1 < \cdots < x_n = 1\) of \([0,1]\). Fix \(j \in \{1,\ldots,n\}\). The open interval \((x_{j-1}, x_j)\) is a nonempty open subset of \([0,1]\), and \((x_{j-1}, x_j) \cap (0,1)\) is still nonempty and open (it is \((x_{j-1},x_j)\) itself unless \(j = 1\) or \(j = n\), in which cases it is \((0,x_1)\) or \((x_{n-1},1)\), again nonempty and open). By (i), \(G\) meets every nonempty open subset of \((0,1)\), so there exists \(y \in G\) with \(x_{j-1} < y < x_j\). Hence

\begin{equation*} \sup_{[x_{j-1},x_j]} f \ge f(y) = 1, \end{equation*}

and since \(f \le 1\) everywhere we get \(\sup_{[x_{j-1},x_j]} f = 1\). Therefore, by the definition 1.3 of the upper Riemann sum,

\begin{equation*} U(f,P,[0,1]) = \sum_{j=1}^{n} (x_j - x_{j-1}) \cdot 1 = 1 . \end{equation*}

As \(P\) was an arbitrary partition, \(U(f,[0,1]) = \inf_P U(f,P,[0,1]) = 1\).

The lower Riemann integral is at most \(\tfrac12\). Again let \(P\) be the partition \(0 = x_0 < x_1 < \cdots < x_n = 1\). Because \(f\) takes only the values \(0\) and \(1\), for each \(j\) we have

\begin{equation*} \inf_{[x_{j-1},x_j]} f = \begin{cases} 1 & \text{if } [x_{j-1},x_j] \subset G,\\ 0 & \text{otherwise.} \end{cases} \end{equation*}

Let \(J = \{ j : [x_{j-1},x_j] \subset G \}\). The intervals \([x_{j-1},x_j]\) for \(j \in J\) have pairwise disjoint interiors and are all contained in \(G\), so by finite additivity and monotonicity of \(\lambda\) (the overlaps are finitely many points, which have measure \(0\)),

\begin{equation*} \sum_{j \in J} (x_j - x_{j-1}) = \lambda\!\left( \bigcup_{j \in J} [x_{j-1},x_j] \right) \le \lambda(G) \le \frac12 . \end{equation*}

Hence, by the definition 1.3 of the lower Riemann sum,

\begin{equation*} L(f,P,[0,1]) = \sum_{j \in J} (x_j - x_{j-1}) \le \frac12 . \end{equation*}

As \(P\) was arbitrary, \(L(f,[0,1]) = \sup_P L(f,P,[0,1]) \le \tfrac12\).

Conclusion. We have \(L(f,[0,1]) \le \tfrac12 < 1 = U(f,[0,1])\). By 1.9, \(f = \chi_G\) is not Riemann integrable on \([0,1]\). This is the desired counterexample.

Remark. Nothing is wrong with \(\chi_G\) from the Lebesgue point of view: \(G\) is an open, hence Borel, set, so \(\chi_G\) is a Borel measurable bounded function and \(\int_{[0,1]} \chi_G \, d\lambda = \lambda(G)\), a perfectly good finite number. The failure above is exactly the phenomenon that motivates Lebesgue integration. Note also that the hypothesis cannot be salvaged by shrinking \(\lambda(G)\): replacing \(2^{-k-2}\) by \(\varepsilon 2^{-k-1}\) in the construction gives, for every \(\varepsilon \in (0,1)\), a dense open \(G \subset (0,1)\) with \(\lambda(G) \le \varepsilon\), for which \(L(\chi_G,[0,1]) \le \varepsilon\) while \(U(\chi_G,[0,1]) = 1\). On the other hand, if \(G\) is a finite union of open intervals then \(\chi_G\) is Riemann integrable, so the statement fails only because open subsets of \((0,1)\) can be complicated.

Throughout, \(\lambda\) denotes Lebesgue measure and \(\mathcal{S}\) denotes the \(\sigma\)-algebra with respect to which \(\mathcal{L}^1(\mathbb{R})\) is being taken, that is, either the Borel subsets of \(\mathbb{R}\) or the Lebesgue measurable subsets of \(\mathbb{R}\) (3.45). The strategy for both parts is the same: the assertion is obvious for continuous functions that vanish outside a bounded set, and by 3.48 such functions are dense in \(\mathcal{L}^1(\mathbb{R})\). The transfer from the dense class to a general \(f\) requires knowing how the \(L^1\)-norm behaves under the relevant change of variable, so we begin with that.

Step 1: the two \(\sigma\)-algebras are invariant under translations and dilations.

Fix \(t \in \mathbb{R}\) and \(c > 0\), and for \(A \subset \mathbb{R}\) write \(t + A = \{t + a : a \in A\}\) and \(cA = \{ca : a \in A\}\).

Write \(|A|\) for the outer measure of an arbitrary set \(A \subset \mathbb{R}\); recall that \(\lambda(A) = |A|\) for every \(A \in \mathcal{S}\).

Translation invariance of outer measure (2.7) says \(|t + A| = |A|\) for every \(A \subset \mathbb{R}\). For dilations: if \(A \subset \bigcup_{k} I_k\) with each \(I_k\) an open interval, then \(cA \subset \bigcup_k cI_k\) and \(cI_k\) is an open interval whose length is \(c\) times the length of \(I_k\); taking the infimum over all such covers in the definition of outer measure (2.2) gives \(|cA| \le c\,|A|\). Applying this inequality with \(cA\) in place of \(A\) and \(1/c\) in place of \(c\) gives \(|A| \le \frac{1}{c}|cA|\), that is, \(c\,|A| \le |cA|\). Hence

\begin{equation*} |cA| = c\,|A| \qquad \text{for all } A \subset \mathbb{R} \text{ and all } c > 0 . \end{equation*}

Now suppose \(A \in \mathcal{S}\). First take \(\mathcal{S}\) to be the Borel \(\sigma\)-algebra. The collection \(\{A \subset \mathbb{R} : t + A \in \mathcal{S}\}\) is a \(\sigma\)-algebra on \(\mathbb{R}\) (it contains \(\mathbb{R}\) because \(t + \mathbb{R} = \mathbb{R}\), and it is closed under complementation and countable unions because \(t + (\mathbb{R} \setminus A) = \mathbb{R} \setminus (t + A)\) and \(t + \bigcup_k A_k = \bigcup_k (t + A_k)\)) that contains every open set, since a translate of an open set is open; hence it contains every Borel set. The same argument, using that \(cU\) is open when \(U\) is open and \(c > 0\), shows \(cA\) is Borel whenever \(A\) is Borel.

Now take \(\mathcal{S}\) to be the \(\sigma\)-algebra of Lebesgue measurable sets. By 2.70, \(A\) Lebesgue measurable means there is a Borel set \(B \subset A\) with \(|A \setminus B| = 0\). Then \(t + B\) is Borel by the previous paragraph, \(t + B \subset t + A\), and

\begin{equation*} (t+A) \setminus (t+B) = t + (A \setminus B), \quad \text{so} \quad \big| (t+A) \setminus (t+B) \big| = |A \setminus B| = 0 \end{equation*}

by 2.7; thus \(t + A\) is Lebesgue measurable by 2.70. Identically, \(cB\) is Borel, \(cB \subset cA\), and \(|(cA) \setminus (cB)| = |c(A \setminus B)| = c\,|A \setminus B| = 0\), so \(cA\) is Lebesgue measurable. In both cases:

\begin{equation*} A \in \mathcal{S} \ \Longrightarrow\ t + A \in \mathcal{S} \text{ and } cA \in \mathcal{S} . \end{equation*}

Step 2: change of variables in the integral.

Claim. Suppose \(h \in \mathcal{L}^1(\mathbb{R})\).

(i) If \(t \in \mathbb{R}\) and \(h^{(t)}(x) = h(x-t)\), then \(h^{(t)} \in \mathcal{L}^1(\mathbb{R})\) and \(\lVert h^{(t)} \rVert_1 = \lVert h \rVert_1\).

(ii) If \(c > 0\) and \(h_{[c]}(x) = h(cx)\), then \(h_{[c]} \in \mathcal{L}^1(\mathbb{R})\) and \(\lVert h_{[c]} \rVert_1 = \frac{1}{c} \lVert h \rVert_1\).

Proof of the claim. Let \(\tau(x) = x - t\), so that \(h^{(t)} = h \circ \tau\) and \(\tau^{-1}(A) = t + A\) for every \(A \subset \mathbb{R}\). For a Borel set \(B \subset \mathbb{R}\) we have \((h \circ \tau)^{-1}(B) = \tau^{-1}\big(h^{-1}(B)\big) = t + h^{-1}(B) \in \mathcal{S}\) by Step 1, so \(h^{(t)}\) is \(\mathcal{S}\)-measurable.

Next let \(u \colon \mathbb{R} \to [0,\infty]\) be any \(\mathcal{S}\)-measurable function and put \(u^{(t)} = u \circ \tau\); the preimage computation just given applies verbatim to \(u\) (for a Borel subset \(B\) of \([0,\infty]\) we again have \((u \circ \tau)^{-1}(B) = t + u^{-1}(B) \in \mathcal{S}\)), so \(u^{(t)}\) is \(\mathcal{S}\)-measurable and \(\int u^{(t)} \, d\lambda\) is defined. If \(A_1, \ldots, A_m\) is an \(\mathcal{S}\)-partition of \(\mathbb{R}\) (3.1), then \(t + A_1, \ldots, t + A_m\) is again an \(\mathcal{S}\)-partition of \(\mathbb{R}\): the sets lie in \(\mathcal{S}\) by Step 1, they are disjoint, and their union is \(t + \mathbb{R} = \mathbb{R}\). Moreover \(\tau\) maps \(t + A_j\) bijectively onto \(A_j\), so

\begin{equation*} \inf_{t + A_j} u^{(t)} = \inf \{ u(\tau(x)) : x \in t + A_j \} = \inf\{u(y) : y \in A_j\} = \inf_{A_j} u , \end{equation*}

while \(\lambda(t + A_j) = \lambda(A_j)\) by Step 1. Hence the lower Lebesgue sums (3.2) satisfy

\begin{equation*} \mathcal{L}\big(u^{(t)}, \{t+A_j\}_{j=1}^m\big) = \sum_{j=1}^m \lambda(t+A_j) \inf_{t+A_j} u^{(t)} = \sum_{j=1}^m \lambda(A_j) \inf_{A_j} u = \mathcal{L}\big(u, \{A_j\}_{j=1}^m\big). \end{equation*}

Since \(\{A_j\} \mapsto \{t + A_j\}\) is a bijection from the set of \(\mathcal{S}\)-partitions of \(\mathbb{R}\) onto itself (its inverse is \(\{A_j\} \mapsto \{-t + A_j\}\)), taking suprema over all \(\mathcal{S}\)-partitions and using the definition 3.3 of the integral of a nonnegative function gives

\begin{equation*} \int u^{(t)} \, d\lambda = \int u \, d\lambda . \end{equation*}

Applying this with \(u = |h|\) and noting \(|h|^{(t)} = |h^{(t)}|\) yields \(\lVert h^{(t)}\rVert_1 = \lVert h \rVert_1 < \infty\), so \(h^{(t)} \in \mathcal{L}^1(\mathbb{R})\). This proves (i). (Applying it instead with \(u = h^+\) and \(u = h^-\), and using \((h^{(t)})^{\pm} = (h^{\pm})^{(t)}\) together with 3.17 and 3.18, also gives \(\int h^{(t)} \, d\lambda = \int h \, d\lambda\), though we do not need this.)

Part (ii) is proved by the identical argument with \(\tau\) replaced by \(\delta(x) = cx\), for which \(\delta^{-1}(A) = \frac{1}{c}A\) and \(\lambda\big(\frac{1}{c}A\big) = \frac{1}{c}\lambda(A)\) by Step 1. The partition \(A_1, \ldots, A_m\) corresponds to the partition \(\frac{1}{c}A_1, \ldots, \frac{1}{c}A_m\), on which \(\inf_{\frac1c A_j} u_{[c]} = \inf_{A_j} u\), so every lower Lebesgue sum gets multiplied by exactly \(\frac1c\); taking suprema gives \(\int u_{[c]}\, d\lambda = \frac{1}{c}\int u \, d\lambda\) for nonnegative \(\mathcal{S}\)-measurable \(u\), and then \(\lVert h_{[c]}\rVert_1 = \frac1c \lVert h \rVert_1\). This completes the proof of the claim.

Step 3: continuous functions vanishing outside a bounded set are bounded and uniformly continuous.

Suppose \(g \colon \mathbb{R} \to \mathbb{R}\) is continuous and \(\{x \in \mathbb{R} : g(x) \ne 0\}\) is bounded, say \(g(x) = 0\) whenever \(|x| \ge M\), where \(M > 0\). Then \(g\) is continuous on the compact interval \([-M-1, M+1]\), hence bounded there, and hence bounded on \(\mathbb{R}\) (it vanishes off \([-M,M]\)); write \(C = \sup_{\mathbb{R}} |g| < \infty\). Also \(g\) is uniformly continuous on \([-M-1,M+1]\). Given \(\eta > 0\), choose \(\delta \in (0,1)\) such that \(|g(x) - g(y)| < \eta\) whenever \(x, y \in [-M-1,M+1]\) with \(|x - y| < \delta\). Now let \(x, y \in \mathbb{R}\) with \(|x-y| < \delta\). If both lie in \([-M-1,M+1]\) we are done. Otherwise one of them, say \(x\), satisfies \(|x| > M+1\); then \(|y| \ge |x| - |x-y| > M + 1 - 1 = M\), so \(g(x) = g(y) = 0\) and \(|g(x)-g(y)| = 0 < \eta\). Thus \(g\) is uniformly continuous on \(\mathbb{R}\).

Note also that such a \(g\) lies in \(\mathcal{L}^1(\mathbb{R})\): it is continuous, hence Borel measurable, and \(|g| \le C \chi_{[-M,M]}\), so \(\lVert g \rVert_1 \le 2CM < \infty\) by 3.8 and 3.4.

Part (a).

By Step 2(i), \(f_t \in \mathcal{L}^1(\mathbb{R})\) for every \(t \in \mathbb{R}\), so \(\lVert f - f_t \rVert_1\) is a finite nonnegative number. Let \(\varepsilon > 0\).

By 3.48 there exists a continuous \(g \colon \mathbb{R} \to \mathbb{R}\) such that \(\{x : g(x) \ne 0\}\) is bounded and

\begin{equation*} \lVert f - g \rVert_1 < \frac{\varepsilon}{3}. \end{equation*}

Fix \(M > 0\) with \(g(x) = 0\) for \(|x| \ge M\), and for \(t \in \mathbb{R}\) put \(g_t(x) = g(x-t)\).

First we control \(\lVert g - g_t \rVert_1\). Suppose \(|t| \le 1\) and \(|x| > M + 1\). Then \(|x| > M\), so \(g(x) = 0\), and \(|x - t| \ge |x| - |t| > M\), so \(g_t(x) = 0\). Thus \(g - g_t\) vanishes outside \([-M-1, M+1]\), and therefore

\begin{equation*} |g - g_t| \le \Big( \sup_{x \in \mathbb{R}} |g(x) - g(x-t)| \Big) \chi_{[-M-1,\,M+1]} . \end{equation*}

By 3.8 and 3.4 this gives

\begin{equation*} \lVert g - g_t \rVert_1 \le (2M+2) \sup_{x \in \mathbb{R}} |g(x) - g(x-t)| . \end{equation*}

By Step 3, \(g\) is uniformly continuous, so there exists \(\delta > 0\), which we may take to satisfy \(\delta \le 1\), such that

\begin{equation*} |g(x) - g(y)| < \frac{\varepsilon}{3(2M+2)} \quad \text{whenever } |x - y| < \delta . \end{equation*}

Consequently, for every \(t\) with \(|t| < \delta\) we have \(|g(x) - g(x-t)| \le \frac{\varepsilon}{3(2M+2)}\) for all \(x\) (the supremum of quantities each less than that bound is at most that bound), and hence

\begin{equation*} \lVert g - g_t \rVert_1 \le (2M+2) \cdot \frac{\varepsilon}{3(2M+2)} = \frac{\varepsilon}{3}. \end{equation*}

Next, \(f_t - g_t\) is precisely the translate \((f-g)_t\) of \(f - g\), so Step 2(i) applied to \(h = f - g \in \mathcal{L}^1(\mathbb{R})\) gives

\begin{equation*} \lVert f_t - g_t \rVert_1 = \lVert f - g \rVert_1 < \frac{\varepsilon}{3} \end{equation*}

for every \(t \in \mathbb{R}\).

Combining the three estimates with the triangle inequality for \(\lVert \cdot \rVert_1\) (3.43), we obtain, for every \(t\) with \(|t| < \delta\),

\begin{equation*} \lVert f - f_t \rVert_1 \le \lVert f - g \rVert_1 + \lVert g - g_t \rVert_1 + \lVert g_t - f_t \rVert_1 < \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon . \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, \(\lim_{t \to 0} \lVert f - f_t \rVert_1 = 0\), as desired.

Part (b).

By Step 2(ii), \(f_t \in \mathcal{L}^1(\mathbb{R})\) for every \(t > 0\). Let \(\varepsilon > 0\).

By 3.48 there exists a continuous \(g \colon \mathbb{R} \to \mathbb{R}\) such that \(\{x : g(x) \ne 0\}\) is bounded and

\begin{equation*} \lVert f - g \rVert_1 < \frac{\varepsilon}{4}. \end{equation*}

Fix \(M > 0\) with \(g(x) = 0\) for \(|x| \ge M\), and for \(t > 0\) put \(g_t(x) = g(tx)\).

First, since \(f_t - g_t\) is the dilate \((f-g)_t\) of \(f - g\), Step 2(ii) applied to \(h = f-g\) gives

\begin{equation*} \lVert f_t - g_t \rVert_1 = \frac{1}{t} \lVert f - g \rVert_1 . \end{equation*}

Hence for every \(t \ge \frac12\),

\begin{equation*} \lVert f_t - g_t \rVert_1 \le 2 \lVert f - g \rVert_1 < \frac{\varepsilon}{2} . \end{equation*}

Second we control \(\lVert g - g_t \rVert_1\) for \(t\) near \(1\). Suppose \(\frac12 \le t \le 2\) and \(|x| > 2M\). Then \(|x| > M\), so \(g(x) = 0\), and \(|tx| \ge \frac12 |x| > M\), so \(g_t(x) = 0\). Thus \(g - g_t\) vanishes outside \([-2M, 2M]\). For \(|x| \le 2M\) we have

\begin{equation*} |x - tx| = |x| \, |t-1| \le 2M |t - 1| . \end{equation*}

By Step 3, \(g\) is uniformly continuous, so there exists \(\eta > 0\) such that

\begin{equation*} |g(u) - g(v)| < \frac{\varepsilon}{16M} \quad \text{whenever } |u - v| < \eta . \end{equation*}

Put \(\rho = \min\{ \tfrac12, \tfrac{\eta}{2M} \} > 0\) and suppose \(|t - 1| < \rho\). Then \(\frac12 < t < \frac32\), and for every \(x\) with \(|x| \le 2M\) we have \(|x - tx| \le 2M|t-1| < \eta\), hence \(|g(x) - g(tx)| < \frac{\varepsilon}{16M}\). Therefore

\begin{equation*} |g - g_t| \le \frac{\varepsilon}{16M} \, \chi_{[-2M,\,2M]} , \end{equation*}

and so by 3.8 and 3.4,

\begin{equation*} \lVert g - g_t \rVert_1 \le 4M \cdot \frac{\varepsilon}{16M} = \frac{\varepsilon}{4}. \end{equation*}

Combining, for every \(t > 0\) with \(|t - 1| < \rho\) (so in particular \(t > \frac12\)), the triangle inequality (3.43) gives

\begin{equation*} \lVert f - f_t \rVert_1 \le \lVert f - g \rVert_1 + \lVert g - g_t \rVert_1 + \lVert g_t - f_t \rVert_1 < \frac{\varepsilon}{4} + \frac{\varepsilon}{4} + \frac{\varepsilon}{2} = \varepsilon . \end{equation*}

Since \(\varepsilon > 0\) was arbitrary, \(\lim_{t \to 1} \lVert f - f_t \rVert_1 = 0\), as desired.

Remark. The dilation constant in Step 2(ii) is what forces the extra care in (b): unlike translation, \(h \mapsto h_{[t]}\) is not an isometry of \(\mathcal{L}^1(\mathbb{R})\), only a bounded operator of norm \(1/t\). Restricting to \(t \ge \frac12\) keeps that factor bounded, which is all that is needed since we are taking \(t \to 1\). Both parts say that the natural actions of the translation group and of the multiplicative group \((0,\infty)\) on \(\mathcal{L}^1(\mathbb{R})\) are strongly continuous; the proofs are the standard \(\varepsilon/3\) argument made possible by 3.48.

Fix positive numbers \(c\) and \(p\), and set

\begin{equation*} E = \{x \in X : |h(x)| \ge c\}. \end{equation*}

First note that everything in sight is measurable. Because \(h\) is \(\mathcal{S}\)-measurable, so is \(|h|\), and hence so is \(|h|^p\) (the map \(t \mapsto |t|^p\) is a continuous function from \(\mathbb{R}\) to \([0,\infty)\), so it is Borel measurable, and the composition of a Borel measurable function with an \(\mathcal{S}\)-measurable function is \(\mathcal{S}\)-measurable by 2.44 and 2.45). Also

\begin{equation*} E = |h|^{-1}\bigl([c, \infty)\bigr) \in \mathcal{S}, \end{equation*}

because \([c,\infty)\) is a Borel subset of \(\mathbb{R}\).

The point of the proof is the pointwise inequality

\begin{equation*} c^p \chi_E \le |h|^p \quad \text{on all of } X. \end{equation*}

Indeed, if \(x \notin E\) then the left side equals \(0\) and the right side is nonnegative. If \(x \in E\) then \(|h(x)| \ge c > 0\), and because \(t \mapsto t^p\) is increasing on \([0,\infty)\) we get \(|h(x)|^p \ge c^p\), which is exactly what the displayed inequality asserts at \(x\).

Both sides above are nonnegative \(\mathcal{S}\)-measurable functions, so 3.8 (integration is order preserving) applies and gives

\begin{equation*} \int c^p \chi_E \, d\mu \le \int |h|^p \, d\mu . \end{equation*}

The left side equals \(c^p \mu(E)\), since \(c^p \chi_E\) is a simple function whose integral is \(c^p \mu(E)\) by 3.15 (the integral of a linear combination of characteristic functions). Hence

\begin{equation*} c^p \, \mu(\{x \in X : |h(x)| \ge c\}) \le \int |h|^p \, d\mu . \end{equation*}

Dividing by the positive number \(c^p\) gives

\begin{equation*} \mu(\{x \in X : |h(x)| \ge c\}) \le \frac{1}{c^p} \int |h|^p \, d\mu , \end{equation*}

as desired.

Note that no integrability hypothesis was needed: if \(\int |h|^p \, d\mu = \infty\) the inequality is trivially true, and the argument above is valid in that case as well because all the integrals involved are integrals of nonnegative functions and thus are defined in \([0,\infty]\). Taking \(p = 1\) recovers Markov’s inequality (4.1), except that here \(h\) need not belong to \(L^1(\mu)\).

Write

\begin{equation*} m = \int h \, d\mu , \end{equation*}

which is a real number because \(h \in \mathcal{L}^1(\mu)\). Let \(c > 0\).

The function \(h - m\) is a real-valued \(\mathcal{S}\)-measurable function on \(X\). Applying Exercise 1 in this section (with \(p = 2\)) to \(h - m\) gives

\begin{equation*} \mu\bigl(\{x \in X : |h(x) - m| \ge c\}\bigr) \le \frac{1}{c^2}\int (h - m)^2 \, d\mu . \end{equation*}

Thus the proof will be complete once we show

\begin{equation*} \int (h-m)^2 \, d\mu = \int h^2 \, d\mu - m^2 . \end{equation*}

If \(\int h^2 \, d\mu = \infty\), then the right side of the inequality we are proving equals \(\infty\) (because \(m^2 < \infty\)) and there is nothing to prove. So assume from now on that

\begin{equation*} \int h^2 \, d\mu < \infty , \end{equation*}

which says precisely that \(h^2 \in \mathcal{L}^1(\mu)\).

The constant function \(m\) on \(X\) satisfies \(\int |m| \, d\mu = |m| \, \mu(X) = |m| < \infty\), using the hypothesis \(\mu(X) = 1\); thus the constant function \(m\) is in \(\mathcal{L}^1(\mu)\), and so is \(-2mh\) (a scalar multiple of \(h\)). Because

\begin{equation*} (h - m)^2 = h^2 - 2mh + m^2 \end{equation*}

pointwise on \(X\), and each of the three functions on the right side is in \(\mathcal{L}^1(\mu)\), the additivity of integration (3.21), applied twice to handle three summands, together with the homogeneity of integration (3.20), gives

\begin{equation*} \begin{aligned} \int (h-m)^2 \, d\mu &= \int h^2 \, d\mu - 2m \int h \, d\mu + m^2 \, \mu(X) \\ &= \int h^2 \, d\mu - 2m^2 + m^2 \\ &= \int h^2 \, d\mu - m^2 , \end{aligned} \end{equation*}

where we again used \(\mu(X) = 1\) and \(\int h \, d\mu = m\).

Combining the last display with the inequality obtained from Exercise 1 gives

\begin{equation*} \mu\left(\left\{x \in X : \left|h(x) - \int h \, d\mu\right| \ge c\right\}\right) \le \frac{1}{c^2}\left(\int h^2 \, d\mu - \left(\int h \, d\mu\right)^2\right), \end{equation*}

as desired.

In probabilistic language, \(\mu\) is a probability measure, \(h\) is a random variable with expectation \(m\), and the quantity \(\int h^2 \, d\mu - m^2 = \int (h-m)^2 \, d\mu\) is the variance of \(h\); the inequality says that the probability that \(h\) deviates from its mean by at least \(c\) is at most the variance divided by \(c^2\). Note also that the computation above shows the right side is always nonnegative, since it equals the integral of the nonnegative function \((h-m)^2\).

For \(c \in (0,\infty)\) let

\begin{equation*} E_c = \{x \in X : |h(x)| \ge c\} , \end{equation*}

which is in \(\mathcal{S}\) because \(|h|\) is \(\mathcal{S}\)-measurable (2.44 and 2.45). Call \(c\) an equality point if

\begin{equation*} \mu(E_c) = \frac{1}{c}\|h\|_1 . \end{equation*}

Step 1: an equality point forces equality throughout the proof of Markov’s inequality.

Suppose \(c \in (0,\infty)\) is an equality point. Since \(\chi_{E_c}|h| \le |h|\) pointwise and \(|h| = \chi_{E_c}|h| + \chi_{X \setminus E_c}|h|\) pointwise, the additivity of integration for nonnegative functions (3.16) gives

\begin{equation*} \|h\|_1 = \int_{E_c} |h| \, d\mu + \int_{X \setminus E_c} |h| \, d\mu . \end{equation*}

On \(E_c\) we have \(|h| \ge c\), so \(\chi_{E_c}|h| \ge c \chi_{E_c}\) pointwise on \(X\), and hence by 3.8

\begin{equation*} \int_{E_c} |h| \, d\mu \ge c \, \mu(E_c) = \|h\|_1 , \end{equation*}

the last equality being the assumption that \(c\) is an equality point. Substituting this into the previous display yields

\begin{equation*} \|h\|_1 \ge \|h\|_1 + \int_{X \setminus E_c} |h| \, d\mu . \end{equation*}

Because \(\|h\|_1 < \infty\), we may cancel it, obtaining

\begin{equation*} \int_{X \setminus E_c} |h| \, d\mu = 0 \qquad \text{and} \qquad \int_{E_c} |h| \, d\mu = \|h\|_1 = c\,\mu(E_c) . \end{equation*}

Step 2: consequences.

Note first that \(\mu(E_c) = \|h\|_1 / c\) satisfies \(0 < \mu(E_c) < \infty\), because \(0 < \|h\|_1 < \infty\).

Next, the function \(\chi_{E_c}(|h| - c)\) is nonnegative (on \(E_c\) we have \(|h| \ge c\), and off \(E_c\) it is \(0\)) and \(\mathcal{S}\)-measurable, and we claim that its integral is \(0\). Indeed, \(\chi_{E_c}|h| = \chi_{E_c}(|h| - c) + c\chi_{E_c}\) pointwise, so 3.16 gives \(\int_{E_c}|h| \, d\mu = \int \chi_{E_c}(|h| - c) \, d\mu + c\,\mu(E_c)\); since \(\int_{E_c}|h| \, d\mu = c\,\mu(E_c)\) by Step 1 and both of these numbers are finite (as noted in the previous paragraph), the subtraction is legitimate and \(\int \chi_{E_c}(|h| - c) \, d\mu = 0\).

Now we use the standard fact that a nonnegative measurable function with integral \(0\) vanishes almost everywhere. To see this for a nonnegative \(\mathcal{S}\)-measurable \(g\) with \(\int g \, d\mu = 0\): by Exercise 1 in this section applied with \(p = 1\) (or by Markov’s inequality),

\begin{equation*} \mu\bigl(\{x : g(x) \ge \tfrac{1}{n}\}\bigr) \le n \int g \, d\mu = 0 \end{equation*}

for every \(n \in \mathbb{Z}^+\), and \(\{x : g(x) > 0\} = \bigcup_{n=1}^{\infty}\{x : g(x) \ge \frac{1}{n}\}\), so countable subadditivity of \(\mu\) (2.58) gives \(\mu(\{x : g(x) > 0\}) = 0\).

Applying this to \(g = \chi_{E_c}(|h| - c)\) shows

\begin{equation*} |h(x)| = c \quad \text{for almost every } x \in E_c . \end{equation*}

Step 3: at most one equality point.

Suppose \(c_1\) and \(c_2\) are equality points with \(c_1 < c_2\). Because \(c_1 < c_2\) we have \(E_{c_2} \subseteq E_{c_1}\). By Step 2 applied to \(c_1\), the set

\begin{equation*} A = \{x \in E_{c_1} : |h(x)| \ne c_1\} \end{equation*}

has \(\mu(A) = 0\). By Step 2 applied to \(c_2\), the set

\begin{equation*} B = \{x \in E_{c_2} : |h(x)| \ne c_2\} \end{equation*}

has \(\mu(B) = 0\). But every \(x \in E_{c_2}\) lies in \(E_{c_1}\), and \(|h(x)|\) cannot equal both \(c_1\) and \(c_2\) (these are distinct numbers); hence \(E_{c_2} \subseteq A \cup B\). Therefore

\begin{equation*} \mu(E_{c_2}) \le \mu(A) + \mu(B) = 0 \end{equation*}

by subadditivity (2.58). This contradicts \(\mu(E_{c_2}) = \|h\|_1 / c_2 > 0\) from Step 2.

Hence no two distinct numbers in \((0,\infty)\) can both be equality points, so there is at most one \(c \in (0,\infty)\) with \(\mu(\{x \in X : |h(x)| \ge c\}) = \frac{1}{c}\|h\|_1\).

Remark: the hypothesis \(\|h\|_1 > 0\) cannot be dropped. If \(h = 0\), then \(\mu(E_c) = 0 = \frac{1}{c}\|h\|_1\) for every \(c \in (0,\infty)\), so every positive number is an equality point.

For a bounded nonempty open interval \(I\) and a number \(\alpha > 0\), let \(\alpha * I\) denote the open interval with the same center as \(I\) and \(\alpha\) times the length of \(I\) (this extends Definition 4.2, which is the case \(\alpha = 3\)). We must show: if \(\alpha \in (0,3)\), then there is a list \(I_1, \dots, I_n\) of bounded nonempty open intervals of \(\mathbb{R}\) such that no disjoint sublist \(I_{k_1}, \dots, I_{k_m}\) satisfies

\begin{equation*} I_1 \cup \cdots \cup I_n \subseteq (\alpha * I_{k_1}) \cup \cdots \cup (\alpha * I_{k_m}) . \end{equation*}

So fix \(\alpha\) with \(0 < \alpha < 3\), and choose \(\varepsilon\) with

\begin{equation*} 0 < \varepsilon < \min\{2, \, 3 - \alpha\} . \end{equation*}

Take \(n = 2\) and

\begin{equation*} I_1 = (0, 2), \qquad I_2 = (2 - \varepsilon, 4 - \varepsilon) . \end{equation*}

Then \(I_1\) and \(I_2\) are bounded nonempty open intervals, each of length \(2\), and, because \(0 < \varepsilon < 2\),

\begin{equation*} I_1 \cap I_2 = (2 - \varepsilon, 2) \ne \emptyset , \end{equation*}

so \(I_1, I_2\) is not a disjoint list. Consequently the only disjoint sublists of \(I_1, I_2\) are the sublist consisting of \(I_1\) alone, the sublist consisting of \(I_2\) alone, and the empty sublist.

Also

\begin{equation*} I_1 \cup I_2 = (0, 4 - \varepsilon) . \end{equation*}

The empty sublist clearly fails, because \(I_1 \cup I_2 \ne \emptyset\). Now consider the two one-element sublists.

The interval \(I_1 = (0,2)\) has center \(1\) and length \(2\), so

\begin{equation*} \alpha * I_1 = (1 - \alpha, 1 + \alpha) . \end{equation*}

If \(I_1 \cup I_2 \subseteq \alpha * I_1\), then in particular every point of \((0, 4 - \varepsilon)\) is less than \(1 + \alpha\), so

\begin{equation*} 4 - \varepsilon \le 1 + \alpha , \qquad \text{that is,} \qquad \alpha \ge 3 - \varepsilon . \end{equation*}

But \(\varepsilon < 3 - \alpha\) gives \(\alpha < 3 - \varepsilon\), a contradiction. Hence \(I_1 \cup I_2 \not\subseteq \alpha * I_1\).

The interval \(I_2 = (2 - \varepsilon, 4 - \varepsilon)\) has center \(3 - \varepsilon\) and length \(2\), so

\begin{equation*} \alpha * I_2 = (3 - \varepsilon - \alpha, \; 3 - \varepsilon + \alpha) . \end{equation*}

If \(I_1 \cup I_2 \subseteq \alpha * I_2\), then every point of \((0, 4-\varepsilon)\) exceeds \(3 - \varepsilon - \alpha\), so

\begin{equation*} 3 - \varepsilon - \alpha \le 0 , \qquad \text{that is,} \qquad \alpha \ge 3 - \varepsilon , \end{equation*}

again contradicting \(\alpha < 3 - \varepsilon\). Hence \(I_1 \cup I_2 \not\subseteq \alpha * I_2\).

Thus no disjoint sublist of \(I_1, I_2\) has the covering property, so the conclusion of the Vitali Covering Lemma fails for this list when \(3\) is replaced by \(\alpha\). Since \(\alpha \in (0,3)\) was arbitrary, the constant \(3\) in 4.4 cannot be replaced by any smaller positive constant.

Remark: the example is exactly the extremal configuration of the greedy algorithm used in the proof of 4.4. Two intervals of equal length overlap in an arbitrarily short segment; only one of them can be selected, and stretching it by the factor \(\alpha\) about its center must reach across essentially two full interval lengths beyond its center, i.e. a total half-length of essentially \(3\) times its own half-length. Letting \(\varepsilon \downarrow 0\) shows the constant \(3\) is sharp.

Let \(I\) and \(J\) be bounded nonempty open intervals of \(\mathbb{R}\) with

\begin{equation*} I \cap J \ne \emptyset \qquad \text{and} \qquad |I| \ge |J| , \end{equation*}

where \(|\cdot|\) denotes length. We must show \(J \subseteq 3 * I\).

Write the intervals in terms of their centers and half-lengths:

\begin{equation*} I = (a - r, \, a + r), \qquad J = (b - s, \, b + s) , \end{equation*}

where \(a, b \in \mathbb{R}\) and \(r, s > 0\) (positivity of \(r\) and \(s\) holds because the intervals are nonempty open intervals; boundedness makes \(r\) and \(s\) finite). Then \(|I| = 2r\) and \(|J| = 2s\), so the hypothesis \(|I| \ge |J|\) says

\begin{equation*} s \le r . \end{equation*}

By Definition 4.2,

\begin{equation*} 3 * I = (a - 3r, \, a + 3r) , \end{equation*}

since \(3 * I\) has the same center \(a\) as \(I\) and length \(3|I| = 6r\), hence half-length \(3r\).

First we bound the distance between the two centers. Because \(I \cap J \ne \emptyset\), there exists \(x \in I \cap J\). Then \(|x - a| < r\) and \(|x - b| < s\), so the triangle inequality gives

\begin{equation*} |a - b| \le |a - x| + |x - b| < r + s . \end{equation*}

Now let \(y \in J\) be arbitrary, so \(|y - b| < s\). Then

\begin{equation*} |y - a| \le |y - b| + |b - a| < s + (r + s) = r + 2s \le r + 2r = 3r , \end{equation*}

where the final inequality used \(s \le r\). Thus \(a - 3r < y < a + 3r\), that is, \(y \in 3 * I\).

Since \(y \in J\) was arbitrary, \(J \subseteq 3 * I\), which is the assertion left as an exercise in the proof of 4.4.

Note where each hypothesis was used: the nonempty intersection kept the centers within \(r + s\) of each other, and the inequality \(|I| \ge |J|\) converted the resulting bound \(r + 2s\) into \(3r\). The computation also explains the appearance of the constant \(3\): in the worst case \(s = r\) and the two intervals barely touch, and then \(J\) reaches exactly \(3r\) away from the center of \(I\). (See Exercise 4 in this section, which shows that no constant smaller than \(3\) works.)

Write \(h = \chi_{[0,1]}\). By Definition 4.6,

\begin{equation*} h^*(b) = \sup_{t > 0} \frac{1}{2t}\int_{b-t}^{b+t} |h| = \sup_{t>0} \psi_b(t), \qquad \text{where } \psi_b(t) = \frac{\ell(b,t)}{2t} \end{equation*}

and

\begin{equation*} \ell(b,t) = \bigl|[0,1] \cap [b-t, \, b+t]\bigr| \end{equation*}

is the Lebesgue measure of the overlap (the integral of \(\chi_{[0,1]}\) over \([b-t,b+t]\) is exactly the measure of that intersection). Explicitly, the intersection of the two intervals \([0,1]\) and \([b-t,b+t]\) is \([\max\{b-t,0\}, \min\{b+t,1\}]\) when this is nonempty, so

\begin{equation*} \ell(b,t) = \max\bigl\{0, \; \min\{b+t, 1\} - \max\{b-t, 0\}\bigr\} . \end{equation*}

A useful preliminary observation: since \([0,1] \cap [b-t,b+t] \subseteq [b-t,b+t]\), we always have \(\ell(b,t) \le 2t\), and hence

\begin{equation*} \psi_b(t) \le 1 \quad \text{for all } b \in \mathbb{R} \text{ and } t > 0 . \end{equation*}

Case 1: \(0 < b < 1\).

Let \(t_0 = \min\{b, 1-b\}\), which is positive. If \(0 < t \le t_0\), then \(b - t \ge 0\) and \(b + t \le 1\), so \([b-t,b+t] \subseteq [0,1]\) and therefore \(\ell(b,t) = 2t\), giving \(\psi_b(t) = 1\). Combined with the preliminary observation \(\psi_b \le 1\), this gives

\begin{equation*} h^*(b) = 1 . \end{equation*}

Case 2: \(b \le 0\).

Here \(b - t < b \le 0\) for every \(t > 0\), so \(\max\{b-t,0\} = 0\) and \(\ell(b,t) = \max\{0, \min\{b+t,1\}\}\). We examine three ranges of \(t\).

If \(0 < t \le -b\) (this range is empty when \(b = 0\)), then \(b + t \le 0\), so \(\ell(b,t) = 0\) and \(\psi_b(t) = 0\).

If \(-b < t \le 1 - b\), then \(0 < b + t \le 1\), so \(\ell(b,t) = b + t\) and

\begin{equation*} \psi_b(t) = \frac{b+t}{2t} = \frac{1}{2} + \frac{b}{2t} . \end{equation*}

Because \(b \le 0\), the function \(t \mapsto \frac{b}{2t}\) is nondecreasing on \((0,\infty)\), so \(\psi_b\) is nondecreasing on this range and attains its largest value there at \(t = 1-b\), namely

\begin{equation*} \psi_b(1-b) = \frac{b + (1-b)}{2(1-b)} = \frac{1}{2(1-b)} . \end{equation*}

If \(t > 1 - b\), then \(b + t > 1\), so \(\ell(b,t) = 1\) and \(\psi_b(t) = \frac{1}{2t} < \frac{1}{2(1-b)}\).

Taking the supremum over the three ranges gives

\begin{equation*} h^*(b) = \frac{1}{2(1-b)} , \end{equation*}

and the supremum is attained at \(t = 1 - b\) (note \(1 - b \ge 1 > 0\)).

Case 3: \(b \ge 1\).

Here \(b + t > 1\) for every \(t > 0\), so \(\min\{b+t,1\} = 1\) and \(\ell(b,t) = \max\{0, \, 1 - \max\{b-t,0\}\}\). Again three ranges.

If \(0 < t \le b - 1\) (empty when \(b = 1\)), then \(b - t \ge 1\), so \(\ell(b,t) = 0\) and \(\psi_b(t) = 0\).

If \(b - 1 < t \le b\), then \(0 \le b - t < 1\), so \(\ell(b,t) = 1 - (b-t)\) and

\begin{equation*} \psi_b(t) = \frac{1 - b + t}{2t} = \frac{1}{2} + \frac{1-b}{2t} . \end{equation*}

Because \(1 - b \le 0\), this is nondecreasing in \(t\), so on this range it is largest at \(t = b\), where

\begin{equation*} \psi_b(b) = \frac{1 - b + b}{2b} = \frac{1}{2b} . \end{equation*}

If \(t > b\), then \(b - t < 0\), so \(\ell(b,t) = 1\) and \(\psi_b(t) = \frac{1}{2t} < \frac{1}{2b}\).

Hence

\begin{equation*} h^*(b) = \frac{1}{2b} , \end{equation*}

attained at \(t = b\).

Combining the three cases gives exactly the formula in Example 4.7:

\begin{equation*} (\chi_{[0,1]})^*(b) = \begin{cases} \dfrac{1}{2(1-b)} & \text{if } b \le 0, \\[2mm] 1 & \text{if } 0 < b < 1, \\[2mm] \dfrac{1}{2b} & \text{if } b \ge 1. \end{cases} \end{equation*}

Two remarks. First, the three formulas do not match up at \(b = 0\) and \(b = 1\): the value is \(\frac{1}{2}\) at each of those two points but \(1\) just inside \((0,1)\), so \(h^*\) is discontinuous there. There is no contradiction with Exercise 9 in this section (each set \(\{b : h^*(b) > c\}\) is open): that exercise says only that \(h^*\) is lower semicontinuous, and the jumps at \(b = 0\) and \(b = 1\) are upward as one moves into \((0,1)\), exactly as lower semicontinuity permits. Second, \(h^*\) is not in \(\mathcal{L}^1(\mathbb{R})\), because \(h^*(b) = \frac{1}{2b}\) for \(b \ge 1\) and \(\int_1^{\infty} \frac{db}{2b} = \infty\); this shows that the Hardy-Littlewood maximal inequality (4.8) cannot be strengthened to the assertion that \(h \in \mathcal{L}^1(\mathbb{R})\) implies \(h^* \in \mathcal{L}^1(\mathbb{R})\).

Let \(E = [0,1] \cup [2,3]\) and \(h = \chi_E\). We will show that

\begin{equation*} h^*(b) = \begin{cases} \dfrac{1}{3-b} & \text{if } b \le -1, \\[2mm] \dfrac{1}{2(1-b)} & \text{if } -1 \le b \le 0, \\[2mm] 1 & \text{if } 0 < b < 1, \\[2mm] \dfrac{2b-1}{2b} & \text{if } 1 \le b \le \tfrac{3}{2}, \\[2mm] \dfrac{5-2b}{2(3-b)} & \text{if } \tfrac{3}{2} \le b \le 2, \\[2mm] 1 & \text{if } 2 < b < 3, \\[2mm] \dfrac{1}{2(b-2)} & \text{if } 3 \le b \le 4, \\[2mm] \dfrac{1}{b} & \text{if } b \ge 4. \end{cases} \end{equation*}

(The two formulas agree at each of the overlapping endpoints \(b = -1\), \(b = \frac{3}{2}\), and \(b = 4\), so the formula is unambiguous.)

Setup. By Definition 4.6,

\begin{equation*} h^*(b) = \sup_{t>0} \psi_b(t), \qquad \psi_b(t) = \frac{\bigl|E \cap [b-t, b+t]\bigr|}{2t} . \end{equation*}

Define \(g : \mathbb{R} \to [0,2]\) by \(g(y) = \bigl|E \cap (-\infty, y]\bigr|\), so that

\begin{equation*} g(y) = \begin{cases} 0 & \text{if } y \le 0, \\ y & \text{if } 0 \le y \le 1, \\ 1 & \text{if } 1 \le y \le 2, \\ y - 1 & \text{if } 2 \le y \le 3, \\ 2 & \text{if } y \ge 3. \end{cases} \end{equation*}

Because \(E \cap (-\infty, b+t]\) is the disjoint union of \(E \cap (-\infty, b-t)\) and \(E \cap [b-t, b+t]\), and because \(|E \cap (-\infty, b-t)| = |E \cap (-\infty, b-t]| = g(b-t)\) (a single point has measure \(0\)), additivity of Lebesgue measure gives

\begin{equation*} \bigl|E \cap [b-t, b+t]\bigr| = g(b+t) - g(b-t), \qquad \text{so } \psi_b(t) = \frac{g(b+t) - g(b-t)}{2t} . \end{equation*}

Note that \(g\) is continuous and nondecreasing, and is affine on each of the five intervals \((-\infty,0]\), \([0,1]\), \([1,2]\), \([2,3]\), \([3,\infty)\). Since \(E \cap [b-t,b+t] \subseteq [b-t,b+t]\), we always have

\begin{equation*} \psi_b(t) \le 1 . \end{equation*}

Reduction to finitely many candidates. Fix \(b\) and let

\begin{equation*} T_b = \{t > 0 : b+t \in \{0,1,2,3\} \text{ or } b-t \in \{0,1,2,3\}\} , \end{equation*}

a set with at most eight elements. List its elements as \(t_1 < \cdots < t_N\) and put \(t_0 = 0\), \(t_{N+1} = \infty\). On each interval \([t_i, t_{i+1}] \cap (0,\infty)\), neither \(b+t\) nor \(b-t\) crosses any of the numbers \(0,1,2,3\), so each of \(t \mapsto g(b+t)\) and \(t \mapsto g(b-t)\) is affine there. Hence there are constants \(\alpha, \beta\) (depending on \(i\)) with

\begin{equation*} \psi_b(t) = \frac{\alpha t + \beta}{2t} = \frac{\alpha}{2} + \frac{\beta}{2t} \end{equation*}

on that interval, and this is a monotone function of \(t\). Because \(g\) is continuous, \(\psi_b\) is continuous on \((0,\infty)\), so its supremum over each bounded such interval is attained at an endpoint. Two ends need separate treatment. On \((0, t_1]\) the function \(\psi_b\) is monotone and bounded above by \(1\), so \(\lim_{t \downarrow 0}\psi_b(t)\) exists in \([0,1]\) and the supremum over \((0,t_1]\) equals \(\max\{\lim_{t \downarrow 0}\psi_b(t), \, \psi_b(t_1)\}\). On \([t_N, \infty)\), the bound \(g(b+t) - g(b-t) \le 2\) gives \(\lim_{t \to \infty}\psi_b(t) = 0\); since \(\psi_b \ge 0\), a monotone function on \([t_N,\infty)\) tending to \(0\) is either nonincreasing, in which case its supremum is \(\psi_b(t_N)\), or nondecreasing, in which case it is identically \(0\) there and again its supremum is \(\psi_b(t_N)\). Therefore

\begin{equation*} h^*(b) = \max\Bigl\{\ \lim_{t \downarrow 0}\psi_b(t), \ \psi_b(t_1), \ \dots, \ \psi_b(t_N)\ \Bigr\} . \end{equation*}

Symmetry. The map \(y \mapsto 3 - y\) carries \(E\) onto \(E\) (it interchanges \([0,1]\) and \([2,3]\)) and preserves Lebesgue measure (it is a reflection followed by a translation, each of which preserves the length of every interval and hence preserves outer measure). Applying it to \(E \cap [3-b-t, \, 3-b+t]\) produces \(E \cap [b-t, b+t]\), so \(\psi_{3-b}(t) = \psi_b(t)\) for all \(t > 0\) and therefore

\begin{equation*} h^*(3-b) = h^*(b) \qquad \text{for all } b \in \mathbb{R} . \end{equation*}

Thus it suffices to compute \(h^*(b)\) for \(b \le \frac{3}{2}\); the values for \(b \ge \frac{3}{2}\) follow by reflection.

Case A: \(0 < b < 1\).

For \(0 < t \le \min\{b, 1-b\}\) we have \([b-t,b+t] \subseteq [0,1] \subseteq E\), so \(\psi_b(t) = 1\). Since \(\psi_b \le 1\) always, \(h^*(b) = 1\). By symmetry, \(h^*(b) = 1\) for \(2 < b < 3\) as well.

Case B: \(b \le 0\).

Here \(b - t < 0\) for all \(t > 0\), so \(g(b-t) = 0\) and \(\psi_b(t) = \frac{g(b+t)}{2t}\). The candidate values of \(t\) are those with \(b + t \in \{0,1,2,3\}\), namely \(t = -b\) (only when \(b<0\)), \(t = 1-b\), \(t = 2-b\), \(t = 3-b\); and \(\lim_{t \downarrow 0}\psi_b(t)\) equals \(0\) if \(b < 0\) and equals \(\frac{1}{2}\) if \(b = 0\). Computing:

\begin{equation*} \psi_b(-b) = 0, \quad \psi_b(1-b) = \frac{1}{2(1-b)}, \quad \psi_b(2-b) = \frac{1}{2(2-b)}, \quad \psi_b(3-b) = \frac{2}{2(3-b)} = \frac{1}{3-b} , \end{equation*}

using \(g(1) = g(2) = 1\) and \(g(3) = 2\). Clearly \(\frac{1}{2(2-b)} < \frac{1}{2(1-b)}\), and the limit at \(0\) is at most \(\frac12 = \frac{1}{2(1-b)}\big|_{b=0}\) and in fact never exceeds \(\frac{1}{2(1-b)}\) (it is \(0\) when \(b<0\), and equals \(\frac12\) when \(b = 0\)). Hence

\begin{equation*} h^*(b) = \max\left\{\frac{1}{2(1-b)}, \ \frac{1}{3-b}\right\} . \end{equation*}

Now \(\frac{1}{2(1-b)} \ge \frac{1}{3-b}\) if and only if \(3 - b \ge 2 - 2b\), i.e. if and only if \(b \ge -1\). Therefore

\begin{equation*} h^*(b) = \frac{1}{2(1-b)} \text{ for } -1 \le b \le 0, \qquad h^*(b) = \frac{1}{3-b} \text{ for } b \le -1 . \end{equation*}

Case C: \(1 \le b \le \frac{3}{2}\).

Since \(b \ge 1\), we have \(b + t > 1\) for all \(t > 0\), so the candidates coming from \(b+t\) are \(t = 2-b\) and \(t = 3-b\); those coming from \(b-t\) are \(t = b-1\) (only when \(b > 1\)) and \(t = b\). Also \(\lim_{t \downarrow 0}\psi_b(t)\) equals \(0\) if \(b > 1\) and equals \(\frac{1}{2}\) if \(b = 1\). Using the formula for \(g\) and the ranges \(2b-1 \in [1,2]\), \(2b \in [2,3]\), \(2b-2 \in [0,1]\), \(2b-3 \in [-1,0]\) valid for \(1 \le b \le \frac32\):

\begin{equation*} \begin{aligned} \psi_b(b-1) &= \frac{g(2b-1) - g(1)}{2(b-1)} = \frac{1 - 1}{2(b-1)} = 0 \qquad (b > 1), \\ \psi_b(b) &= \frac{g(2b) - g(0)}{2b} = \frac{(2b-1) - 0}{2b} = \frac{2b-1}{2b}, \\ \psi_b(2-b) &= \frac{g(2) - g(2b-2)}{2(2-b)} = \frac{1 - (2b-2)}{2(2-b)} = \frac{3-2b}{2(2-b)}, \\ \psi_b(3-b) &= \frac{g(3) - g(2b-3)}{2(3-b)} = \frac{2 - 0}{2(3-b)} = \frac{1}{3-b} . \end{aligned} \end{equation*}

Write \(A = \frac{2b-1}{2b} = 1 - \frac{1}{2b}\), \(B = \frac{3-2b}{2(2-b)}\), \(C = \frac{1}{3-b}\).

First, \(A \ge C\): indeed \(A \ge C\) is equivalent to \((2b-1)(3-b) \ge 2b\), i.e. to \(2b^2 - 5b + 3 \le 0\), i.e. to \((2b-3)(b-1) \le 0\), which holds exactly when \(1 \le b \le \frac{3}{2}\).

Second, \(A \ge B\): the function \(b \mapsto A = 1 - \frac{1}{2b}\) is increasing, while \(B = \frac{3-2b}{4-2b}\) has derivative \(\frac{-2(4-2b) + 2(3-2b)}{(4-2b)^2} = \frac{-2}{(4-2b)^2} < 0\), so \(B\) is decreasing; since \(A = B = \frac{1}{2}\) at \(b = 1\), we get \(B \le \frac12 \le A\) on \([1, \frac{3}{2}]\).

Finally the remaining candidates are \(0\) and (when \(b=1\)) \(\frac{1}{2} = A(1)\), neither of which exceeds \(A\). Hence

\begin{equation*} h^*(b) = \frac{2b-1}{2b} \qquad \text{for } 1 \le b \le \tfrac{3}{2} . \end{equation*}

Case D: \(b \ge \frac{3}{2}\), by symmetry.

For \(\frac{3}{2} \le b \le 2\) we have \(1 \le 3 - b \le \frac32\), so by Case C,

\begin{equation*} h^*(b) = h^*(3-b) = \frac{2(3-b) - 1}{2(3-b)} = \frac{5-2b}{2(3-b)} . \end{equation*}

For \(2 < b < 3\) we have \(0 < 3-b < 1\), so \(h^*(b) = 1\) by Case A. For \(3 \le b \le 4\) we have \(-1 \le 3-b \le 0\), so by Case B,

\begin{equation*} h^*(b) = h^*(3-b) = \frac{1}{2\bigl(1 - (3-b)\bigr)} = \frac{1}{2(b-2)} . \end{equation*}

For \(b \ge 4\) we have \(3 - b \le -1\), so by Case B,

\begin{equation*} h^*(b) = h^*(3-b) = \frac{1}{3 - (3-b)} = \frac{1}{b} . \end{equation*}

This establishes the formula displayed at the start.

Sanity checks. The maximum value is \(1\), as it must be since \(0 \le h \le 1\). At \(b = \frac{3}{2}\), the midpoint of the gap, \(h^*(\frac32) = \frac{2}{3}\), achieved with \(t = \frac32\): the interval \((0,3)\) has length \(3\) and meets \(E\) in a set of measure \(2\). At \(b = 1\) and \(b = 2\) the value is \(\frac12\), and at \(b = 0\) and \(b = 3\) it is also \(\frac12\), while \(h^* = 1\) on the open intervals \((0,1)\) and \((2,3)\); thus \(h^*\) is discontinuous exactly at \(0, 1, 2, 3\) and is lower semicontinuous, consistent with Exercise 9 in this section. For large \(|b|\) we get \(h^*(b) \approx \frac{2}{2|b|} = \frac{1}{|b|}\), which is the total measure \(|E| = 2\) divided by the length \(2|b|\) of the smallest centered interval that swallows \(E\); as with Example 4.7, this decay rate shows \(h^* \notin \mathcal{L}^1(\mathbb{R})\).

Because \(h \ge 0\) we have \(|h| = h\). Define

\begin{equation*} F(x) = \int_{-\infty}^{x} h = \begin{cases} 0 & \text{if } x \le 0,\\ \frac{x^2}{2} & \text{if } 0 \le x \le 1,\\ \frac12 & \text{if } x \ge 1,\end{cases} \end{equation*}

so that, in the notation of 4.6,

\begin{equation*} h^{*}(b) = \sup_{t > 0} V_b(t), \qquad V_b(t) := \frac{1}{2t}\int_{b-t}^{b+t} h = \frac{F(b+t) - F(b-t)}{2t}. \end{equation*}

We compute \(\sup_{t>0} V_b(t)\) by splitting the range of \(t\) according to where \(b - t\) and \(b + t\) sit relative to \(0\) and \(1\).

Case 1: \(b \le 0\). Here \(b - t < 0\) for all \(t > 0\), so \(F(b-t) = 0\) and \(V_b(t) = F(b+t)/(2t)\).

If \(t \le -b\) then \(b + t \le 0\) and \(V_b(t) = 0\).

If \(-b \le t \le 1 - b\) then \(b + t \in [0,1]\), so with \(v = b + t \in [0,1]\) (thus \(t = v - b\)) we get

\begin{equation*} V_b(t) = \frac{v^2}{4(v - b)}, \qquad \frac{d}{dv}\Big(\frac{v^2}{v-b}\Big) = \frac{v(v - 2b)}{(v-b)^2} \ge 0 \end{equation*}

because \(v \ge 0\) and \(v - 2b \ge 0\) (as \(b \le 0\)). Hence \(V_b\) increases on this range, with maximum at \(v = 1\), i.e. \(t = 1 - b\), where the value is \(\frac{1}{4(1-b)}\).

If \(t \ge 1 - b\) then \(b + t \ge 1\), so \(V_b(t) = \frac{1/2}{2t} = \frac{1}{4t}\), which is decreasing and again has largest value \(\frac{1}{4(1-b)}\) at \(t = 1-b\).

Therefore \(h^{*}(b) = \frac{1}{4(1-b)}\) for \(b \le 0\).

Case 2: \(0 < b < 1\). For \(0 < t \le \min\{b, 1-b\}\) both endpoints \(b \pm t\) lie in \([0,1]\), so

\begin{equation*} V_b(t) = \frac{(b+t)^2 - (b-t)^2}{4t} = \frac{4bt}{4t} = b . \end{equation*}

So \(V_b \equiv b\) on that initial range. Now we look beyond it.

Suppose first that \(b \le \frac12\), so \(\min\{b,1-b\} = b\). For \(b \le t \le 1 - b\) we have \(b - t \le 0 \le b + t \le 1\), so \(V_b(t) = \frac{(b+t)^2}{4t}\), and

\begin{equation*} \frac{d}{dt}\Big(\frac{(b+t)^2}{4t}\Big) = \frac{(b+t)(t-b)}{4t^2} \ge 0 \quad \text{for } t \ge b, \end{equation*}

so \(V_b\) increases on \([b, 1-b]\), reaching \(\frac{1}{4(1-b)}\) at \(t = 1-b\). For \(t \ge 1-b\) we have \(b+t \ge 1\) and \(b - t \le 0\), so \(V_b(t) = \frac{1}{4t}\), decreasing. Hence for \(0 < b \le \frac12\),

\begin{equation*} h^{*}(b) = \max\Big\{b, \tfrac{1}{4(1-b)}\Big\} = \frac{1}{4(1-b)}, \end{equation*}

the last equality because \(1 - 4b(1-b) = (2b-1)^2 \ge 0\) gives \(\frac{1}{4(1-b)} \ge b\).

Now suppose \(\frac12 \le b < 1\), so \(\min\{b,1-b\} = 1-b\). For \(1 - b \le t \le b\) we have \(b - t \in [0,1]\) and \(b + t \ge 1\), so writing \(u = b - t \in [0, 2b-1]\),

\begin{equation*} V_b(t) = \frac{1 - (b-t)^2}{4t} = \frac{1-u^2}{4(b-u)}, \qquad \frac{d}{du}\Big(\frac{1-u^2}{b-u}\Big) = \frac{u^2 - 2bu + 1}{(b-u)^2}. \end{equation*}

The quadratic \(u^2 - 2bu + 1\) has discriminant \(4b^2 - 4 < 0\), hence is strictly positive; so the expression is strictly increasing in \(u\), i.e. strictly decreasing in \(t\). Thus on \([1-b, b]\) the function \(V_b\) decreases from its value \(b\) at \(t = 1 - b\). For \(t \ge b\) we get \(V_b(t) = \frac{1}{4t} \le \frac{1}{4b} \le b\) (the last step because \(b \ge \frac12\)). Hence \(h^{*}(b) = b\) for \(\frac12 \le b < 1\).

Case 3: \(b \ge 1\). Now \(b + t > 1\) for every \(t>0\), so \(F(b+t) = \frac12\).

If \(t \le b - 1\) then \(b - t \ge 1\) and \(V_b(t) = 0\).

If \(b - 1 \le t \le b\) then \(b - t \in [0,1]\) and, putting \(u = b - t \in [0,1]\),

\begin{equation*} V_b(t) = \frac{1 - u^2}{4(b-u)} =: g(u), \qquad g^{\prime}(u) = \frac{u^2 - 2bu + 1}{4(b-u)^2}. \end{equation*}

The roots of \(u^2 - 2bu + 1\) are \(b \pm \sqrt{b^2-1}\), and \(u_0 := b - \sqrt{b^2-1}\) lies in \((0,1]\) (indeed \(u_0 \le 1\) is equivalent to \(b - 1 \le \sqrt{b^2-1}\), which holds since \((b-1)^2 \le (b-1)(b+1)\)). The other root \(b + \sqrt{b^2-1}\) is at least \(1\), so on \([0,1]\) the quadratic is positive to the left of \(u_0\) and negative to its right; thus \(g^{\prime} > 0\) on \([0,u_0)\) and \(g^{\prime} < 0\) on \((u_0, 1]\), so \(g\) attains its maximum over \([0,1]\) at \(u_0\). Since \(b - u_0 = \sqrt{b^2-1}\) and

\begin{equation*} 1 - u_0^2 = 1 - \big(2b^2 - 1 - 2b\sqrt{b^2-1}\big) = 2\sqrt{b^2-1}\,\big(b - \sqrt{b^2-1}\big), \end{equation*}

we get

\begin{equation*} g(u_0) = \frac{2\sqrt{b^2-1}\,(b-\sqrt{b^2-1})}{4\sqrt{b^2-1}} = \frac{b - \sqrt{b^2-1}}{2}. \end{equation*}

One caveat about the endpoint: the admissible values of \(t\) are \(t > 0\), so \(u = b - t\) ranges over \([0, b) \cap [0,1]\). When \(b > 1\) this is all of \([0,1]\) and the maximum \(g(u_0)\) is attained (at \(t = \sqrt{b^2-1} > 0\)). When \(b = 1\) we have \(u_0 = 1\), which corresponds to the forbidden value \(t = 0\); but \(g\) is increasing on \([0,1)\), so the supremum of \(g\) over \([0,1)\) equals \(\lim_{u \uparrow 1} g(u) = \lim_{u \uparrow 1}\frac{1+u}{4} = \frac12\), which is again the value given by the formula \(\frac{b - \sqrt{b^2-1}}{2}\) at \(b = 1\). So in both cases the supremum over this range of \(t\) equals \(\frac{b-\sqrt{b^2-1}}{2}\) (attained when \(b > 1\), approached as \(t \downarrow 0\) when \(b = 1\)).

If \(t \ge b\) then \(b - t \le 0\) and \(V_b(t) = \frac{1}{4t} \le \frac{1}{4b}\). Since \(b - \sqrt{b^2-1} = \frac{1}{b+\sqrt{b^2-1}} \ge \frac{1}{2b}\), we have \(g(u_0) \ge \frac{1}{4b}\), so this range contributes nothing new. Hence \(h^{*}(b) = \frac{b - \sqrt{b^2-1}}{2}\) for \(b \ge 1\).

Conclusion. Combining the three cases (and noting that at \(b = \frac12\) both formulas give \(\frac12\)):

\begin{equation*} h^{*}(b) = \begin{cases} \dfrac{1}{4(1-b)} & \text{if } b \le \frac12,\\[2mm] b & \text{if } \frac12 \le b < 1,\\[2mm] \dfrac{b - \sqrt{b^2-1}}{2} = \dfrac{1}{2\big(b + \sqrt{b^2-1}\big)} & \text{if } b \ge 1.\end{cases} \end{equation*}

Note that \(h^{*}\) is continuous except at \(b = 1\), where it drops from the left-hand limit \(1\) to the value \(h^{*}(1) = \frac12\); this is consistent with the lower semicontinuity implied by Exercise 9, since the limit from the right at \(1\) also equals \(\frac12\).

Fix \(c \in \mathbb{R}\) and let \(E = \{b \in \mathbb{R} : h^{*}(b) > c\}\). Let \(b \in E\); we produce \(\delta > 0\) with \((b - \delta, b+\delta) \subseteq E\), which shows \(E\) is open.

Because \(h^{*}(b) = \sup_{t>0} \frac{1}{2t}\int_{b-t}^{b+t}|h| > c\), there exists \(t > 0\) such that

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t} |h| > c . \end{equation*}

Write \(\alpha = \int_{b-t}^{b+t}|h| \in [0,\infty]\); the displayed inequality says \(\alpha > 2tc\).

Now choose \(t^{\prime} > t\) with \(2t^{\prime}c < \alpha\). Such a \(t^{\prime}\) exists: if \(c \le 0\) then \(2t^{\prime}c \le 0 \le \alpha\) for every \(t^{\prime} > t\), with strict inequality unless \(c = 0\) and \(\alpha = 0\), which cannot happen since \(\alpha > 2tc = 0\) in that case; if \(c > 0\) and \(\alpha = \infty\) take \(t^{\prime} = t+1\); and if \(c > 0\) and \(\alpha < \infty\) then \(\frac{\alpha}{2c} > t\), so any \(t^{\prime}\) in the nonempty interval \(\big(t, \frac{\alpha}{2c}\big)\) works.

Put \(\delta = t^{\prime} - t > 0\) and suppose \(|b^{\prime} - b| < \delta\). Then

\begin{equation*} b^{\prime} - t^{\prime} = b^{\prime} - b - t^{\prime} + b < \delta - t^{\prime} + b = b - t \quad\text{and}\quad b^{\prime} + t^{\prime} = b^{\prime} - b + t^{\prime} + b > -\delta + t^{\prime} + b = b + t, \end{equation*}

so \((b - t, b+t) \subseteq (b^{\prime} - t^{\prime}, b^{\prime} + t^{\prime})\). Since \(|h| \ge 0\), monotonicity of the integral gives

\begin{equation*} \int_{b^{\prime}-t^{\prime}}^{b^{\prime}+t^{\prime}} |h| \ \ge\ \int_{b-t}^{b+t} |h| \ = \ \alpha . \end{equation*}

Therefore

\begin{equation*} h^{*}(b^{\prime}) \ \ge\ \frac{1}{2t^{\prime}}\int_{b^{\prime}-t^{\prime}}^{b^{\prime}+t^{\prime}}|h| \ \ge\ \frac{\alpha}{2t^{\prime}} \ > \ c, \end{equation*}

the last inequality because \(2t^{\prime}c < \alpha\). Hence \(b^{\prime} \in E\).

Thus every point of \(E\) is an interior point, so \(E\) is open. In particular, taking \(c\) to range over \(\mathbb{R}\) shows that \(h^{*}\) is a Borel measurable function, as asserted in the text following 4.7.

The statement is true; here is a proof.

First note that an increasing function \(h \colon \mathbb{R} \to [0,\infty)\) is Borel measurable (for each \(c \in \mathbb{R}\) the set \(\{x : h(x) > c\}\) is an interval, since \(x \in \{h > c\}\) and \(y > x\) imply \(h(y) \ge h(x) > c\)), so \(h^{*}\) is defined. Also \(|h| = h\) because \(h \ge 0\).

Suppose \(b_1 < b_2\) and let \(s = b_2 - b_1 > 0\). Fix \(t > 0\). Translation invariance of Lebesgue measure (2.7) gives translation invariance of the integral: for every nonnegative Borel measurable \(g\),

\begin{equation*} \int_{b_1 - t}^{b_1+t} g(x + s)\,dx = \int_{b_2-t}^{b_2+t} g(y)\,dy \end{equation*}

(substitute \(y = x+s\); formally, \(x \mapsto g(x+s)\) is the composition of \(g\) with a translation, and the two integrals agree first for characteristic functions of Borel sets by 2.7, hence for simple functions by linearity, hence for all nonnegative Borel measurable functions by the definition of the integral as a supremum over simple functions below the integrand).

Since \(h\) is increasing and \(s > 0\), we have \(h(x) \le h(x+s)\) for every \(x \in \mathbb{R}\). Hence, by monotonicity of the integral and the identity above,

\begin{equation*} \int_{b_1-t}^{b_1+t} h(x)\,dx \ \le\ \int_{b_1-t}^{b_1+t} h(x+s)\,dx \ = \ \int_{b_2-t}^{b_2+t} h(y)\,dy . \end{equation*}

Dividing by \(2t\),

\begin{equation*} \frac{1}{2t}\int_{b_1-t}^{b_1+t} |h| \ \le\ \frac{1}{2t}\int_{b_2-t}^{b_2+t}|h| \ \le\ h^{*}(b_2). \end{equation*}

This holds for every \(t > 0\), so taking the supremum over \(t > 0\) on the left gives \(h^{*}(b_1) \le h^{*}(b_2)\).

Therefore \(h^{*}\) is increasing.

Define

\begin{equation*} h = \sum_{n=1}^{\infty} n\, \chi_{[n,\, n + n^{-3}]}. \end{equation*}

Because \(n + n^{-3} < n+1\) for every positive integer \(n\), the intervals \([n, n+n^{-3}]\) are pairwise disjoint, so at each \(x \in \mathbb{R}\) at most one term of the sum is nonzero; thus \(h(x) \in [0,\infty)\) for every \(x\). Indeed at each fixed \(x\) the partial sums are eventually constant, so they converge pointwise to \(h\); each partial sum is a finite linear combination of characteristic functions of Borel sets and hence is a Borel measurable real-valued function. Therefore \(h\) is Borel measurable by 2.48.

Step 1: \(h \in \mathcal{L}^1(\mathbb{R})\) and \(h\) is bounded on bounded sets. By the Monotone Convergence Theorem (3.11) applied to the partial sums,

\begin{equation*} \int_{-\infty}^{\infty} h = \sum_{n=1}^{\infty} n \cdot |[n, n+n^{-3}]| = \sum_{n=1}^{\infty} \frac{n}{n^{3}} = \sum_{n=1}^{\infty}\frac{1}{n^{2}} < \infty . \end{equation*}

Write \(\|h\|_1 = \int |h| < \infty\). Moreover, if \(R > 0\) then \(h(x) = 0\) for \(x < 1\) and \(h(x) \le \lceil R \rceil\) for \(x \in [-R, R]\), since the only spikes meeting \([-R,R]\) are those with \(n \le R\). So \(M_R := \sup_{[-R,R]} h < \infty\) for every \(R > 0\).

Step 2: \(h^{*}(b) < \infty\) for every \(b \in \mathbb{R}\). Fix \(b \in \mathbb{R}\) and put \(R = |b| + 1\). If \(0 < t \le 1\) then \([b-t, b+t] \subseteq [-R, R]\), so

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t} h \ \le\ \frac{1}{2t}\cdot 2t \cdot M_R = M_R . \end{equation*}

If \(t \ge 1\) then

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t} h \ \le\ \frac{1}{2t}\,\|h\|_1 \ \le\ \frac{\|h\|_1}{2}. \end{equation*}

Taking the supremum over all \(t > 0\) gives \(h^{*}(b) \le \max\{M_R, \tfrac12\|h\|_1\} < \infty\).

Step 3: \(\sup_{b \in \mathbb{R}} h^{*}(b) = \infty\). Let \(n\) be a positive integer and take \(b = n\), \(t = n^{-3}\). Then \([n, n+n^{-3}] \subseteq [b-t, b+t]\), so

\begin{equation*} h^{*}(n) \ \ge\ \frac{1}{2n^{-3}}\int_{n - n^{-3}}^{n + n^{-3}} h \ \ge\ \frac{n^{3}}{2}\cdot n \cdot n^{-3} \ = \ \frac{n}{2}. \end{equation*}

Since \(n/2 \to \infty\), we conclude \(\sup\{h^{*}(b) : b \in \mathbb{R}\} = \infty\).

Thus \(h\) has all the required properties.

Let \(h \in \mathcal{L}^1(\mathbb{R})\) and set

\begin{equation*} E = \{b \in \mathbb{R} : h^{*}(b) = \infty\}. \end{equation*}

Since \(h^{*}\) takes values in \([0,\infty]\), we have

\begin{equation*} E = \bigcap_{n=1}^{\infty}\{b \in \mathbb{R} : h^{*}(b) > n\}, \end{equation*}

and each set on the right is open by Exercise 9 in this section; hence \(E\) is a Borel set, in particular Lebesgue measurable. (Even without this, the argument below works verbatim with \(|\cdot|\) read as outer measure, which is monotone by 2.5.)

For each positive integer \(n\) we have \(E \subseteq \{b \in \mathbb{R} : h^{*}(b) > n\}\), so the Hardy–Littlewood maximal inequality (4.8) gives

\begin{equation*} |E| \ \le\ \big|\{b \in \mathbb{R} : h^{*}(b) > n\}\big| \ \le\ \frac{3}{n}\,\|h\|_1 . \end{equation*}

Because \(h \in \mathcal{L}^1(\mathbb{R})\), the number \(\|h\|_1\) is finite, so \(\frac{3}{n}\|h\|_1 \to 0\) as \(n \to \infty\). As \(|E|\) is a fixed number in \([0,\infty]\) bounded above by every term of a sequence tending to \(0\), we conclude \(|E| = 0\).

Two elementary properties of \(h \mapsto h^{*}\). Let \(f, g \colon \mathbb{R} \to [0,\infty]\) be Borel measurable.

(i) If \(f \le g\) pointwise then \(f^{*} \le g^{*}\); indeed \(\frac{1}{2t}\int_{b-t}^{b+t} f \le \frac{1}{2t}\int_{b-t}^{b+t} g \le g^{*}(b)\) for every \(t > 0\), and we take the supremum over \(t\).

(ii) For \(r \in \mathbb{R}\) let \(g_r(x) = g(x - r)\). Then \(g_r^{*}(b) = g^{*}(b - r)\) for every \(b\), because translation invariance of Lebesgue measure (2.7) gives \(\int_{b-t}^{b+t} g_r = \int_{b-r-t}^{b-r+t} g\) for every \(t > 0\).

A single bad point. Define \(g \colon \mathbb{R} \to [0,\infty)\) by

\begin{equation*} g(x) = \begin{cases} \dfrac{1}{x\,(\log x)^{2}} & \text{if } 0 < x < \frac12,\\[2mm] 0 & \text{otherwise.}\end{cases} \end{equation*}

Then \(g\) is Borel measurable (it is continuous on the open set \((0,\tfrac12)\) and zero elsewhere) and nonnegative. Since \(\frac{d}{dx}\big(\!-\!\frac{1}{\log x}\big) = \frac{1}{x(\log x)^{2}}\) for \(0 < x < 1\), and \(-\frac{1}{\log x} \to 0\) as \(x \downarrow 0\), we get for \(0 < t < \frac12\)

\begin{equation*} \int_{0}^{t} g = -\frac{1}{\log t} = \frac{1}{|\log t|}, \end{equation*}

and in particular \(\|g\|_1 = \int_{0}^{1/2} g = \frac{1}{\log 2} < \infty\), so \(g \in \mathcal{L}^1(\mathbb{R})\). Now for \(0 < t < \frac12\),

\begin{equation*} \frac{1}{2t}\int_{-t}^{t} g = \frac{1}{2t}\int_{0}^{t} g = \frac{1}{2t\,|\log t|} \xrightarrow[\ t \downarrow 0\ ]{} \infty , \end{equation*}

because \(t|\log t| \to 0\) as \(t \downarrow 0\). Hence \(g^{*}(0) = \infty\).

Spreading it over \(\mathbb{Q}\). Let \(r_1, r_2, r_3, \dots\) be an enumeration of \(\mathbb{Q}\) (possible since \(\mathbb{Q}\) is countable), and define \(\tilde h \colon \mathbb{R} \to [0,\infty]\) by

\begin{equation*} \tilde h(x) = \sum_{k=1}^{\infty} 2^{-k} g(x - r_k). \end{equation*}

Each summand \(x \mapsto 2^{-k} g(x - r_k)\) is Borel measurable, because for \(c \in \mathbb{R}\) the set \(\{x : 2^{-k}g(x - r_k) > c\}\) is the translate by \(r_k\) of the Borel set \(\{y : g(y) > 2^{k}c\}\), and translates of Borel sets are Borel. The partial sums \(s_m = \sum_{k=1}^{m} 2^{-k}g(\cdot - r_k)\) are therefore Borel measurable, and since all terms are nonnegative we have \(\tilde h = \sup_m s_m\) pointwise; hence \(\tilde h\) is Borel measurable by 2.53. By the Monotone Convergence Theorem (3.11) applied to the increasing sequence \(s_1 \le s_2 \le \cdots\),

\begin{equation*} \int_{-\infty}^{\infty} \tilde h = \sum_{k=1}^{\infty} 2^{-k}\int_{-\infty}^{\infty} g(x - r_k)\,dx = \sum_{k=1}^{\infty} 2^{-k}\,\|g\|_1 = \|g\|_1 = \frac{1}{\log 2} < \infty , \end{equation*}

using translation invariance again. Since \(\tilde h\) is nonnegative with finite integral, the set \(N = \{x : \tilde h(x) = \infty\}\) has measure \(0\) (if \(|N| > 0\) then \(\int \tilde h \ge c\,|N| \to \infty\) as \(c \to \infty\), a contradiction). Define

\begin{equation*} h(x) = \begin{cases} \tilde h(x) & \text{if } x \notin N,\\ 0 & \text{if } x \in N.\end{cases} \end{equation*}

Then \(h \colon \mathbb{R} \to \mathbb{R}\) is Borel measurable, \(h = \tilde h\) except on the set \(N\) of measure \(0\), and \(\int |h| = \int \tilde h < \infty\); thus \(h \in \mathcal{L}^1(\mathbb{R})\).

Conclusion. First we check that \(h^{*} = \tilde h^{*}\). Fix \(b \in \mathbb{R}\) and \(t > 0\) and put \(A = (b-t, b+t)\). Since \(h \ge 0\) we have \(|h| = h\), and \(h\) and \(\tilde h\) agree on \(A \setminus N\) while \(A \cap N\) has measure \(0\). The integral of a nonnegative measurable function over a set of measure \(0\) is \(0\): any simple function \(\sum_j c_j \chi_{E_j}\) lying below it has integral \(\sum_j c_j |E_j \cap A \cap N| = 0\) over that set, and the integral is the supremum of such numbers. Hence

\begin{equation*} \int_{A} \tilde h = \int_{A \setminus N} \tilde h + \int_{A \cap N} \tilde h = \int_{A\setminus N} h + 0 = \int_{A} h = \int_{b-t}^{b+t}|h| . \end{equation*}

As this holds for every \(b\) and every \(t > 0\), taking suprema over \(t\) gives \(h^{*} = \tilde h^{*}\). Fix \(b \in \mathbb{Q}\) and choose \(k\) with \(r_k = b\). Since \(\tilde h \ge 2^{-k} g(\cdot - r_k)\) pointwise, properties (i) and (ii) give

\begin{equation*} h^{*}(b) = \tilde h^{*}(b) \ \ge\ 2^{-k}\,\big(g(\cdot - r_k)\big)^{*}(b) = 2^{-k}\, g^{*}(b - r_k) = 2^{-k}\, g^{*}(0) = \infty . \end{equation*}

(Here we also used the obvious identity \((\lambda f)^{*} = \lambda f^{*}\) for \(\lambda > 0\).)

Hence \(h \in \mathcal{L}^1(\mathbb{R})\) and \(h^{*}(b) = \infty\) for every \(b \in \mathbb{Q}\), as required. Note that this is consistent with Exercise 12: the set \(\{b : h^{*}(b) = \infty\}\) contains \(\mathbb{Q}\) but still has measure \(0\).

Fix \(c > 0\) and set

\begin{equation*} E_c = \{b \in \mathbb{R} : h^{*}(b) \ge c\}, \qquad U_s = \{b \in \mathbb{R} : h^{*}(b) > s\} \ \ (s > 0). \end{equation*}

Each \(U_s\) is open by Exercise 9 in this section, and

\begin{equation*} E_c = \bigcap_{n > 1/c} U_{c - 1/n}, \end{equation*}

so \(E_c\) is a Borel set; in particular it is Lebesgue measurable. (As in Exercise 12, the argument below is equally valid with \(|\cdot|\) interpreted as outer measure, using monotonicity 2.5.)

Let \(n\) be any integer with \(n > 1/c\), so that \(c - \frac1n > 0\). If \(h^{*}(b) \ge c\) then \(h^{*}(b) > c - \frac1n\), so

\begin{equation*} E_c \subseteq U_{c - 1/n}. \end{equation*}

The Hardy–Littlewood maximal inequality (4.8), applied with the positive number \(c - \frac1n\) in place of \(c\), gives

\begin{equation*} \big|U_{c-1/n}\big| \ \le\ \frac{3}{c - \frac1n}\,\|h\|_1 . \end{equation*}

Hence, by monotonicity of Lebesgue measure,

\begin{equation*} |E_c| \ \le\ \frac{3\,\|h\|_1}{c - \frac1n} \qquad \text{for every integer } n > \tfrac1c . \end{equation*}

Since \(h \in \mathcal{L}^1(\mathbb{R})\) we have \(\|h\|_1 < \infty\), and \(\frac{3\|h\|_1}{c - 1/n} \to \frac{3\|h\|_1}{c}\) as \(n \to \infty\). Therefore

\begin{equation*} |E_c| \ \le\ \lim_{n \to \infty} \frac{3\,\|h\|_1}{c - \frac1n} \ = \ \frac{3}{c}\,\|h\|_1 , \end{equation*}

which is the desired inequality.

Let \(E\) be the set of \(b \in \mathbb{R}\) for which

\begin{equation*} \lim_{t \downarrow 0} \frac{1}{2t} \int_{b-t}^{b+t} |f - f(b)| = 0 . \end{equation*}

By the first version of the Lebesgue Differentiation Theorem (4.10), \(\mathbb{R} \setminus E\) has Lebesgue measure \(0\). We show that the desired conclusion holds at every \(b \in E\); this suffices.

So fix \(b \in E\) and fix \(t > 0\). Write \(I = [b-t,\, b+t]\), so that \(0 < |I| = 2t < \infty\) and hence \(f_I\) is defined, with

\begin{equation*} f_I = \frac{1}{2t}\int_{b-t}^{b+t} f . \end{equation*}

Note that \(f(b) \in \mathbb{R}\), because by definition (3.40 and 3.45) the elements of \(L^1(\mathbb{R})\) are real-valued functions. Note also that \(\int_{b-t}^{b+t}|f| \le \int_{-\infty}^{\infty}|f| < \infty\) and \(\int_{b-t}^{b+t}|f(b)| = 2t\,|f(b)| < \infty\), so every integral written below is the integral of a function that is integrable on \(I\); in particular all the quantities below are finite real numbers and the manipulations are legitimate.

First we estimate how far the average \(f_I\) is from the value \(f(b)\). Because \(\frac{1}{2t}\int_{b-t}^{b+t} f(b) = f(b)\), additivity of integration (3.21) gives

\begin{equation*} |f_I - f(b)| = \left| \frac{1}{2t}\int_{b-t}^{b+t} \bigl(f - f(b)\bigr) \right| \le \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)| , \end{equation*}

where the inequality holds because the absolute value of an integral is at most the integral of the absolute value (3.23).

Now the triangle inequality gives \(|f(x) - f_I| \le |f(x) - f(b)| + |f(b) - f_I|\) for every \(x \in I\). Integrating this over \(I\) (integration is order preserving, by 3.22, and additive, by 3.21) and using \(\int_I |f(b) - f_I| = 2t\,|f(b) - f_I|\), then dividing by \(2t\), we obtain

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t} |f - f_I| \le \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)| + |f(b) - f_I| \le 2 \cdot \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)| , \end{equation*}

where the last inequality uses the estimate of the previous paragraph.

Because \(b \in E\), the right side of the last display has limit \(2 \cdot 0 = 0\) as \(t \downarrow 0\). Hence

\begin{equation*} \lim_{t \downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t} |f - f_{[b-t,\, b+t]}| = 0 , \end{equation*}

as desired. Because this holds for every \(b \in E\) and \(\mathbb{R}\setminus E\) is a set of measure \(0\), the conclusion holds for almost every \(b \in \mathbb{R}\).

As in the previous exercise, let \(E\) denote the set of \(b \in \mathbb{R}\) such that

\begin{equation*} \lim_{t \downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t} |f - f(b)| = 0 . \end{equation*}

The first version of the Lebesgue Differentiation Theorem (4.10) tells us that \(\mathbb{R}\setminus E\) has measure \(0\). We prove that the limit in the statement of this exercise equals \(0\) for every \(b \in E\).

Fix \(b \in E\). For each \(t > 0\) set

\begin{equation*} \Phi(t) = \sup\left\{ \frac{1}{|I|}\int_I |f - f_I| \; : \; I \text{ is an interval of length } t \text{ containing } b \right\} . \end{equation*}

The set over which this supremum is taken is nonempty (it contains the value coming from \(I = [b,\, b+t]\), for instance), and each \(I\) occurring in it has \(0 < |I| = t < \infty\), so \(f_I\) is defined for each such \(I\).

Now fix \(t > 0\) and let \(I\) be any interval of length \(t\) with \(b \in I\). Every \(x \in I\) satisfies \(|x - b| \le t\), because \(x\) and \(b\) both lie in \(I\) and \(I\) has length \(t\). Hence

\begin{equation*} I \subseteq [b-t,\, b+t] . \end{equation*}

Repeating the computation from the previous exercise with the interval \(I\) in place of \([b-t,\,b+t]\), we get

\begin{equation*} |f_I - f(b)| = \left|\frac{1}{|I|}\int_I \bigl(f - f(b)\bigr)\right| \le \frac{1}{|I|}\int_I |f - f(b)| \end{equation*}

by 3.23, and therefore, by the triangle inequality applied pointwise and then integrated over \(I\),

\begin{equation*} \frac{1}{|I|}\int_I |f - f_I| \le \frac{1}{|I|}\int_I |f - f(b)| + |f(b) - f_I| \le \frac{2}{|I|}\int_I |f - f(b)| . \end{equation*}

Because \(|f - f(b)| \ge 0\) and \(I \subseteq [b-t,\,b+t]\), we have \(\int_I |f - f(b)| \le \int_{b-t}^{b+t} |f - f(b)|\). Since \(|I| = t\), combining this with the previous display yields

\begin{equation*} \frac{1}{|I|}\int_I |f - f_I| \le \frac{2}{t}\int_{b-t}^{b+t} |f - f(b)| = 4 \cdot \frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)| . \end{equation*}

The right side of the last display does not depend on \(I\); it depends only on \(b\) and \(t\). Thus it is an upper bound for the set whose supremum defines \(\Phi(t)\), and hence

\begin{equation*} 0 \le \Phi(t) \le 4 \cdot \frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)| . \end{equation*}

Because \(b \in E\), the right side has limit \(0\) as \(t \downarrow 0\). By the squeeze principle, \(\lim_{t \downarrow 0} \Phi(t) = 0\), which is exactly the assertion of the exercise at the point \(b\).

Because \(\mathbb{R}\setminus E\) has measure \(0\), this holds for almost every \(b \in \mathbb{R}\).

The hypothesis here is that \(f^2 \in L^1(\mathbb{R})\), not that \(f \in L^1(\mathbb{R})\), so 4.10 does not apply directly to \(f\) and we must build the conclusion out of functions to which it does apply. We first record a localized form of 4.10, which frees us from having to integrate over all of \(\mathbb{R}\).

Localization Lemma. Suppose \(h : \mathbb{R} \to \mathbb{R}\) is Lebesgue measurable and \(\int_{-n}^{n} |h| < \infty\) for every \(n \in \mathbb{Z}^+\). Then

\begin{equation*} \lim_{t \downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t} |h - h(b)| = 0 \end{equation*}

for almost every \(b \in \mathbb{R}\); in particular \(\frac{1}{2t}\int_{b-t}^{b+t} h \to h(b)\) as \(t \downarrow 0\) for almost every \(b \in \mathbb{R}\).

Proof of the lemma. For \(n \in \mathbb{Z}^+\) let \(h_n = h \chi_{(-n,\, n)}\). Then \(h_n\) is Lebesgue measurable and \(\int |h_n| = \int_{-n}^{n}|h| < \infty\), so \(h_n \in L^1(\mathbb{R})\). By 4.10 there is a set \(N_n \subseteq \mathbb{R}\) with \(|N_n| = 0\) such that

\begin{equation*} \lim_{t\downarrow 0} \frac{1}{2t}\int_{b-t}^{b+t}|h_n - h_n(b)| = 0 \quad \text{for all } b \in \mathbb{R}\setminus N_n . \end{equation*}

Let \(N = \bigcup_{n=1}^{\infty} N_n\); then \(|N| = 0\) by countable subadditivity (2.8). Suppose \(b \in \mathbb{R}\setminus N\). Choose \(n \in \mathbb{Z}^+\) with \(|b| < n\) and put \(\delta = n - |b| > 0\). If \(0 < t < \delta\) then \((b-t,\, b+t) \subseteq (-n,\, n)\), so \(h = h_n\) on \((b-t,\,b+t)\) and \(h(b) = h_n(b)\); hence

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t}|h - h(b)| = \frac{1}{2t}\int_{b-t}^{b+t}|h_n - h_n(b)| . \end{equation*}

Because \(b \notin N_n\), the right side has limit \(0\) as \(t \downarrow 0\), and therefore so does the left side. The last assertion of the lemma follows because

\begin{equation*} \left|\frac{1}{2t}\int_{b-t}^{b+t} h - h(b)\right| = \left|\frac{1}{2t}\int_{b-t}^{b+t}\bigl(h - h(b)\bigr)\right| \le \frac{1}{2t}\int_{b-t}^{b+t}|h - h(b)| , \end{equation*}

using 3.23. This completes the proof of the lemma.

Now we turn to the exercise. For \(q \in \mathbb{Q}\) define \(h_q : \mathbb{R}\to\mathbb{R}\) by

\begin{equation*} h_q(x) = \bigl(f(x) - q\bigr)^2 . \end{equation*}

Each \(h_q\) is Lebesgue measurable, because sums and products of measurable functions are measurable (2.46). Furthermore, for each \(n \in \mathbb{Z}^+\) the inequality \((\alpha - \beta)^2 \le 2\alpha^2 + 2\beta^2\) gives

\begin{equation*} \int_{-n}^{n} h_q \le 2\int_{-n}^{n} f^2 + 2q^2 \cdot 2n \le 2\int_{-\infty}^{\infty} f^2 + 4nq^2 < \infty , \end{equation*}

because \(f^2 \in L^1(\mathbb{R})\). Since \(h_q \ge 0\), we have \(|h_q| = h_q\), so the display above says \(\int_{-n}^{n}|h_q| < \infty\) for every \(n \in \mathbb{Z}^+\); thus \(h_q\) satisfies the hypothesis of the Localization Lemma. Let \(N_q\) be a set of measure \(0\) such that

\begin{equation*} \lim_{t\downarrow 0}\frac{1}{2t}\int_{b-t}^{b+t} h_q = h_q(b) = \bigl(f(b) - q\bigr)^2 \quad \text{for all } b \in \mathbb{R}\setminus N_q , \end{equation*}

which the lemma provides. Let

\begin{equation*} N = \bigcup_{q \in \mathbb{Q}} N_q . \end{equation*}

Because \(\mathbb{Q}\) is countable, \(|N| = 0\).

Suppose now \(b \in \mathbb{R}\setminus N\); we show the conclusion of the exercise holds at \(b\). Let \(\varepsilon > 0\). Because \(f(b) \in \mathbb{R}\) and \(\mathbb{Q}\) is dense in \(\mathbb{R}\), there exists \(q \in \mathbb{Q}\) with \(|f(b) - q| < \varepsilon\). For every \(x \in \mathbb{R}\),

\begin{equation*} |f(x) - f(b)|^2 \le \bigl(|f(x) - q| + |q - f(b)|\bigr)^2 \le 2\bigl(f(x)-q\bigr)^2 + 2\bigl(q - f(b)\bigr)^2 , \end{equation*}

using \((\alpha+\beta)^2 \le 2\alpha^2 + 2\beta^2\). Integrating over \((b-t,\,b+t)\) and dividing by \(2t\) gives

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)|^2 \le 2 \cdot \frac{1}{2t}\int_{b-t}^{b+t} h_q + 2\bigl(q - f(b)\bigr)^2 . \end{equation*}

Letting \(t \downarrow 0\) and using \(b \notin N_q\), we obtain

\begin{equation*} \limsup_{t\downarrow 0}\; \frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)|^2 \le 2\bigl(f(b)-q\bigr)^2 + 2\bigl(q-f(b)\bigr)^2 = 4\bigl(f(b)-q\bigr)^2 < 4\varepsilon^2 . \end{equation*}

The number \(\varepsilon > 0\) was arbitrary and the left side of the last display does not depend on \(\varepsilon\); hence that limit superior is \(0\). Since \(\frac{1}{2t}\int_{b-t}^{b+t}|f-f(b)|^2 \ge 0\) for every \(t > 0\), the corresponding limit inferior is also \(0\), and therefore

\begin{equation*} \lim_{t\downarrow 0}\frac{1}{2t}\int_{b-t}^{b+t}|f - f(b)|^2 = 0 . \end{equation*}

As \(|N| = 0\), this holds for almost every \(b \in \mathbb{R}\).

Suppose \(f : \mathbb{R}\to\mathbb{R}\) is Lebesgue measurable and

\begin{equation*} \int_{-\infty}^{x} |f| < \infty \quad \text{for every } x \in \mathbb{R} . \end{equation*}

Then \(f\chi_{(-\infty,\, x)} \in L^1(\mathbb{R})\) for every \(x\), so the function \(g : \mathbb{R}\to\mathbb{R}\) defined by

\begin{equation*} g(x) = \int_{-\infty}^{x} f \end{equation*}

is well defined and real valued. We must show that \(g\) is differentiable at \(b\) with \(g^{\prime}(b) = f(b)\) for almost every \(b \in \mathbb{R}\).

Fix \(n \in \mathbb{Z}^+\) and define \(f_n : \mathbb{R}\to\mathbb{R}\) by

\begin{equation*} f_n = f \chi_{(-\infty,\, n)} . \end{equation*}

Then \(f_n\) is Lebesgue measurable and

\begin{equation*} \int_{-\infty}^{\infty} |f_n| = \int_{-\infty}^{n} |f| < \infty , \end{equation*}

so \(f_n \in L^1(\mathbb{R})\). Define \(g_n : \mathbb{R}\to\mathbb{R}\) by \(g_n(x) = \int_{-\infty}^{x} f_n\). By the Lebesgue Differentiation Theorem in the form 4.19, applied to \(f_n\), there is a set \(N_n \subseteq \mathbb{R}\) with \(|N_n| = 0\) such that

\begin{equation*} g_n^{\prime}(b) = f_n(b) \quad \text{for every } b \in \mathbb{R}\setminus N_n . \end{equation*}

Now observe that if \(x < n\) then \(f = f_n\) on \((-\infty,\, x)\), and hence

\begin{equation*} g_n(x) = \int_{-\infty}^{x} f_n = \int_{-\infty}^{x} f = g(x) . \end{equation*}

Thus \(g\) and \(g_n\) agree on the open interval \((-\infty,\, n)\). Suppose \(b < n\). If \(|t|\) is small enough that \(b + t < n\) and \(t \ne 0\), then

\begin{equation*} \frac{g(b+t) - g(b)}{t} = \frac{g_n(b+t) - g_n(b)}{t} , \end{equation*}

because both \(b\) and \(b+t\) lie in \((-\infty,\, n)\). Hence the two difference quotients in the definition of the derivative (4.16) agree for all \(t\) near \(0\), so one of the limits exists if and only if the other does, and then they are equal. In other words: if \(b < n\) and \(g_n\) is differentiable at \(b\), then \(g\) is differentiable at \(b\) and \(g^{\prime}(b) = g_n^{\prime}(b)\). Moreover \(f_n(b) = f(b)\) for \(b < n\), directly from the definition of \(f_n\).

Combining the two previous paragraphs: if \(b \in (-\infty,\, n)\setminus N_n\), then \(g\) is differentiable at \(b\) and

\begin{equation*} g^{\prime}(b) = g_n^{\prime}(b) = f_n(b) = f(b) . \end{equation*}

Let \(N = \bigcup_{n=1}^{\infty} N_n\). By countable subadditivity of outer measure (2.8), \(|N| = 0\). Suppose \(b \in \mathbb{R}\setminus N\). Choose \(n \in \mathbb{Z}^+\) with \(n > b\). Because \(b \notin N_n\) and \(b < n\), the previous paragraph gives \(g^{\prime}(b) = f(b)\).

Hence \(g^{\prime}(b) = f(b)\) for every \(b \in \mathbb{R}\setminus N\), and thus for almost every \(b \in \mathbb{R}\). This proves that 4.19 holds under the weakened hypothesis.

Recall from 4.6 that the Hardy–Littlewood maximal function of \(f\) is

\begin{equation*} f^*(b) = \sup_{t > 0} \frac{1}{2t}\int_{b-t}^{b+t}|f| , \end{equation*}

with values in \([0,\infty]\). Note that no integrability hypothesis is assumed on \(f\), so we must handle the points near which \(|f|\) fails to be integrable; at those points the inequality will turn out to hold trivially because \(f^*\) is infinite there.

Let

\begin{equation*} \mathcal{F} = \left\{ (p,q) \; : \; p, q \in \mathbb{Q},\; p < q, \; \int_p^q |f| < \infty \right\} , \end{equation*}

which is a subset of \(\mathbb{Q}\times\mathbb{Q}\) and hence countable. Let

\begin{equation*} L = \bigcup_{(p,q) \in \mathcal{F}} (p,q) . \end{equation*}

So \(L\) is an open subset of \(\mathbb{R}\). We treat \(\mathbb{R}\setminus L\) and \(L\) separately.

Step 1: the inequality holds at every \(b \in \mathbb{R}\setminus L\). Suppose \(b \notin L\) and let \(t > 0\). If we had \(\int_{b-t}^{b+t}|f| < \infty\), then choosing rational numbers \(p, q\) with \(b - t < p < b < q < b + t\) would give \(\int_p^q |f| \le \int_{b-t}^{b+t}|f| < \infty\) (the inequality holds because \(|f| \ge 0\) and \((p,q) \subseteq (b-t,\,b+t)\)), so \((p,q) \in \mathcal{F}\) and therefore \(b \in (p,q) \subseteq L\), a contradiction. Hence

\begin{equation*} \int_{b-t}^{b+t}|f| = \infty \quad \text{for every } t > 0 , \end{equation*}

and thus \(f^*(b) = \infty\). Because \(f\) is real valued, \(|f(b)| < \infty = f^*(b)\). So the desired inequality holds at every point of \(\mathbb{R}\setminus L\), with no exceptional set at all.

Step 2: the inequality holds at almost every \(b \in L\). Fix \((p,q) \in \mathcal{F}\) and define

\begin{equation*} h = |f| \chi_{(p,\, q)} . \end{equation*}

Then \(h\) is Lebesgue measurable, \(h \ge 0\), and \(\int_{-\infty}^{\infty} h = \int_p^q |f| < \infty\), so \(h \in L^1(\mathbb{R})\). By the first version of the Lebesgue Differentiation Theorem (4.10) there is a set \(N_{p,q}\) with \(|N_{p,q}| = 0\) such that

\begin{equation*} \lim_{t\downarrow 0}\frac{1}{2t}\int_{b-t}^{b+t}|h - h(b)| = 0 \quad \text{for all } b \in \mathbb{R}\setminus N_{p,q} . \end{equation*}

Because

\begin{equation*} \left|\frac{1}{2t}\int_{b-t}^{b+t} h \;-\; h(b)\right| \le \frac{1}{2t}\int_{b-t}^{b+t}|h - h(b)| \end{equation*}

by 3.23, we get

\begin{equation*} \lim_{t\downarrow 0}\frac{1}{2t}\int_{b-t}^{b+t} h = h(b) \quad \text{for all } b \in \mathbb{R}\setminus N_{p,q} . \end{equation*}

Now suppose \(b \in (p,q)\setminus N_{p,q}\), and let \(\delta = \min\{b - p,\; q - b\} > 0\). If \(0 < t < \delta\) then \((b-t,\,b+t) \subseteq (p,q)\), so \(h = |f|\) on \((b-t,\,b+t)\); also \(h(b) = |f(b)|\). Consequently

\begin{equation*} \frac{1}{2t}\int_{b-t}^{b+t}|f| = \frac{1}{2t}\int_{b-t}^{b+t} h \longrightarrow h(b) = |f(b)| \quad \text{as } t \downarrow 0 . \end{equation*}

By the definition of \(f^*\), every \(t > 0\) satisfies \(\frac{1}{2t}\int_{b-t}^{b+t}|f| \le f^*(b)\). A limit of quantities each of which is at most \(f^*(b)\) is itself at most \(f^*(b)\), so the display above yields

\begin{equation*} |f(b)| = \lim_{t\downarrow 0}\frac{1}{2t}\int_{b-t}^{b+t}|f| \le f^*(b) . \end{equation*}

Finally, let

\begin{equation*} N = \bigcup_{(p,q)\in\mathcal{F}} N_{p,q} . \end{equation*}

Because \(\mathcal{F}\) is countable, \(|N| = 0\) by countable subadditivity of outer measure (2.8). If \(b \in \mathbb{R}\setminus N\), then either \(b \notin L\), in which case Step 1 gives \(|f(b)| \le f^*(b)\), or \(b \in L\), in which case \(b \in (p,q)\) for some \((p,q) \in \mathcal{F}\) and \(b \notin N_{p,q}\), so Step 2 gives \(|f(b)| \le f^*(b)\).

Hence \(|f(b)| \le f^*(b)\) for every \(b \in \mathbb{R}\setminus N\), and therefore for almost every \(b \in \mathbb{R}\).

Suppose \(h \in L^1(\mathbb{R})\) and \(\int_{-\infty}^{s} h = 0\) for every \(s \in \mathbb{R}\). Define \(g : \mathbb{R}\to\mathbb{R}\) by

\begin{equation*} g(x) = \int_{-\infty}^{x} h , \end{equation*}

which makes sense because \(h \in L^1(\mathbb{R})\). The hypothesis says precisely that

\begin{equation*} g(x) = 0 \quad \text{for every } x \in \mathbb{R} , \end{equation*}

so \(g\) is the zero function. In particular \(g\) is differentiable at every \(b \in \mathbb{R}\) with

\begin{equation*} g^{\prime}(b) = \lim_{t\to 0}\frac{g(b+t) - g(b)}{t} = \lim_{t\to 0}\frac{0 - 0}{t} = 0 . \end{equation*}

On the other hand, the Lebesgue Differentiation Theorem in its second version (4.19), applied to \(h \in L^1(\mathbb{R})\) and to this same function \(g\), states that

\begin{equation*} g^{\prime}(b) = h(b) \quad \text{for almost every } b \in \mathbb{R} . \end{equation*}

Comparing the two previous displays, we get \(h(b) = g^{\prime}(b) = 0\) for almost every \(b \in \mathbb{R}\). That is, \(h(s) = 0\) for almost every \(s \in \mathbb{R}\), as desired.

Recall from 4.22 that the density of a set \(E \subseteq \mathbb{R}\) at \(0\) is

\begin{equation*} \lim_{t\downarrow 0}\frac{|E \cap (-t,\, t)|}{2t} \end{equation*}

if this limit exists, and is undefined otherwise. We produce a Borel set \(E\) for which the ratio above oscillates between \(\frac{1}{5}\) and \(\frac{4}{5}\) as \(t \downarrow 0\).

Define

\begin{equation*} E = \bigcup_{k=0}^{\infty} \Bigl( \bigl[-4^{-2k},\, -4^{-2k-1}\bigr] \cup \bigl[4^{-2k-1},\, 4^{-2k}\bigr] \Bigr) . \end{equation*}

Thus \(E\) is symmetric about \(0\), and \(E \cap (0,\infty)\) consists of the closed intervals

\begin{equation*} [4^{-1},\, 4^{0}],\quad [4^{-3},\, 4^{-2}],\quad [4^{-5},\, 4^{-4}],\quad \dots , \end{equation*}

which pile up at \(0\) with gaps between them. Being a countable union of closed intervals, \(E\) is a Borel set.

These intervals are pairwise disjoint: for \(j \ge 0\) the interval indexed by \(j+1\) has right endpoint \(4^{-2j-2}\), which is strictly less than the left endpoint \(4^{-2j-1}\) of the interval indexed by \(j\). So countable additivity of Lebesgue measure applies to any subfamily of them.

We first record the basic length computation. For \(j \ge 0\),

\begin{equation*} \bigl| [4^{-2j-1},\, 4^{-2j}] \bigr| = 4^{-2j} - 4^{-2j-1} = \tfrac{3}{4}\,4^{-2j} = \tfrac{3}{4}\,16^{-j} . \end{equation*}

Case 1: \(t = t_m := 4^{-2m}\) for \(m \in \mathbb{Z}^+\). Then

\begin{equation*} E \cap (0,\, t_m) = \bigl[4^{-2m-1},\, 4^{-2m}\bigr) \cup \bigcup_{j > m} \bigl[4^{-2j-1},\, 4^{-2j}\bigr] , \end{equation*}

because the intervals of \(E\) lying in \((0,\infty)\) that meet \((0, 4^{-2m})\) are exactly \([4^{-2j-1}, 4^{-2j}]\) for \(j \ge m\) (those with \(j < m\) lie entirely in \([4^{-2m+1},\infty)\), to the right of \(t_m\)), and the one with \(j = m\) has its right endpoint at \(t_m\), which is excluded. Since a single point has measure \(0\), countable additivity gives

\begin{equation*} |E \cap (0,\, t_m)| = \sum_{j \ge m} \tfrac{3}{4}\, 16^{-j} = \tfrac{3}{4}\cdot \frac{16^{-m}}{1 - \frac{1}{16}} = \tfrac{3}{4}\cdot\frac{16}{15}\cdot 16^{-m} = \tfrac{4}{5}\, 4^{-2m} = \tfrac{4}{5}\, t_m . \end{equation*}

By the symmetry of \(E\) about \(0\) (and since \(\{0\}\) has measure \(0\)),

\begin{equation*} \frac{|E \cap (-t_m,\, t_m)|}{2t_m} = \frac{2 \cdot \frac{4}{5}t_m}{2t_m} = \frac{4}{5} . \end{equation*}

Case 2: \(t = s_m := 4^{-2m-1}\) for \(m \in \mathbb{Z}^+\). The point \(s_m\) is the left endpoint of the interval \([4^{-2m-1}, 4^{-2m}]\) of \(E\), so that interval meets \((0, s_m)\) in the empty set, and the intervals with \(j < m\) lie even further to the right. The intervals with \(j \ge m+1\) have right endpoint \(4^{-2j} \le 4^{-2m-2} < s_m\), so they lie entirely inside \((0, s_m)\). Hence \(E \cap (0, s_m)\) is exactly the disjoint union of \([4^{-2j-1}, 4^{-2j}]\) over \(j \ge m+1\), and

\begin{equation*} |E \cap (0,\, s_m)| = \sum_{j \ge m+1} \tfrac{3}{4}\,16^{-j} = \tfrac{4}{5}\,16^{-(m+1)} = \tfrac{4}{5}\cdot \frac{4^{-2m}}{16} = \tfrac{1}{5}\cdot \frac{4^{-2m}}{4} = \tfrac{1}{5}\, s_m , \end{equation*}

where the second equality uses the computation of Case 1 with \(m\) replaced by \(m+1\). Again by symmetry,

\begin{equation*} \frac{|E \cap (-s_m,\, s_m)|}{2s_m} = \frac{2\cdot\frac{1}{5}s_m}{2s_m} = \frac{1}{5} . \end{equation*}

Both \(t_m \downarrow 0\) and \(s_m \downarrow 0\) as \(m \to \infty\), yet along the first sequence the ratio \(\frac{|E\cap(-t,t)|}{2t}\) is constantly \(\frac{4}{5}\) and along the second it is constantly \(\frac{1}{5}\). Because \(\frac{4}{5} \ne \frac{1}{5}\), the limit

\begin{equation*} \lim_{t\downarrow 0}\frac{|E \cap (-t,\, t)|}{2t} \end{equation*}

does not exist. Therefore the density of the Borel set \(E\) at \(0\) is not defined.

This is consistent with the Lebesgue Density Theorem (4.24): that theorem asserts that a Lebesgue measurable set has density \(1\) at almost every one of its points and density \(0\) at almost every point of its complement, and here the single point \(0\) (a set of measure \(0\)) is exceptional.

Recall (4.22) that the density of a set \(A \subseteq \mathbb{R}\) at \(0\) is

\begin{equation*} \lim_{t \downarrow 0} \frac{|A \cap (-t, t)|}{2t}, \end{equation*}

if this limit exists.

Two remarks explain the shape of the example below; neither is needed for the proof. First, one cannot simply demand that the set occupy exactly the proportion \(\frac{1}{3}\) of every interval \((0,t)\): by 4.20 there is no Lebesgue measurable \(A \subseteq [0,1]\) with \(|A \cap [0,b]| = \frac{b}{2}\) for all \(b \in [0,1]\), and the proof of 4.20 works verbatim with \(\frac{b}{2}\) replaced by \(cb\) for any \(c \in (0,1)\), since the derivative it produces is then the constant \(c\), a value never taken by a characteristic function. Second, a set built on a geometric scale, occupying a fixed proportion of each interval \(\big[2^{-k-1}, 2^{-k}\big)\), has a ratio \(\frac{|A \cap (0,t)|}{t}\) that is unchanged under \(t \mapsto t/2\) and is not constant in \(t\), hence has no limit as \(t \downarrow 0\); that is the mechanism behind Exercise 7. The construction below instead puts blocks on the harmonic scale \(\frac{1}{k}\), with the \(k\)-th block of length \(\frac{1}{3k(k+1)}\), small compared with its distance \(\frac{1}{k+1}\) from the origin. Truncating in the middle of a block then perturbs the ratio by an amount tending to \(0\), and the limit exists.

Construction. For \(k \in \mathbb{Z}^{+}\) let

\begin{equation*} J_k = \Big[\tfrac{1}{k+1}, \; \tfrac{1}{k+1} + \tfrac{1}{3k(k+1)}\Big), \end{equation*}

which is the left third of the interval \(\big[\frac{1}{k+1}, \frac{1}{k}\big)\), because

\begin{equation*} \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}. \end{equation*}

In particular the intervals \(J_1, J_2, \ldots\) are disjoint, \(J_k \subseteq \big[\frac{1}{k+1}, \frac{1}{k}\big) \subseteq (0,1)\), and \(|J_k| = \frac{1}{3k(k+1)}\). Put

\begin{equation*} F = \bigcup_{k=1}^{\infty} J_k \quad \text{and} \quad E = F \cup (-F), \end{equation*}

where \(-F = \{-x : x \in F\}\). Each \(J_k\) is an interval, hence a Borel set, so \(F\) is a Borel set (a countable union of Borel sets), and \(-F\) is Borel because \(x \mapsto -x\) is a homeomorphism of \(\mathbb{R}\). Thus \(E\) is a Borel subset of \(\mathbb{R}\).

The density of \(E\) at \(0\) equals \(\frac{1}{3}\). Because \(F \subseteq (0,1)\) and \(-F \subseteq (-1,0)\), the point \(0\) lies in neither, so for every \(t > 0\) the set \(E \cap (-t,t)\) is the union of the two disjoint Borel sets \(F \cap (0,t)\) and \((-F) \cap (-t,0)\). Since outer measure restricted to the Borel sets is a measure (2.68), it is additive on that pair, so

\begin{equation*} |E \cap (-t,t)| = |F \cap (0,t)| + |(-F) \cap (-t, 0)| = 2\,|F \cap (0,t)|, \end{equation*}

the second equality because Lebesgue measure is invariant under \(x \mapsto -x\): reflection maps an open interval onto an open interval of the same length, so it carries any cover of a set by open intervals to a cover of the reflected set by open intervals of the same total length, and hence preserves outer measure (this is the argument used for translation invariance in 2.7). Hence

\begin{equation*} \frac{|E \cap (-t,t)|}{2t} = \frac{|F \cap (0,t)|}{t}, \end{equation*}

so it suffices to show that \(\frac{|F \cap (0,t)|}{t} \to \frac{1}{3}\) as \(t \downarrow 0\).

First compute \(|F \cap (0,t)|\) at the special values \(t = \frac{1}{n}\), where \(n \in \mathbb{Z}^{+}\). If \(k \geq n\) then \(J_k \subseteq \big[\frac{1}{k+1}, \frac{1}{k}\big) \subseteq \big(0, \frac{1}{n}\big)\), while if \(k < n\) then every element of \(J_k\) is at least \(\frac{1}{k+1} \geq \frac{1}{n}\), so \(J_k\) is disjoint from \(\big(0,\frac{1}{n}\big)\). Hence \(F \cap \big(0,\frac{1}{n}\big) = \bigcup_{k \geq n} J_k\), a disjoint union of Borel sets, and countable additivity (2.68) together with the formula for the outer measure of an interval (2.14) gives

\begin{equation*} \Big|F \cap \big(0, \tfrac{1}{n}\big)\Big| = \sum_{k=n}^{\infty} |J_k| = \frac{1}{3}\sum_{k=n}^{\infty}\Big(\frac{1}{k} - \frac{1}{k+1}\Big) = \frac{1}{3n}, \end{equation*}

the last equality by telescoping.

Now let \(t \in (0,1]\) and put \(n = \lfloor 1/t \rfloor\), so that \(n \in \mathbb{Z}^{+}\) and \(\frac{1}{n+1} < t \leq \frac{1}{n}\). Since \(\big(0,\frac{1}{n+1}\big) \subseteq (0,t) \subseteq \big(0,\frac{1}{n}\big]\) and the single point \(\frac{1}{n}\) has outer measure \(0\), monotonicity of outer measure gives

\begin{equation*} \frac{1}{3(n+1)} = \Big|F \cap \big(0,\tfrac{1}{n+1}\big)\Big| \leq |F \cap (0,t)| \leq \Big|F \cap \big(0,\tfrac{1}{n}\big)\Big| = \frac{1}{3n}. \end{equation*}

Combining this with \(\frac{1}{n+1} < t \leq \frac{1}{n}\) we obtain

\begin{equation*} \frac{n}{3(n+1)} = \frac{1/\big(3(n+1)\big)}{1/n} \leq \frac{|F \cap (0,t)|}{t} \leq \frac{1/(3n)}{1/(n+1)} = \frac{n+1}{3n}. \end{equation*}

Both outer bounds are functions of \(n\) alone, the left one increasing in \(n\) and the right one decreasing in \(n\), and both tend to \(\frac{1}{3}\) as \(n \to \infty\). Given \(\varepsilon > 0\), pick \(N \in \mathbb{Z}^{+}\) with \(\frac{N}{3(N+1)} > \frac{1}{3} - \varepsilon\) and \(\frac{N+1}{3N} < \frac{1}{3} + \varepsilon\). If \(0 < t \leq \frac{1}{N}\) then \(n = \lfloor 1/t \rfloor \geq N\), so the displayed bounds give \(\big|\frac{|F \cap (0,t)|}{t} - \frac{1}{3}\big| < \varepsilon\). Therefore

\begin{equation*} \lim_{t \downarrow 0} \frac{|F \cap (0,t)|}{t} = \frac{1}{3}, \end{equation*}

and consequently the density of the Borel set \(E\) at \(0\) equals \(\frac{1}{3}\).

To avoid a clash with the variable \(t\) appearing in the definition of density (4.22), write the given number as \(c\); thus \(c \in [0,1]\) is given and we must produce a Borel set \(E \subseteq \mathbb{R}\) whose density at \(0\) equals \(c\), meaning

\begin{equation*} \lim_{s \downarrow 0} \frac{|E \cap (-s, s)|}{2s} = c. \end{equation*}

The construction is the one used in Exercise 8, with the proportion \(\frac{1}{3}\) replaced by \(c\). For \(k \in \mathbb{Z}^{+}\) let

\begin{equation*} J_k = \Big[\tfrac{1}{k+1}, \; \tfrac{1}{k+1} + \tfrac{c}{k(k+1)}\Big). \end{equation*}

Because \(\frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)}\) and \(0 \leq c \leq 1\), the interval \(J_k\) is the left-hand fraction \(c\) of the interval \(\big[\frac{1}{k+1}, \frac{1}{k}\big)\); in particular

\begin{equation*} J_k \subseteq \Big[\tfrac{1}{k+1}, \tfrac{1}{k}\Big) \subseteq (0,1), \qquad |J_k| = \frac{c}{k(k+1)}, \end{equation*}

and the intervals \(J_1, J_2, \ldots\) are disjoint because the intervals \(\big[\frac{1}{k+1}, \frac{1}{k}\big)\) are disjoint. (If \(c = 0\) then each \(J_k\) is empty; if \(c = 1\) then \(J_k = \big[\frac{1}{k+1}, \frac{1}{k}\big)\). Both extreme cases are allowed in what follows.) Set

\begin{equation*} F = \bigcup_{k=1}^{\infty} J_k, \qquad E = F \cup (-F), \end{equation*}

where \(-F = \{-x : x \in F\}\). Each \(J_k\) is an interval and hence Borel, so \(F\) is Borel as a countable union of Borel sets; \(-F\) is Borel because \(x \mapsto -x\) is a homeomorphism of \(\mathbb{R}\) and so maps Borel sets to Borel sets. Thus \(E\) is a Borel subset of \(\mathbb{R}\).

Step 1: reduction to a one-sided statement. Since \(F \subseteq (0,1)\) and \(-F \subseteq (-1,0)\), the point \(0\) belongs to neither, so for each \(s > 0\) the sets \(F \cap (0,s)\) and \((-F) \cap (-s,0)\) are disjoint Borel sets whose union is \(E \cap (-s,s)\). Outer measure restricted to the Borel sets is a measure (2.68), hence additive on that pair. Moreover Lebesgue measure is invariant under the reflection \(x \mapsto -x\): reflection maps an open interval onto an open interval of the same length, so it carries any cover of a set by open intervals to a cover of the reflected set by open intervals of the same total length, and therefore preserves outer measure (this is the argument used for translation invariance in 2.7). Since \((-F) \cap (-s,0)\) is the reflection of \(F \cap (0,s)\), we get

\begin{equation*} |E \cap (-s,s)| = |F \cap (0,s)| + |F \cap (0,s)| = 2\,|F \cap (0,s)|, \end{equation*}

so that

\begin{equation*} \frac{|E \cap (-s,s)|}{2s} = \frac{|F \cap (0,s)|}{s}. \end{equation*}

It therefore suffices to prove

\begin{equation*} \lim_{s \downarrow 0} \frac{|F \cap (0,s)|}{s} = c. \end{equation*}

Step 2: the measure of \(F\) at the scales \(\frac{1}{n}\). Fix \(n \in \mathbb{Z}^{+}\). Since \(J_k \subseteq \big(0, \frac{1}{n}\big)\) for \(k \geq n\), while \(J_k \subseteq \big[\frac{1}{k+1}, \frac{1}{k}\big)\) is disjoint from \(\big(0, \frac{1}{n}\big)\) for \(k < n\), we have \(F \cap \big(0,\frac{1}{n}\big) = \bigcup_{k \geq n} J_k\). The \(J_k\) are disjoint Borel sets, so countable additivity of outer measure on Borel sets (2.68), together with the fact that the outer measure of an interval is its length (2.14), gives

\begin{equation*} \Big|F \cap \big(0,\tfrac{1}{n}\big)\Big| = \sum_{k=n}^{\infty} \frac{c}{k(k+1)} = c \sum_{k=n}^{\infty} \Big(\frac{1}{k} - \frac{1}{k+1}\Big) = \frac{c}{n}, \end{equation*}

the last equality by telescoping.

Step 3: squeeze at intermediate scales. Let \(s \in (0,1]\) and choose \(n \in \mathbb{Z}^{+}\) with

\begin{equation*} \frac{1}{n+1} < s \leq \frac{1}{n} \end{equation*}

(such an \(n\) exists and is unique: take \(n = \lfloor 1/s \rfloor\), which lies in \(\mathbb{Z}^{+}\) because \(1/s \geq 1\)). Then \(\big(0,\frac{1}{n+1}\big) \subseteq (0,s) \subseteq \big(0,\frac{1}{n}\big]\), and the single point \(\frac{1}{n}\) has outer measure \(0\), so monotonicity of outer measure and Step 2 give

\begin{equation*} \frac{c}{n+1} = \Big|F \cap \big(0,\tfrac{1}{n+1}\big)\Big| \leq |F \cap (0,s)| \leq \Big|F \cap \big(0,\tfrac{1}{n}\big)\Big| = \frac{c}{n}. \end{equation*}

Dividing by \(s\) and using \(\frac{1}{n+1} < s \leq \frac{1}{n}\) (so that \(\frac{1}{s} < n+1\) and \(\frac{1}{s} \geq n\)) yields

\begin{equation*} \frac{cn}{n+1} = \frac{c/(n+1)}{1/n} \leq \frac{|F \cap (0,s)|}{s} \leq \frac{c/n}{1/(n+1)} = \frac{c(n+1)}{n}. \end{equation*}

As \(s \downarrow 0\) we have \(n = \lfloor 1/s \rfloor \to \infty\), and

\begin{equation*} \frac{cn}{n+1} \to c, \qquad \frac{c(n+1)}{n} \to c. \end{equation*}

Because \(c \geq 0\), the lower bound \(\frac{cn}{n+1}\) increases with \(n\) and the upper bound \(\frac{c(n+1)}{n}\) decreases with \(n\). Hence, given \(\varepsilon > 0\), choose \(N \in \mathbb{Z}^{+}\) with \(\frac{cN}{N+1} > c - \varepsilon\) and \(\frac{c(N+1)}{N} < c + \varepsilon\); every \(s \in \big(0,\frac{1}{N}\big]\) has \(n = \lfloor 1/s \rfloor \geq N\) and therefore satisfies \(c - \varepsilon < \frac{cn}{n+1} \leq \frac{|F \cap (0,s)|}{s} \leq \frac{c(n+1)}{n} < c + \varepsilon\). Thus

\begin{equation*} \Big| \frac{|F \cap (0,s)|}{s} - c \Big| < \varepsilon \quad \text{for all } s \in \big(0, \tfrac{1}{N}\big]. \end{equation*}

Therefore \(\lim_{s \downarrow 0} \frac{|F \cap (0,s)|}{s} = c\), and by Step 1 the density of \(E\) at \(0\) equals \(c\), as required.

Remarks. The two extreme cases are consistent with the general computation: for \(c = 0\) we get \(F = \varnothing\) and \(E = \varnothing\), whose density at \(0\) is \(0\); for \(c = 1\) we get \(F = (0,1)\) and \(E = (-1,0) \cup (0,1)\), whose density at \(0\) is \(1\) (of course \(E = \mathbb{R}\) would serve equally well). Note also that no set can have density at \(0\) outside \([0,1]\), since \(0 \leq \frac{|E \cap (-s,s)|}{2s} \leq 1\) for every \(s > 0\); so this exercise shows that every conceivable value of the density is attained.

The essential point in the construction is that the length \(\frac{c}{k(k+1)}\) of the \(k\)-th block is small compared with its distance \(\frac{1}{k+1}\) from the origin: their ratio is \(\frac{c}{k}\), which tends to \(0\). That is what makes the squeeze in Step 3 close up, since truncating in the middle of a block then changes \(\frac{|F \cap (0,s)|}{s}\) by an amount tending to \(0\). Blocks placed on a geometric scale, say occupying the fraction \(c\) of each interval \(\big[2^{-k-1}, 2^{-k}\big)\), would instead give a ratio invariant under \(s \mapsto s/2\) and not constant in \(s\), hence with no limit as \(s \downarrow 0\); that is how one answers Exercise 7.

The Lebesgue Density Theorem (4.24) says that every Lebesgue measurable set \(E\) has density \(1\) at almost every element of \(E\) and density \(0\) at almost every element of \(\mathbb{R} \setminus E\). This exercise says that the exceptional sets of measure \(0\) can be removed only in the two trivial cases: a set that is its own set of density points, and whose complement is likewise, must be \(\varnothing\) or \(\mathbb{R}\).

The strategy: build from \(E\) an everywhere-differentiable function whose derivative is exactly \(\chi_E\), and then use the fact that a derivative cannot omit an intermediate value.

Step 1: an antiderivative of \(\chi_E\). Define \(g : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} g(x) = \begin{cases} \ \ \, |E \cap [0,x]| & \text{if } x \geq 0, \\[2pt] -|E \cap [x,0]| & \text{if } x < 0. \end{cases} \end{equation*}

The Lebesgue measurable sets form a \(\sigma\)-algebra on which outer measure is a measure (2.72), so each set \(E \cap [u,v]\) is Lebesgue measurable, and \(|E \cap [u,v]| \leq v - u < \infty\) by monotonicity of outer measure and 2.14. Thus \(g\) is a well-defined real-valued function. Below, every use of the word additivity means countable additivity of the measure supplied by 2.72, applied to disjoint Lebesgue measurable sets; all the sets involved have finite measure, so the subtractions performed are legitimate.

If \(u < v\), then

\begin{equation*} g(v) - g(u) = |E \cap [u,v]|. \end{equation*}

Indeed, if \(0 \leq u < v\) this is additivity applied to the disjoint measurable sets \(E \cap [0,u]\) and \(E \cap (u,v]\), whose union is \(E \cap [0,v]\), together with \(|E \cap (u,v]| = |E \cap [u,v]|\) (the single point \(u\) has measure \(0\)); if \(u < v < 0\) it is the same computation for \(E \cap [u,0] = (E \cap [u,v)) \cup (E \cap [v,0])\); and if \(u < 0 \leq v\) then \(g(v) - g(u) = |E \cap [0,v]| + |E \cap [u,0]| = |E \cap [u,v]|\), again because \(\{0\}\) is a null set.

In particular \(0 \leq g(v) - g(u) \leq v - u\) whenever \(u < v\), so \(g\) is nondecreasing and Lipschitz with constant \(1\); in particular \(g\) is continuous.

Step 2: \(g\) is differentiable at every point, with \(g^{\prime} = \chi_E\). Let \(b \in \mathbb{R}\) and let \(t > 0\). Using Step 1 twice,

\begin{equation*} \frac{g(b+t) - g(b)}{t} = \frac{|E \cap [b, b+t]|}{t}, \qquad \frac{g(b) - g(b-t)}{t} = \frac{|E \cap [b-t, b]|}{t}. \end{equation*}

Both quantities lie between \(0\) and \(1\).

First suppose \(b \in E\), so the density of \(E\) at \(b\) equals \(1\). Since \([b, b+t] \subseteq [b-t, b+t]\) and \(E\) is measurable,

\begin{equation*} 0 \leq 1 - \frac{|E \cap [b,b+t]|}{t} = \frac{\big|[b,b+t] \setminus E\big|}{t} \leq \frac{\big|[b-t,b+t] \setminus E\big|}{t} = \frac{2t - |E \cap (b-t,b+t)|}{t}, \end{equation*}

where the first equality is additivity applied to the disjoint measurable sets \(E \cap [b,b+t]\) and \([b,b+t] \setminus E\), whose union is \([b,b+t]\) (of measure \(t\) by 2.14); the middle inequality is monotonicity, since \([b,b+t] \setminus E \subseteq [b-t,b+t] \setminus E\); and the last equality is additivity applied to \(E \cap [b-t,b+t]\) and \([b-t,b+t] \setminus E\) inside \([b-t,b+t]\) (of measure \(2t\) by 2.14), together with the fact that discarding the two endpoints changes no measure, so \(|E \cap [b-t,b+t]| = |E \cap (b-t,b+t)|\). The right-hand side equals

\begin{equation*} 2\Big(1 - \frac{|E \cap (b-t,b+t)|}{2t}\Big), \end{equation*}

which tends to \(0\) as \(t \downarrow 0\) because the density of \(E\) at \(b\) is \(1\). The identical bound applies with \([b,b+t]\) replaced by \([b-t,b]\), since that interval is also contained in \([b-t,b+t]\). Hence both one-sided difference quotients displayed above tend to \(1\) as \(t \downarrow 0\); that is, \(g\) is differentiable at \(b\) and \(g^{\prime}(b) = 1\).

Now suppose \(b \in \mathbb{R} \setminus E\), so the density of \(E\) at \(b\) equals \(0\). Then

\begin{equation*} 0 \leq \frac{|E \cap [b,b+t]|}{t} = \frac{|E \cap (b,b+t)|}{t} \leq \frac{|E \cap (b-t,b+t)|}{t} = 2 \cdot \frac{|E \cap (b-t,b+t)|}{2t} \longrightarrow 0 \end{equation*}

as \(t \downarrow 0\), and the same computation applies to \(|E \cap [b-t,b]|\). Hence \(g\) is differentiable at \(b\) and \(g^{\prime}(b) = 0\).

Combining the two cases: \(g\) is differentiable at every point of \(\mathbb{R}\) and

\begin{equation*} g^{\prime}(b) = \chi_E(b) \quad \text{for every } b \in \mathbb{R}. \end{equation*}

Note that this is an assertion at every point of \(\mathbb{R}\), not merely at almost every point; that is precisely what the strong hypothesis on \(E\) buys us, and it is what makes the next step work. Compare 4.19, which for an arbitrary \(f \in L^1(\mathbb{R})\) gives only the corresponding almost-everywhere statement (and which does not apply directly here, because \(\chi_E\) need not be in \(L^1(\mathbb{R})\)).

Step 3: a derivative cannot skip the value \(\frac{1}{2}\). Suppose, for contradiction, that \(E \neq \varnothing\) and \(E \neq \mathbb{R}\). Choose \(p \in E\) and \(q \in \mathbb{R} \setminus E\).

We may assume \(p < q\). Indeed, if \(q < p\), replace \(E\) by \(-E = \{-x : x \in E\}\). Lebesgue measure is invariant under the reflection \(x \mapsto -x\), so for every \(b \in \mathbb{R}\) and \(t > 0\) we have \(|(-E) \cap (-b-t, -b+t)| = |E \cap (b-t,b+t)|\); hence the density of \(-E\) at \(-b\) equals the density of \(E\) at \(b\), and therefore \(-E\) satisfies the same hypotheses as \(E\). Moreover \(-p \in -E\) and \(-q \notin -E\) with \(-p < -q\), while \(-E = \varnothing\) exactly when \(E = \varnothing\) and \(-E = \mathbb{R}\) exactly when \(E = \mathbb{R}\). So it does no harm to assume \(p < q\).

Define \(\varphi : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} \varphi(x) = g(x) - \tfrac{1}{2} x . \end{equation*}

By Step 2, \(\varphi\) is differentiable at every point, with \(\varphi^{\prime}(x) = \chi_E(x) - \frac{1}{2}\). In particular

\begin{equation*} \varphi^{\prime}(p) = \tfrac{1}{2} > 0 \qquad \text{and} \qquad \varphi^{\prime}(q) = -\tfrac{1}{2} < 0 . \end{equation*}

Since \(g\) is continuous, so is \(\varphi\); as \([p,q]\) is compact, \(\varphi\) attains a maximum on \([p,q]\), say at \(c \in [p,q]\).

The maximum is not attained at \(p\): because

\begin{equation*} \lim_{t \downarrow 0} \frac{\varphi(p+t) - \varphi(p)}{t} = \varphi^{\prime}(p) = \tfrac{1}{2} > 0, \end{equation*}

there exists \(t \in (0, q-p)\) with \(\frac{\varphi(p+t) - \varphi(p)}{t} > 0\), hence \(\varphi(p+t) > \varphi(p)\) with \(p + t \in (p,q)\).

The maximum is not attained at \(q\): because

\begin{equation*} \lim_{t \uparrow 0} \frac{\varphi(q+t) - \varphi(q)}{t} = \varphi^{\prime}(q) = -\tfrac{1}{2} < 0, \end{equation*}

there exists \(t \in (p-q, 0)\) with \(\frac{\varphi(q+t) - \varphi(q)}{t} < 0\); since \(t < 0\), this gives \(\varphi(q+t) > \varphi(q)\) with \(q + t \in (p,q)\).

Hence \(c \in (p,q)\), an interior maximum point. Since \(\varphi\) is differentiable at \(c\) we get, considering the two one-sided difference quotients,

\begin{equation*} \varphi^{\prime}( c) = \lim_{t \downarrow 0} \frac{\varphi(c+t) - \varphi( c)}{t} \leq 0 \qquad \text{and} \qquad \varphi^{\prime}( c) = \lim_{t \uparrow 0} \frac{\varphi(c+t) - \varphi( c)}{t} \geq 0, \end{equation*}

because the numerators are \(\leq 0\) for all \(t\) with \(c + t \in [p,q]\) (as \(c\) is a maximum point of \(\varphi\) on \([p,q]\), and \(c\) is interior so both signs of small \(t\) are allowed), while \(t > 0\) in the first limit and \(t < 0\) in the second. Therefore \(\varphi^{\prime}( c) = 0\), which means

\begin{equation*} g^{\prime}( c) = \tfrac{1}{2}. \end{equation*}

But Step 2 says \(g^{\prime}( c) = \chi_E( c) \in \{0, 1\}\), a contradiction.

Consequently no such pair \(p, q\) exists, which means \(E = \varnothing\) or \(E = \mathbb{R}\).

Remarks. (a) Step 3 is just the proof of Darboux’s theorem (derivatives have the intermediate value property) specialized to the value \(\frac{1}{2}\); the point is that \(g^{\prime}\) takes the values \(1\) (at \(p\)) and \(0\) (at \(q\)) but, by Step 2, never takes any value strictly between them.

(b) A soft version of the conclusion is already visible without Darboux’s theorem: if \(|E| = 0\), then \(\frac{|E \cap (b-t,b+t)|}{2t} = 0\) for all \(b\) and all \(t > 0\), so the density of \(E\) is \(0\) at every point, so no point can be an element of \(E\) (each element of \(E\) must have density \(1\)), giving \(E = \varnothing\). Symmetrically, if \(|\mathbb{R} \setminus E| = 0\) then the density of \(E\) is \(1\) everywhere, forcing \(\mathbb{R} \setminus E = \varnothing\). The content of the exercise is to rule out the remaining possibility, that \(E\) and its complement both have positive measure, and for that the differentiability argument above does the work.

(c) In the language of point-set topology, the measurable sets each of whose points is a density point of the set form a topology on \(\mathbb{R}\), called the density topology; it is finer than the usual topology, and this exercise says that \(\mathbb{R}\) is connected in it.

Suppose \(A \subseteq X\) and \(B \subseteq Y\) are nonempty and \(A \times B \in \mathcal{S} \otimes \mathcal{T}\).

Because \(B\) is nonempty, we may choose \(b \in B\). By 5.6, the cross section \([A \times B]^b\) belongs to \(\mathcal{S}\) (this is where the hypothesis \(A \times B \in \mathcal{S} \otimes \mathcal{T}\) is used). On the other hand, Example 5.5 tells us that

\begin{equation*} [A \times B]^b = \begin{cases} A & \text{if } b \in B,\\ \varnothing & \text{if } b \notin B.\end{cases} \end{equation*}

Since \(b \in B\), we have \([A \times B]^b = A\). Hence \(A \in \mathcal{S}\).

For the reader who prefers to see the cross section computed directly: \(x \in [A \times B]^b\) means \((x, b) \in A \times B\), which holds if and only if \(x \in A\) and \(b \in B\); as \(b \in B\), this is equivalent to \(x \in A\).

Similarly, because \(A\) is nonempty we may choose \(a \in A\). By 5.6 we have \([A \times B]_a \in \mathcal{T}\), and by Example 5.5,

\begin{equation*} [A \times B]_a = \begin{cases} B & \text{if } a \in A,\\ \varnothing & \text{if } a \notin A,\end{cases} \end{equation*}

so \([A \times B]_a = B\). Hence \(B \in \mathcal{T}\).

Thus \(A \in \mathcal{S}\) and \(B \in \mathcal{T}\), as desired.

The hypothesis that \(A\) and \(B\) are nonempty cannot be dropped: if \(A\) is an arbitrary (possibly non-measurable) subset of \(X\) and \(B = \varnothing\), then \(A \times B = \varnothing \in \mathcal{S} \otimes \mathcal{T}\), even though \(A\) need not be in \(\mathcal{S}\).

For \(E \subseteq X \times X\) write

\begin{equation*} \Delta(E) = \{x \in X : (x,x) \in E\}. \end{equation*}

We use the standard two-step technique described just before 5.10: we show that all measurable rectangles have the desired property, and that the collection of sets with the desired property is a \(\sigma\)-algebra.

Let

\begin{equation*} \mathcal{E} = \{E \subseteq X \times X : \Delta(E) \in \mathcal{S}\}. \end{equation*}

Measurable rectangles belong to \(\mathcal{E}\). Indeed, suppose \(A, B \in \mathcal{S}\). Then for \(x \in X\) we have \((x,x) \in A \times B\) if and only if \(x \in A\) and \(x \in B\). Hence

\begin{equation*} \Delta(A \times B) = A \cap B, \end{equation*}

which is in \(\mathcal{S}\) because \(\sigma\)-algebras are closed under finite intersections. Thus \(A \times B \in \mathcal{E}\).

\(\mathcal{E}\) contains \(\varnothing\), since \(\Delta(\varnothing) = \varnothing \in \mathcal{S}\).

\(\mathcal{E}\) is closed under complementation. Suppose \(E \in \mathcal{E}\). For \(x \in X\) we have \((x,x) \in (X \times X) \setminus E\) if and only if \((x,x) \notin E\). Hence

\begin{equation*} \Delta\bigl((X \times X) \setminus E\bigr) = X \setminus \Delta(E), \end{equation*}

which is in \(\mathcal{S}\) because \(\Delta(E) \in \mathcal{S}\) and \(\mathcal{S}\) is closed under complementation. Thus \((X \times X) \setminus E \in \mathcal{E}\).

\(\mathcal{E}\) is closed under countable unions. Suppose \(E_1, E_2, \ldots \in \mathcal{E}\). For \(x \in X\) we have \((x,x) \in \bigcup_{k=1}^{\infty} E_k\) if and only if \((x,x) \in E_k\) for some \(k\). Hence

\begin{equation*} \Delta\Bigl(\bigcup_{k=1}^{\infty} E_k\Bigr) = \bigcup_{k=1}^{\infty} \Delta(E_k), \end{equation*}

which is in \(\mathcal{S}\) because each \(\Delta(E_k) \in \mathcal{S}\) and \(\mathcal{S}\) is closed under countable unions. Thus \(\bigcup_{k=1}^{\infty} E_k \in \mathcal{E}\).

The three previous paragraphs show that \(\mathcal{E}\) is a \(\sigma\)-algebra on \(X \times X\), and the paragraph before them shows that \(\mathcal{E}\) contains every measurable rectangle in \(\mathcal{S} \otimes \mathcal{S}\). Because \(\mathcal{S} \otimes \mathcal{S}\) is by definition the smallest \(\sigma\)-algebra on \(X \times X\) containing all the measurable rectangles, we conclude that

\begin{equation*} \mathcal{S} \otimes \mathcal{S} \subseteq \mathcal{E}. \end{equation*}

In other words, if \(E \in \mathcal{S} \otimes \mathcal{S}\), then \(\{x \in X : (x,x) \in E\} \in \mathcal{S}\), as desired.

Equivalently, one can phrase the argument as follows: the map \(\varphi : X \to X \times X\) defined by \(\varphi(x) = (x,x)\) satisfies \(\varphi^{-1}(A \times B) = A \cap B \in \mathcal{S}\) for all \(A, B \in \mathcal{S}\), and \(\{E \subseteq X \times X : \varphi^{-1}(E) \in \mathcal{S}\}\) is a \(\sigma\)-algebra (because taking inverse images commutes with complementation and unions); hence \(\varphi^{-1}(E) \in \mathcal{S}\) for every \(E \in \mathcal{S} \otimes \mathcal{S}\), and \(\varphi^{-1}(E) = \Delta(E)\).

By 2.67 there exists a set \(A \subseteq \mathbb{R}\) that is not a Borel set. Fix such an \(A\) and define

\begin{equation*} E = \{(x, x) : x \in A\}, \end{equation*}

so that \(E\) is the part of the diagonal of \(\mathbb{R} \times \mathbb{R}\) sitting over \(A\).

First we compute the cross sections. Let \(a \in \mathbb{R}\). By definition (5.3),

\begin{equation*} [E]_a = \{y \in \mathbb{R} : (a, y) \in E\}. \end{equation*}

Now \((a,y) \in E\) means \((a,y) = (x,x)\) for some \(x \in A\), which forces \(x = a = y\) and \(a \in A\). Hence

\begin{equation*} [E]_a = \begin{cases} \{a\} & \text{if } a \in A,\\ \varnothing & \text{if } a \notin A.\end{cases} \end{equation*}

In either case \([E]_a\) is a closed subset of \(\mathbb{R}\) and therefore \([E]_a \in \mathcal{B}\).

Because \(E\) is a subset of the diagonal, it is symmetric in the two coordinates in the following sense: \((x,y) \in E\) if and only if \((y,x) \in E\). Consequently the same computation gives

\begin{equation*} [E]^a = \{x \in \mathbb{R} : (x, a) \in E\} = \begin{cases} \{a\} & \text{if } a \in A,\\ \varnothing & \text{if } a \notin A,\end{cases} \end{equation*}

so \([E]^a \in \mathcal{B}\) as well. Thus every cross section of \(E\), of either kind, is a Borel subset of \(\mathbb{R}\).

Now suppose, seeking a contradiction, that \(E \in \mathcal{B} \otimes \mathcal{B}\). Apply Exercise 5A.2 with \(X = \mathbb{R}\) and \(\mathcal{S} = \mathcal{B}\): that exercise states that if \(E \in \mathcal{B} \otimes \mathcal{B}\), then

\begin{equation*} \{x \in \mathbb{R} : (x,x) \in E\} \in \mathcal{B}. \end{equation*}

But by the definition of \(E\) we have \((x,x) \in E\) if and only if \(x \in A\), and hence

\begin{equation*} \{x \in \mathbb{R} : (x,x) \in E\} = A. \end{equation*}

Thus \(A \in \mathcal{B}\), contradicting the choice of \(A\) as a non-Borel set. Therefore \(E \notin \mathcal{B} \otimes \mathcal{B}\).

This \(E\) has all cross sections Borel yet is not in \(\mathcal{B} \otimes \mathcal{B}\), which is what we wanted. In particular the converse of 5.6 is false: measurability of all the cross sections of a set does not imply measurability of the set in the product \(\sigma\)-algebra.

Define \(F, G : X \times Y \to \mathbb{R}\) by

\begin{equation*} F(x,y) = f(x) \quad\text{and}\quad G(x,y) = g(y) \end{equation*}

for \((x,y) \in X \times Y\). Then \(h = F G\), so by 2.46(a) (products of \(\mathcal{S} \otimes \mathcal{T}\)-measurable functions are \(\mathcal{S} \otimes \mathcal{T}\)-measurable) it suffices to prove that \(F\) and \(G\) are \((\mathcal{S} \otimes \mathcal{T})\)-measurable.

Suppose \(D \subseteq \mathbb{R}\) is a Borel set. Then

\begin{equation*} F^{-1}(D) = \{(x,y) \in X \times Y : f(x) \in D\} = f^{-1}(D) \times Y. \end{equation*}

To verify this equality: \((x,y) \in F^{-1}(D)\) if and only if \(F(x,y) = f(x) \in D\), which holds if and only if \(x \in f^{-1}(D)\) (with \(y \in Y\) arbitrary). Because \(f\) is \(\mathcal{S}\)-measurable, \(f^{-1}(D) \in \mathcal{S}\); also \(Y \in \mathcal{T}\). Hence \(f^{-1}(D) \times Y\) is a measurable rectangle in \(\mathcal{S} \otimes \mathcal{T}\) and thus belongs to \(\mathcal{S} \otimes \mathcal{T}\) (by 5.2). Because \(D\) was an arbitrary Borel subset of \(\mathbb{R}\), we conclude that \(F\) is \((\mathcal{S} \otimes \mathcal{T})\)-measurable.

Similarly, for a Borel set \(D \subseteq \mathbb{R}\),

\begin{equation*} G^{-1}(D) = \{(x,y) \in X \times Y : g(y) \in D\} = X \times g^{-1}(D), \end{equation*}

which is a measurable rectangle in \(\mathcal{S} \otimes \mathcal{T}\) because \(X \in \mathcal{S}\) and \(g^{-1}(D) \in \mathcal{T}\) (the latter because \(g\) is \(\mathcal{T}\)-measurable). Hence \(G\) is \((\mathcal{S} \otimes \mathcal{T})\)-measurable.

Now 2.46(a), applied to the measurable space \((X \times Y, \mathcal{S} \otimes \mathcal{T})\) and the two \((\mathcal{S} \otimes \mathcal{T})\)-measurable functions \(F\) and \(G\), shows that \(FG\) is \((\mathcal{S} \otimes \mathcal{T})\)-measurable. Because

\begin{equation*} (FG)(x,y) = F(x,y)\, G(x,y) = f(x)\, g(y) = h(x,y) \end{equation*}

for all \((x,y) \in X \times Y\), we have \(h = FG\). Hence \(h\) is \((\mathcal{S} \otimes \mathcal{T})\)-measurable, as desired.

Let \(\mathcal{A}\) denote the collection of all finite unions of intervals of \(\mathbb{R}\), where as in Example 5.11 the word interval includes open, closed, half-open, bounded, and unbounded intervals, as well as sets consisting of a single point and the empty set (note that \(\varnothing = (0,0)\) and \(\mathbb{R} = (-\infty, \infty)\) are intervals).

Call a set \(I \subseteq \mathbb{R}\) order convex if

\begin{equation*} x, z \in I \text{ and } x < y < z \implies y \in I. \end{equation*}

Step 0 below shows that the order convex subsets of \(\mathbb{R}\) are exactly the intervals; this characterization is what makes the rest of the proof work, so we prove it rather than assume it.

Step 0: a set \(I \subseteq \mathbb{R}\) is an interval if and only if \(I\) is order convex.

Every interval is order convex: each of the forms \((\alpha, \beta)\), \([\alpha, \beta]\), \([\alpha, \beta)\), \((\alpha, \beta]\) (allowing \(\alpha = -\infty\) and \(\beta = \infty\) for the open ends), along with singletons and \(\varnothing\), is defined by a pair of inequalities that any \(y\) strictly between two of its elements also satisfies.

Conversely, suppose \(I\) is order convex. If \(I = \varnothing\) then \(I = (0,0)\) is an interval. So assume \(I \ne \varnothing\) and set

\begin{equation*} \alpha = \inf I \in [-\infty, \infty), \qquad \beta = \sup I \in (-\infty, \infty], \end{equation*}

where \(\alpha = -\infty\) if \(I\) is unbounded below and \(\beta = \infty\) if \(I\) is unbounded above. We claim

\begin{equation*} (\alpha, \beta) \subseteq I \subseteq [\alpha, \beta] \cap \mathbb{R}. \end{equation*}

The right inclusion is immediate from the definitions of infimum and supremum. For the left inclusion, suppose \(\alpha < y < \beta\). Because \(y > \alpha = \inf I\), the number \(y\) is not a lower bound for \(I\), so there exists \(x \in I\) with \(x < y\). Because \(y < \beta = \sup I\), the number \(y\) is not an upper bound for \(I\), so there exists \(z \in I\) with \(y < z\). Now \(x, z \in I\) and \(x < y < z\), so order convexity gives \(y \in I\).

Hence \(I\) is obtained from \((\alpha, \beta)\) by possibly adjoining the finite endpoints among \(\alpha\) and \(\beta\) that happen to lie in \(I\), which makes \(I\) equal to \((\alpha, \beta)\), \([\alpha, \beta)\), \((\alpha, \beta]\), or \([\alpha, \beta]\) (an infinite endpoint is never adjoined, as \(I \subseteq \mathbb{R}\)). Each of these is an interval in the sense of Example 5.11; note that when \(\alpha = \beta\) the only possibility is \(I = [\alpha, \alpha] = \{\alpha\}\), a single point, which Example 5.11 explicitly counts as an interval. This proves Step 0.

From now on we use interval and order convex interchangeably, as Step 0 permits.

Step 1: \(\mathcal{A}\) is closed under finite intersections.

If \(I\) and \(J\) are intervals, then \(I \cap J\) is an interval: by Step 0 it is enough to check that \(I \cap J\) is order convex, so suppose \(x, z \in I \cap J\) and \(x < y < z\); then \(y \in I\) (because \(x, z \in I\) and \(I\) is order convex) and \(y \in J\) (because \(x, z \in J\) and \(J\) is order convex), so \(y \in I \cap J\). Now if \(I_1, \ldots, I_n\) and \(J_1, \ldots, J_m\) are intervals, then

\begin{equation*} \Bigl(\bigcup_{j=1}^{n} I_j\Bigr) \cap \Bigl(\bigcup_{k=1}^{m} J_k\Bigr) = \bigcup_{j=1}^{n} \bigcup_{k=1}^{m} (I_j \cap J_k), \end{equation*}

which is a finite union of intervals and therefore is in \(\mathcal{A}\). Thus \(\mathcal{A}\) is closed under the intersection of two of its elements, and hence, by induction, under the intersection of finitely many of its elements.

Step 2: the complement of an interval is the union of two intervals.

Let \(I\) be an interval and define

\begin{equation*} L = \{x \in \mathbb{R} : x < y \text{ for every } y \in I\}, \qquad U = \{x \in \mathbb{R} : x > y \text{ for every } y \in I\}. \end{equation*}

We claim \(\mathbb{R} \setminus I = L \cup U\).

For the inclusion \(L \cup U \subseteq \mathbb{R} \setminus I\): if \(x \in L\) and we had \(x \in I\), then taking \(y = x\) in the definition of \(L\) would give \(x < x\), a contradiction; hence \(x \notin I\). The same argument applies to \(U\).

For the inclusion \(\mathbb{R} \setminus I \subseteq L \cup U\): suppose \(x \notin I\) but \(x \notin L\) and \(x \notin U\). From \(x \notin L\) we get some \(y \in I\) with \(y \le x\); since \(x \notin I\) we have \(y \ne x\), so \(y < x\). From \(x \notin U\) we get some \(z \in I\) with \(z \ge x\), and again \(z \ne x\), so \(x < z\). Thus \(y, z \in I\) and \(y < x < z\), which by the order convexity of \(I\) forces \(x \in I\), a contradiction. Hence \(x \in L \cup U\).

Finally, \(L\) and \(U\) are intervals. The set \(L\) is closed downward: if \(x < x^{\prime}\) and \(x^{\prime} \in L\), then \(x < x^{\prime} < y\) for every \(y \in I\), so \(x \in L\). A downward closed set is order convex, because if \(x, z \in L\) and \(x < y < z\), then \(y < z\) and \(z \in L\) give \(y \in L\); hence \(L\) is an interval by Step 0. Symmetrically, \(U\) is closed upward, hence order convex, hence an interval. (Concretely, if \(I\) is nonempty then \(L\) is \((-\infty, \inf I)\) or \((-\infty, \inf I]\) and \(U\) is \((\sup I, \infty)\) or \([\sup I, \infty)\), where these are interpreted as \(\varnothing\) when \(\inf I = -\infty\) or \(\sup I = \infty\); if \(I = \varnothing\) then \(L = U = \mathbb{R}\).)

Step 3: \(\mathcal{A}\) is closed under complementation.

Suppose \(E \in \mathcal{A}\), say \(E = I_1 \cup \cdots \cup I_n\) with each \(I_j\) an interval. By De Morgan’s Laws,

\begin{equation*} \mathbb{R} \setminus E = \bigcap_{j=1}^{n} (\mathbb{R} \setminus I_j). \end{equation*}

By Step 2, each \(\mathbb{R} \setminus I_j\) is a union of two intervals and hence is in \(\mathcal{A}\). By Step 1, the intersection of these finitely many elements of \(\mathcal{A}\) is in \(\mathcal{A}\). Hence \(\mathbb{R} \setminus E \in \mathcal{A}\).

Because \(\mathcal{A}\) is obviously closed under finite unions and contains \(\varnothing\), and we have now shown that it is closed under complementation, \(\mathcal{A}\) is an algebra on \(\mathbb{R}\), as asserted in Example 5.11.

Let \(\mathcal{A}\) denote the collection of all countable unions of intervals of \(\mathbb{R}\) (as in Example 5.11, single-point sets count as intervals). We exhibit a set in \(\mathcal{A}\) whose complement is not in \(\mathcal{A}\).

The set \(\mathbb{Q}\) of rational numbers is in \(\mathcal{A}\). Indeed, \(\mathbb{Q}\) is countable, and

\begin{equation*} \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \{q\}, \end{equation*}

which exhibits \(\mathbb{Q}\) as a countable union of the intervals \(\{q\}\) (each a set consisting of a single point).

We claim that \(\mathbb{R} \setminus \mathbb{Q}\), the set of irrational numbers, is not in \(\mathcal{A}\).

First observe that if \(I\) is an interval with \(I \subseteq \mathbb{R} \setminus \mathbb{Q}\), then \(I\) contains at most one point. To see this, suppose \(x, z \in I\) with \(x < z\). Because \(\mathbb{Q}\) is dense in \(\mathbb{R}\), there exists \(q \in \mathbb{Q}\) with \(x < q < z\). Every interval \(I\) has the property that \(x, z \in I\) and \(x < y < z\) imply \(y \in I\), because each of the possible forms \((\alpha, \beta)\), \([\alpha, \beta)\), \((\alpha, \beta]\), \([\alpha, \beta]\) (with \(\alpha = -\infty\) or \(\beta = \infty\) allowed at an open end), together with singletons and \(\varnothing\), is cut out by inequalities that \(y\) inherits from \(x\) and \(z\). Applying this with \(y = q\) gives \(q \in I\), contradicting \(I \cap \mathbb{Q} = \varnothing\). Hence \(I\) cannot contain two distinct points.

Now suppose, seeking a contradiction, that \(\mathbb{R} \setminus \mathbb{Q} \in \mathcal{A}\), say

\begin{equation*} \mathbb{R} \setminus \mathbb{Q} = \bigcup_{k=1}^{\infty} I_k \end{equation*}

with each \(I_k\) an interval (a countable union can always be written as a union indexed by \(\mathbb{Z}^{+}\), repeating terms or using \(\varnothing\) if the index set is finite). Each \(I_k\) satisfies \(I_k \subseteq \mathbb{R} \setminus \mathbb{Q}\), so by the previous paragraph each \(I_k\) has at most one element. Therefore \(\bigcup_{k=1}^{\infty} I_k\) is a countable union of sets each having at most one element, and hence is a countable set.

But \(\mathbb{R} \setminus \mathbb{Q}\) is uncountable: if it were countable, then \(\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})\) would be the union of two countable sets and thus countable, contradicting the uncountability of \(\mathbb{R}\).

This contradiction shows that \(\mathbb{R} \setminus \mathbb{Q} \notin \mathcal{A}\). Since \(\mathbb{Q} \in \mathcal{A}\), the collection \(\mathcal{A}\) is not closed under complementation, as asserted in Example 5.12. In particular \(\mathcal{A}\) is neither an algebra nor a \(\sigma\)-algebra on \(\mathbb{R}\), even though it is closed under countable unions.

Recall from 5.10 that \(\mathcal{A}\) is an algebra on \(W\) if \(\varnothing \in \mathcal{A}\), and \(W \setminus E \in \mathcal{A}\) whenever \(E \in \mathcal{A}\), and \(E \cup F \in \mathcal{A}\) whenever \(E, F \in \mathcal{A}\). As usual, closure under finite intersections means that \(E_1 \cap \cdots \cap E_n \in \mathcal{A}\) whenever \(n \in \mathbb{Z}^{+}\) and \(E_1, \ldots, E_n \in \mathcal{A}\); by induction this is equivalent to closure under the intersection of two elements of \(\mathcal{A}\).

First suppose \(\mathcal{A}\) is an algebra on \(W\). Then \(\mathcal{A}\) is closed under complementation, by the second bullet point of 5.10. To see that \(\mathcal{A}\) is closed under finite intersections, suppose \(E, F \in \mathcal{A}\). Then \(W \setminus E \in \mathcal{A}\) and \(W \setminus F \in \mathcal{A}\) (closure under complementation), hence \((W \setminus E) \cup (W \setminus F) \in \mathcal{A}\) (closure under finite unions), hence, using De Morgan’s Laws,

\begin{equation*} E \cap F = W \setminus \bigl((W \setminus E) \cup (W \setminus F)\bigr) \in \mathcal{A} \end{equation*}

(closure under complementation again). Induction now gives \(E_1 \cap \cdots \cap E_n \in \mathcal{A}\) for all \(E_1, \ldots, E_n \in \mathcal{A}\): the case \(n = 1\) is trivial, and if \(E_1 \cap \cdots \cap E_{n-1} \in \mathcal{A}\) then intersecting this set with \(E_n\) stays in \(\mathcal{A}\) by what we just proved. Thus \(\mathcal{A}\) is closed under finite intersections and under complementation. (Note that this direction does not use the hypothesis that \(\mathcal{A}\) is nonempty, which is automatic here because \(\varnothing \in \mathcal{A}\).)

Conversely, suppose \(\mathcal{A}\) is nonempty and closed under finite intersections and under complementation. We verify the three conditions in 5.10.

Because \(\mathcal{A}\) is nonempty, there exists some \(E \in \mathcal{A}\). Closure under complementation gives \(W \setminus E \in \mathcal{A}\), and then closure under finite intersections gives

\begin{equation*} \varnothing = E \cap (W \setminus E) \in \mathcal{A}. \end{equation*}

Thus the first condition of 5.10 holds. (This is exactly where the hypothesis that \(\mathcal{A}\) is nonempty is needed: the empty collection is closed under complementation and under finite intersections but is not an algebra.)

The second condition of 5.10, closure under complementation, holds by hypothesis.

For the third condition, suppose \(E, F \in \mathcal{A}\). Then \(W \setminus E \in \mathcal{A}\) and \(W \setminus F \in \mathcal{A}\), so \((W \setminus E) \cap (W \setminus F) \in \mathcal{A}\), so by De Morgan’s Laws

\begin{equation*} E \cup F = W \setminus \bigl((W \setminus E) \cap (W \setminus F)\bigr) \in \mathcal{A}. \end{equation*}

Thus \(\mathcal{A}\) is closed under the union of two of its elements, and hence (by induction, as above) under finite unions.

The three conditions of 5.10 are satisfied, so \(\mathcal{A}\) is an algebra on \(W\). This completes the proof of both implications.

We prove (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a).

(a) \(\Rightarrow\) (b). Suppose \(\mu\) is \(\sigma\)-finite. By the definition of \(\sigma\)-finite (5.18), there is a sequence \(E_1, E_2, \ldots\) of sets in \(\mathcal{S}\) with

\begin{equation*} X = \bigcup_{k=1}^{\infty} E_k \quad\text{and}\quad \mu(E_k) < \infty \text{ for every } k \in \mathbf{Z}^+ . \end{equation*}

For \(k \in \mathbf{Z}^+\) define

\begin{equation*} X_k = E_1 \cup E_2 \cup \cdots \cup E_k . \end{equation*}

Each \(X_k\) is in \(\mathcal{S}\) because a \(\sigma\)-algebra is closed under finite unions, and clearly \(X_1 \subseteq X_2 \subseteq \cdots\) is an increasing sequence. Furthermore

\begin{equation*} \bigcup_{k=1}^{\infty} X_k = \bigcup_{k=1}^{\infty} E_k = X , \end{equation*}

because each \(E_k \subseteq X_k\) and each \(X_k \subseteq \bigcup_{j=1}^{\infty} E_j\). Finally, finite subadditivity of a measure (a consequence of countable subadditivity 2.58, applied to the sequence \(E_1, \ldots, E_k, \varnothing, \varnothing, \ldots\)) gives

\begin{equation*} \mu(X_k) \le \mu(E_1) + \cdots + \mu(E_k) < \infty \end{equation*}

for every \(k \in \mathbf{Z}^+\), since each of the finitely many summands is finite. Thus (b) holds.

(b) \(\Rightarrow\) (c). Suppose \(X_1 \subseteq X_2 \subseteq \cdots\) is an increasing sequence in \(\mathcal{S}\) with \(X = \bigcup_{k=1}^{\infty} X_k\) and \(\mu(X_k) < \infty\) for each \(k\). Define

\begin{equation*} Y_1 = X_1 \quad\text{and}\quad Y_k = X_k \setminus X_{k-1} \text{ for } k \ge 2 . \end{equation*}

Each \(Y_k\) is in \(\mathcal{S}\), because a \(\sigma\)-algebra is closed under complementation and finite intersections and \(X_k \setminus X_{k-1} = X_k \cap (X \setminus X_{k-1})\).

The sequence \(Y_1, Y_2, \ldots\) is disjoint: if \(j < k\), then \(Y_j \subseteq X_j \subseteq X_{k-1}\) (using that the sequence \(X_1 \subseteq X_2 \subseteq \cdots\) is increasing and \(j \le k-1\)), while \(Y_k \cap X_{k-1} = \varnothing\); hence \(Y_j \cap Y_k = \varnothing\).

The union of the \(Y_k\) is \(X\): certainly \(\bigcup_{k=1}^{\infty} Y_k \subseteq \bigcup_{k=1}^{\infty} X_k = X\). Conversely, if \(x \in X\), then \(x \in X_k\) for some \(k \in \mathbf{Z}^+\); taking \(m\) to be the smallest such \(k\), we have \(x \in X_m\) and (if \(m \ge 2\)) \(x \notin X_{m-1}\), so \(x \in Y_m\). Thus \(X = \bigcup_{k=1}^{\infty} Y_k\).

Finally \(Y_k \subseteq X_k\), so \(\mu(Y_k) \le \mu(X_k) < \infty\) by 2.57(a). Thus the disjoint sequence \(Y_1, Y_2, \ldots\) witnesses (c).

(c) \(\Rightarrow\) (a). Suppose \(X_1, X_2, \ldots\) is a disjoint sequence of sets in \(\mathcal{S}\) with \(X = \bigcup_{k=1}^{\infty} X_k\) and \(\mu(X_k) < \infty\) for every \(k \in \mathbf{Z}^+\). Then in particular \(X_1, X_2, \ldots\) is a sequence of sets in \(\mathcal{S}\) whose union is \(X\) and each of which has finite measure; disjointness is simply not used. This is exactly the definition (5.18) of \(\mu\) being \(\sigma\)-finite. Thus (a) holds.

Having shown (a) \(\Rightarrow\) (b) \(\Rightarrow\) (c) \(\Rightarrow\) (a), we conclude that (a), (b), and (c) are equivalent.

Say \(\mu\) is a measure on the measurable space \((X, \mathcal{S})\) and \(\nu\) is a measure on the measurable space \((Y, \mathcal{T})\), both \(\sigma\)-finite; then \(\mu \times \nu\) is defined on \(\mathcal{S} \otimes \mathcal{T}\) by 5.25, and it is a measure by 5.27. So only \(\sigma\)-finiteness remains to be proved.

Because \(\mu\) is \(\sigma\)-finite, there is a sequence \(A_1, A_2, \ldots\) of sets in \(\mathcal{S}\) with

\begin{equation*} X = \bigcup_{j=1}^{\infty} A_j \quad\text{and}\quad \mu(A_j) < \infty \text{ for every } j \in \mathbf{Z}^+ . \end{equation*}

Because \(\nu\) is \(\sigma\)-finite, there is a sequence \(B_1, B_2, \ldots\) of sets in \(\mathcal{T}\) with

\begin{equation*} Y = \bigcup_{k=1}^{\infty} B_k \quad\text{and}\quad \nu(B_k) < \infty \text{ for every } k \in \mathbf{Z}^+ . \end{equation*}

Consider the family of measurable rectangles

\begin{equation*} \{ A_j \times B_k : j, k \in \mathbf{Z}^+ \} . \end{equation*}

Each \(A_j \times B_k\) belongs to \(\mathcal{S} \otimes \mathcal{T}\), being a measurable rectangle (5.2). The family is indexed by \(\mathbf{Z}^+ \times \mathbf{Z}^+\), which is a countable set, so we may list these sets as a single sequence \(E_1, E_2, \ldots\) of sets in \(\mathcal{S} \otimes \mathcal{T}\).

The union of this family is all of \(X \times Y\). Indeed, each \(A_j \times B_k \subseteq X \times Y\); conversely, if \((x, y) \in X \times Y\), then \(x \in A_j\) for some \(j \in \mathbf{Z}^+\) and \(y \in B_k\) for some \(k \in \mathbf{Z}^+\), so \((x, y) \in A_j \times B_k\). Hence

\begin{equation*} X \times Y = \bigcup_{j=1}^{\infty} \bigcup_{k=1}^{\infty} (A_j \times B_k) . \end{equation*}

Finally, by 5.26 (product measure of a measurable rectangle is the product of the measures),

\begin{equation*} (\mu \times \nu)(A_j \times B_k) = \mu(A_j)\, \nu(B_k) < \infty \end{equation*}

for all \(j, k \in \mathbf{Z}^+\), because both factors are finite real numbers.

Thus \(X \times Y\) is the union of the countable family \(E_1, E_2, \ldots\) of sets in \(\mathcal{S} \otimes \mathcal{T}\), each of finite \(\mu \times \nu\) measure. By the definition of \(\sigma\)-finite (5.18), the measure \(\mu \times \nu\) is \(\sigma\)-finite.

Throughout, \(\mu \times \nu\) is defined on \(\mathcal{S} \otimes \mathcal{T}\) by 5.25 (legitimate because both measure spaces are \(\sigma\)-finite) and is a measure by 5.27; thus the standard properties 2.57, 2.59, and 2.60 apply to \(\mu \times \nu\) as well as to \(\omega\).

Because \(\mu\) is \(\sigma\)-finite, Exercise 5A.8 provides an increasing sequence \(X_1 \subseteq X_2 \subseteq \cdots\) of sets in \(\mathcal{S}\) with \(X = \bigcup_{n=1}^{\infty} X_n\) and \(\mu(X_n) < \infty\) for every \(n \in \mathbf{Z}^+\). Similarly, there is an increasing sequence \(Y_1 \subseteq Y_2 \subseteq \cdots\) of sets in \(\mathcal{T}\) with \(Y = \bigcup_{n=1}^{\infty} Y_n\) and \(\nu(Y_n) < \infty\) for every \(n \in \mathbf{Z}^+\). For \(n \in \mathbf{Z}^+\) set

\begin{equation*} W_n = X_n \times Y_n . \end{equation*}

Then \(W_n\) is a measurable rectangle in \(\mathcal{S} \otimes \mathcal{T}\), and \(W_1 \subseteq W_2 \subseteq \cdots\) is an increasing sequence because both defining sequences are increasing. Moreover \(\bigcup_{n=1}^{\infty} W_n = X \times Y\): if \((x,y) \in X \times Y\), then \(x \in X_j\) and \(y \in Y_k\) for some \(j, k \in \mathbf{Z}^+\), and hence \((x,y) \in W_n\) with \(n = \max\{j, k\}\). Finally, by hypothesis and by 5.26,

\begin{equation*} \omega(W_n) = \mu(X_n)\, \nu(Y_n) = (\mu \times \nu)(W_n) < \infty \end{equation*}

for every \(n \in \mathbf{Z}^+\).

Fix \(n \in \mathbf{Z}^+\) and define

\begin{equation*} \mathcal{M}_n = \{ E \in \mathcal{S} \otimes \mathcal{T} : \omega(E \cap W_n) = (\mu \times \nu)(E \cap W_n) \} . \end{equation*}

We show that \(\mathcal{M}_n = \mathcal{S} \otimes \mathcal{T}\) by means of the Monotone Class Theorem (5.17).

Step 1: \(\mathcal{M}_n\) contains every measurable rectangle. Suppose \(A \in \mathcal{S}\) and \(B \in \mathcal{T}\). Then

\begin{equation*} (A \times B) \cap W_n = (A \times B) \cap (X_n \times Y_n) = (A \cap X_n) \times (B \cap Y_n), \end{equation*}

which is again a measurable rectangle in \(\mathcal{S} \otimes \mathcal{T}\). Hence, using the hypothesis on \(\omega\) and then 5.26,

\begin{equation*} \omega\bigl((A \times B) \cap W_n\bigr) = \mu(A \cap X_n)\, \nu(B \cap Y_n) = (\mu \times \nu)\bigl((A \times B) \cap W_n\bigr) . \end{equation*}

Thus \(A \times B \in \mathcal{M}_n\).

Step 2: \(\mathcal{M}_n\) contains the algebra \(\mathcal{A}\) of finite unions of measurable rectangles. Let \(\mathcal{A}\) denote the collection of finite unions of measurable rectangles in \(\mathcal{S} \otimes \mathcal{T}\); by 5.13(a) this is an algebra on \(X \times Y\). Suppose \(E \in \mathcal{A}\). By 5.13(b) we may write \(E = E_1 \cup \cdots \cup E_m\) where \(E_1, \ldots, E_m\) are disjoint measurable rectangles in \(\mathcal{S} \otimes \mathcal{T}\). The sets \(E_1 \cap W_n, \ldots, E_m \cap W_n\) are then disjoint sets in \(\mathcal{S} \otimes \mathcal{T}\) whose union is \(E \cap W_n\), so finite additivity of the measures \(\omega\) and \(\mu \times \nu\) (a measure is countably additive, and finite additivity follows by padding a finite disjoint list with copies of \(\varnothing\)) together with Step 1 gives

\begin{equation*} \omega(E \cap W_n) = \sum_{i=1}^{m} \omega(E_i \cap W_n) = \sum_{i=1}^{m} (\mu \times \nu)(E_i \cap W_n) = (\mu \times \nu)(E \cap W_n) . \end{equation*}

Thus \(E \in \mathcal{M}_n\), and hence \(\mathcal{A} \subseteq \mathcal{M}_n\).

Step 3: \(\mathcal{M}_n\) is a monotone class on \(X \times Y\). First suppose \(E_1 \subseteq E_2 \subseteq \cdots\) is an increasing sequence of sets in \(\mathcal{M}_n\), and let \(E = \bigcup_{k=1}^{\infty} E_k\), which is in \(\mathcal{S} \otimes \mathcal{T}\). Then \(E_1 \cap W_n \subseteq E_2 \cap W_n \subseteq \cdots\) is an increasing sequence with union \(E \cap W_n\), so 2.59, applied to \(\omega\) and then to \(\mu \times \nu\), gives

\begin{equation*} \omega(E \cap W_n) = \lim_{k \to \infty} \omega(E_k \cap W_n) = \lim_{k \to \infty} (\mu \times \nu)(E_k \cap W_n) = (\mu \times \nu)(E \cap W_n) . \end{equation*}

Hence \(E \in \mathcal{M}_n\).

Now suppose \(E_1 \supseteq E_2 \supseteq \cdots\) is a decreasing sequence of sets in \(\mathcal{M}_n\), and let \(E = \bigcap_{k=1}^{\infty} E_k \in \mathcal{S} \otimes \mathcal{T}\). Then \(E_1 \cap W_n \supseteq E_2 \cap W_n \supseteq \cdots\) is a decreasing sequence with intersection \(E \cap W_n\). Here is where the truncation to \(W_n\) matters: by 2.57(a),

\begin{equation*} \omega(E_1 \cap W_n) \le \omega(W_n) < \infty \quad\text{and}\quad (\mu \times \nu)(E_1 \cap W_n) \le (\mu \times \nu)(W_n) < \infty , \end{equation*}

so 2.60 applies to both measures and yields

\begin{equation*} \omega(E \cap W_n) = \lim_{k \to \infty} \omega(E_k \cap W_n) = \lim_{k \to \infty} (\mu \times \nu)(E_k \cap W_n) = (\mu \times \nu)(E \cap W_n) . \end{equation*}

Hence \(E \in \mathcal{M}_n\). Thus \(\mathcal{M}_n\) is a monotone class.

Step 4: conclusion for fixed \(n\). By Steps 2 and 3, \(\mathcal{M}_n\) is a monotone class containing the algebra \(\mathcal{A}\), so \(\mathcal{M}_n\) contains the smallest monotone class containing \(\mathcal{A}\), which by the Monotone Class Theorem (5.17) equals the smallest \(\sigma\)-algebra containing \(\mathcal{A}\). That smallest \(\sigma\)-algebra is \(\mathcal{S} \otimes \mathcal{T}\): indeed \(\mathcal{A}\) contains all measurable rectangles, so any \(\sigma\)-algebra containing \(\mathcal{A}\) contains \(\mathcal{S} \otimes \mathcal{T}\) by 5.2, while conversely \(\mathcal{A} \subseteq \mathcal{S} \otimes \mathcal{T}\) since \(\mathcal{S} \otimes \mathcal{T}\) is a \(\sigma\)-algebra containing the measurable rectangles. Because also \(\mathcal{M}_n \subseteq \mathcal{S} \otimes \mathcal{T}\) by definition, we conclude

\begin{equation*} \mathcal{M}_n = \mathcal{S} \otimes \mathcal{T} . \end{equation*}

In other words, \(\omega(E \cap W_n) = (\mu \times \nu)(E \cap W_n)\) for every \(E \in \mathcal{S} \otimes \mathcal{T}\) and every \(n \in \mathbf{Z}^+\).

Step 5: removing the truncation. Suppose \(E \in \mathcal{S} \otimes \mathcal{T}\). Then \(E \cap W_1 \subseteq E \cap W_2 \subseteq \cdots\) is an increasing sequence of sets in \(\mathcal{S} \otimes \mathcal{T}\) whose union is

\begin{equation*} \bigcup_{n=1}^{\infty} (E \cap W_n) = E \cap \Bigl( \bigcup_{n=1}^{\infty} W_n \Bigr) = E \cap (X \times Y) = E . \end{equation*}

Applying 2.59 to \(\omega\) and to \(\mu \times \nu\), and using Step 4,

\begin{equation*} \omega(E) = \lim_{n \to \infty} \omega(E \cap W_n) = \lim_{n \to \infty} (\mu \times \nu)(E \cap W_n) = (\mu \times \nu)(E) . \end{equation*}

Because \(E \in \mathcal{S} \otimes \mathcal{T}\) was arbitrary, \(\omega = \mu \times \nu\), as desired.

Throughout, let

\begin{equation*} f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}\quad\text{for }(x,y)\in[0,1]^2\setminus\{(0,0)\}, \qquad f(0,0)=0 . \end{equation*}

(The value assigned at the origin is irrelevant for everything below.) Note the antisymmetry \(f(y,x)=-f(x,y)\).

(a) The key computation is the pair of antiderivatives

\begin{equation*} \frac{\partial}{\partial y}\Bigl(\frac{y}{x^2+y^2}\Bigr) =\frac{(x^2+y^2)-y\cdot 2y}{(x^2+y^2)^2}=\frac{x^2-y^2}{(x^2+y^2)^2}, \qquad \frac{\partial}{\partial x}\Bigl(\frac{-x}{x^2+y^2}\Bigr) =\frac{x^2-y^2}{(x^2+y^2)^2}, \end{equation*}

valid at every point where \((x,y)\neq(0,0)\).

Fix \(x\in(0,1]\). Then \(y\mapsto f(x,y)\) is continuous on \([0,1]\) (the denominator is at least \(x^4>0\) there), hence Riemann integrable, and by 3.34 its Lebesgue integral equals its Riemann integral. The Fundamental Theorem of Calculus and the first antiderivative above give

\begin{equation*} \int_{[0,1]}f(x,y)\,d\lambda(y)=\left[\frac{y}{x^2+y^2}\right]_{y=0}^{y=1}=\frac{1}{1+x^2}. \end{equation*}

For \(x=0\) we have \(f(0,y)=-1/y^2\) for \(y\in(0,1]\) and \(f(0,0)=0\); thus \(f(0,\cdot)\) is Borel measurable, with \(f(0,\cdot)^+=0\) and \(f(0,\cdot)^-=y^{-2}\chi_{(0,1]}\). For each \(k\in\mathbb{Z}^+\) the integrand \(y^{-2}\) is continuous on \([1/k,1]\), so 3.34 gives

\begin{equation*} \int_{[0,1]}f(0,\cdot)^-\,d\lambda\ \ge\ \int_{[1/k,1]}\frac{dy}{y^2}=k-1 , \end{equation*}

and letting \(k\to\infty\) shows \(\int_{[0,1]}f(0,\cdot)^-\,d\lambda=\infty\). Since \(\int_{[0,1]}f(0,\cdot)^+\,d\lambda=0\) is finite, the integral is defined by 3.18 and equals \(0-\infty=-\infty\).

Thus the inner integral, as a function of \(x\), is the function \(g:[0,1]\to[-\infty,\infty]\) given by \(g(0)=-\infty\) and \(g(x)=1/(1+x^2)\) for \(x>0\). This \(g\) is \(\mathcal{B}\)-measurable, since for \(a\in\mathbb{R}\) the set \(g^{-1}((a,\infty])\) equals \((0,1]\) intersected with an interval (the map \(x\mapsto 1/(1+x^2)\) is decreasing on \((0,1]\), and \(0\) never belongs to this set because \(g(0)=-\infty\)); so 2.52 applies. Moreover \(g^-=\infty\cdot\chi_{\{0\}}\), and \(\int_{[0,1]}g^-\,d\lambda=\infty\cdot\lambda(\{0\})=\infty\cdot 0=0\), so 3.18 gives \(\int_{[0,1]}g\,d\lambda=\int_{[0,1]}g^+\,d\lambda\). Now \(g^+\) differs from the continuous function \(x\mapsto 1/(1+x^2)\) only at the point \(0\); since the integral of a nonnegative measurable function over a set of measure \(0\) is \(0\), additivity of integration over the decomposition \([0,1]=\{0\}\cup(0,1]\) shows that \(g^+\) and that continuous function have the same integral over \([0,1]\). Hence, using 3.34 again,

\begin{equation*} \int_{[0,1]}g\,d\lambda=\int_{[0,1]}g^+\,d\lambda=\int_0^1\frac{dx}{1+x^2}=\arctan 1-\arctan 0=\frac{\pi}{4}. \end{equation*}

This proves the first equation:

\begin{equation*} \int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}\,d\lambda(y)\,d\lambda(x)=\frac{\pi}{4}. \end{equation*}

For the second equation, fix \(y\in(0,1]\). Again \(x\mapsto f(x,y)\) is continuous on \([0,1]\), and the second antiderivative above gives

\begin{equation*} \int_{[0,1]}f(x,y)\,d\lambda(x)=\left[\frac{-x}{x^2+y^2}\right]_{x=0}^{x=1}=-\frac{1}{1+y^2}. \end{equation*}

For \(y=0\) we get \(f(x,0)=1/x^2\) for \(x\in(0,1]\), so by the same estimate as before this inner integral equals \(+\infty\). Hence the inner integral, as a function of \(y\), is \(h:[0,1]\to[-\infty,\infty]\) with \(h(0)=\infty\) and \(h(y)=-1/(1+y^2)\) for \(y>0\); as above \(h\) is \(\mathcal{B}\)-measurable, \(h^+=\infty\cdot\chi_{\{0\}}\) has integral \(\infty\cdot 0=0\), and \(h^-\) differs only at the point \(0\) from the continuous function \(y\mapsto 1/(1+y^2)\). So 3.18 and 3.34 give

\begin{equation*} \int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}\,d\lambda(x)\,d\lambda(y)=0-\int_0^1\frac{dy}{1+y^2}=-\frac{\pi}{4}. \end{equation*}

(Alternatively this follows from the first equation and the antisymmetry \(f(y,x)=-f(x,y)\), which interchanges the roles of the two variables.)

(b) First note that \(f\) is \(\mathcal{B}\otimes\mathcal{B}\)-measurable on \([0,1]^2\), where \(\mathcal{B}\) denotes the Borel subsets of \([0,1]\). Indeed, the coordinate functions \(\pi_1(x,y)=x\) and \(\pi_2(x,y)=y\) are \(\mathcal{B}\otimes\mathcal{B}\)-measurable, because for each \(a\in\mathbb{R}\) the sets

\begin{equation*} \pi_1^{-1}\bigl((a,\infty)\bigr)=\bigl((a,\infty)\cap[0,1]\bigr)\times[0,1], \qquad \pi_2^{-1}\bigl((a,\infty)\bigr)=[0,1]\times\bigl((a,\infty)\cap[0,1]\bigr) \end{equation*}

are measurable rectangles, so 2.39 applies. Hence by 2.46 the functions \((x,y)\mapsto x^2-y^2\) and \((x,y)\mapsto(x^2+y^2)^2\) are \(\mathcal{B}\otimes\mathcal{B}\)-measurable. The denominator vanishes at the origin, so to stay within 2.46(b) define, for \(k\in\mathbb{Z}^+\),

\begin{equation*} f_k(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2+\frac{1}{k}} . \end{equation*}

The denominator here is at least \(1/k>0\) everywhere, so each \(f_k\) is a real-valued \(\mathcal{B}\otimes\mathcal{B}\)-measurable function by 2.46. If \((x,y)\neq(0,0)\) then \((x^2+y^2)^2>0\) and \(f_k(x,y)\to f(x,y)\); and \(f_k(0,0)=0=f(0,0)\) for every \(k\). Thus \(f_k\to f\) pointwise on all of \([0,1]^2\), so \(f\) is \(\mathcal{B}\otimes\mathcal{B}\)-measurable by 2.48.

Tonelli’s Theorem (5.28) is not contradicted, because it requires the function being integrated to be nonnegative, and \(f\) takes both positive values (where \(|x|>|y|\)) and negative values (where \(|x|<|y|\)). The other hypotheses of 5.28 do hold here: \(\lambda\) restricted to \([0,1]\) is a finite measure, hence \(\sigma\)-finite. So this exercise shows that the nonnegativity hypothesis in Tonelli’s Theorem cannot be dropped.

Fubini’s Theorem (5.32) is not contradicted, because its hypothesis \(\int_{[0,1]^2}|f|\,d(\lambda\times\lambda)<\infty\) fails. To see this directly, apply Tonelli’s Theorem to the nonnegative measurable function \(|f|\):

\begin{equation*} \int_{[0,1]^2}|f|\,d(\lambda\times\lambda)=\int_{[0,1]}\int_{[0,1]}|f(x,y)|\,d\lambda(y)\,d\lambda(x). \end{equation*}

Fix \(x\in(0,1]\). On \(\{y:0\le y\le x\}\) we have \(f(x,y)\ge 0\), so

\begin{equation*} \int_{[0,1]}|f(x,y)|\,d\lambda(y)\ge\int_{[0,x]}f(x,y)\,d\lambda(y) =\left[\frac{y}{x^2+y^2}\right]_{y=0}^{y=x}=\frac{x}{2x^2}=\frac{1}{2x}. \end{equation*}

Therefore

\begin{equation*} \int_{[0,1]^2}|f|\,d(\lambda\times\lambda)\ge\int_{(0,1]}\frac{1}{2x}\,d\lambda(x) \ge\frac{1}{2}\int_{[1/k,\,1]}\frac{dx}{x}=\frac{\ln k}{2} \end{equation*}

for every \(k\in\mathbb{Z}^+\) (the middle integrals being computed by 3.34, the integrand being continuous on \([1/k,1]\)), and letting \(k\to\infty\) gives \(\int_{[0,1]^2}|f|\,d(\lambda\times\lambda)=\infty\). So \(f\notin\mathcal{L}^1(\lambda\times\lambda)\) and Fubini’s Theorem does not apply.

(Conversely, once one knows the two iterated integrals in (a) are unequal, Fubini’s Theorem itself forces \(\int|f|\,d(\lambda\times\lambda)=\infty\); the computation above verifies this independently.)

(a) Define

\begin{equation*} x_{m,n}= \begin{cases} 1 & \text{if }n=m,\\ -1 & \text{if }n=2m,\\ 0 & \text{otherwise.} \end{cases} \end{equation*}

Thus each row of the array consists of a \(+1\) on the diagonal and a \(-1\) placed twice as far out; the \(-1\) of row \(m\) lands in an even column, so the rows cancel but the odd columns never receive a compensating \(-1\).

Row sums. Fix \(m\in\mathbb{Z}^+\). Since \(m\ge 1\), we have \(2m\neq m\), so the sequence \(n\mapsto x_{m,n}\) has exactly two nonzero terms, namely \(x_{m,m}=1\) and \(x_{m,2m}=-1\). Hence

\begin{equation*} \sum_{n=1}^{\infty}x_{m,n}=1+(-1)=0 \qquad\text{for every }m\in\mathbb{Z}^+, \end{equation*}

and therefore

\begin{equation*} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}x_{m,n}=\sum_{m=1}^{\infty}0=0. \end{equation*}

Column sums. Fix \(n\in\mathbb{Z}^+\). The term \(x_{m,n}\) is nonzero only if \(m=n\) (giving \(+1\)) or \(2m=n\) (giving \(-1\), which requires \(n\) to be even, with \(m=n/2\)); these two indices are distinct because \(n\neq n/2\) for \(n\ge 1\). Hence

\begin{equation*} \sum_{m=1}^{\infty}x_{m,n}= \begin{cases} 1 & \text{if \(n\) is odd},\\ 1+(-1)=0 & \text{if \(n\) is even.} \end{cases} \end{equation*}

The partial sums of \(\sum_{n}\bigl(\sum_{m}x_{m,n}\bigr)\) therefore count the odd integers up to \(n\) and increase without bound, so

\begin{equation*} \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}x_{m,n}=\infty . \end{equation*}

(b) Put this in the measure-theoretic setting. Let \(X=Y=\mathbb{Z}^+\), let \(\mathcal{S}=\mathcal{T}\) be the \(\sigma\)-algebra of all subsets of \(\mathbb{Z}^+\), and let \(\mu=\nu\) be counting measure. These measure spaces are \(\sigma\)-finite, since \(\mathbb{Z}^+=\bigcup_{k=1}^{\infty}\{1,\dots,k\}\) and each \(\{1,\dots,k\}\) has finite measure. Define \(f:\mathbb{Z}^+\times\mathbb{Z}^+\to\mathbb{R}\) by \(f(m,n)=x_{m,n}\). Every subset of \(\mathbb{Z}^+\times\mathbb{Z}^+\) belongs to \(\mathcal{S}\otimes\mathcal{T}\), because each singleton \(\{(m,n)\}=\{m\}\times\{n\}\) is a measurable rectangle and every subset of the countable set \(\mathbb{Z}^+\times\mathbb{Z}^+\) is a countable union of singletons; hence \(f\) is \(\mathcal{S}\otimes\mathcal{T}\)-measurable. Integration with respect to counting measure is summation (3.6, and 3.42 for the real-valued case), so the two iterated sums in (a) are exactly the two iterated integrals appearing in 5.28 and 5.32. All the inner integrals here are defined: for fixed \(m\) the function \(n\mapsto x_{m,n}\) has only two nonzero values, and for fixed \(n\) the function \(m\mapsto x_{m,n}\) has at most two nonzero values, so in each case the function lies in \(\mathcal{L}^1\) of counting measure.

Tonelli’s Theorem (5.28) is not contradicted, because it requires the integrand to be nonnegative; our \(f\) takes the value \(-1\). The same remark applies to its corollary 5.31 on double sums, whose hypothesis is precisely that the numbers \(x_{m,n}\) be nonnegative. Note that the \(\sigma\)-finiteness hypothesis does hold here, so nonnegativity is the only hypothesis that fails; the example thus shows that nonnegativity cannot be dropped from 5.31.

Fubini’s Theorem (5.32) is not contradicted, because its hypothesis \(\int_{\mathbb{Z}^+\times\mathbb{Z}^+}|f|\,d(\mu\times\nu)<\infty\) fails. Indeed, \(|f|\) is nonnegative and measurable, so Tonelli’s Theorem applies to it and gives

\begin{equation*} \int_{\mathbb{Z}^+\times\mathbb{Z}^+}|f|\,d(\mu\times\nu) =\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}|x_{m,n}| =\sum_{m=1}^{\infty}2=\infty , \end{equation*}

since each row has exactly two nonzero entries, each of absolute value \(1\). So \(f\) is not integrable with respect to \(\mu\times\nu\) and Fubini’s Theorem says nothing about it.

Recall from 5.34 that

\begin{equation*} U_f=\{(x,t)\in X\times(0,\infty):0<t<f(x)\}. \end{equation*}

Note that no measure is involved in this exercise; only the \(\sigma\)-algebras matter.

First suppose \(f\) is an \(\mathcal{S}\)-measurable function. We claim that

\begin{equation*} U_f=\bigcup_{r\in\mathbb{Q}\cap(0,\infty)}f^{-1}\bigl((r,\infty]\bigr)\times(0,r]. \end{equation*}

To prove the inclusion \(\supseteq\), suppose \(r\in\mathbb{Q}\cap(0,\infty)\), \(x\in f^{-1}((r,\infty])\), and \(t\in(0,r]\). Then \(0<t\le r<f(x)\), so \((x,t)\in U_f\). To prove the inclusion \(\subseteq\), suppose \((x,t)\in U_f\), so \(0<t<f(x)\). Because \(\mathbb{Q}\) is dense in \(\mathbb{R}\) there exists \(r\in\mathbb{Q}\) with \(t<r<f(x)\) (if \(f(x)=\infty\), take any rational \(r>t\)). Then \(r>t>0\), so \(r\in\mathbb{Q}\cap(0,\infty)\), and \(x\in f^{-1}((r,\infty])\) and \(t\in(0,r]\). This proves the claim.

Now for each \(r\in\mathbb{Q}\cap(0,\infty)\) the set \((r,\infty]\) is a Borel subset of \([-\infty,\infty]\) (see 2.50), so \(f^{-1}((r,\infty])\in\mathcal{S}\) because \(f\) is \(\mathcal{S}\)-measurable (2.51); also \((0,r]\) is a Borel subset of \((0,\infty)\), so \((0,r]\in\mathcal{B}\). Hence each set \(f^{-1}((r,\infty])\times(0,r]\) is a measurable rectangle in \(\mathcal{S}\otimes\mathcal{B}\). The displayed union is over the countable set \(\mathbb{Q}\cap(0,\infty)\), and \(\mathcal{S}\otimes\mathcal{B}\) is closed under countable unions. Thus \(U_f\in\mathcal{S}\otimes\mathcal{B}\). (This is the same conclusion reached, by a slightly different decomposition, in the first paragraph of the proof of 5.35; the argument there also uses no properties of the measure.)

Conversely, suppose \(U_f\in\mathcal{S}\otimes\mathcal{B}\). Fix \(t\in(0,\infty)\). Because cross sections of measurable sets are measurable (5.6), the cross section

\begin{equation*} [U_f]^t=\{x\in X:(x,t)\in U_f\} \end{equation*}

belongs to \(\mathcal{S}\). Since \(t>0\), the condition \((x,t)\in U_f\) says exactly that \(t<f(x)\). In other words,

\begin{equation*} f^{-1}\bigl((t,\infty]\bigr)=\{x\in X:f(x)>t\}=[U_f]^t\in\mathcal{S} \qquad\text{for every }t\in(0,\infty). \end{equation*}

Now let \(a\in\mathbb{R}\) be arbitrary; we check that \(f^{-1}((a,\infty])\in\mathcal{S}\) in all three possible cases.

• If \(a>0\), then \(f^{-1}((a,\infty])=[U_f]^a\in\mathcal{S}\) by the previous paragraph.

• If \(a<0\), then \(f^{-1}((a,\infty])=X\in\mathcal{S}\), because \(f\) takes values in \([0,\infty]\).

• If \(a=0\), then \(f(x)>0\) if and only if \(f(x)>1/k\) for some \(k\in\mathbb{Z}^+\), so

\begin{equation*} f^{-1}\bigl((0,\infty]\bigr)=\bigcup_{k=1}^{\infty}f^{-1}\bigl((\tfrac{1}{k},\infty]\bigr) =\bigcup_{k=1}^{\infty}[U_f]^{1/k}, \end{equation*}

which is a countable union of elements of \(\mathcal{S}\) and hence belongs to \(\mathcal{S}\).

Thus \(f^{-1}((a,\infty])\in\mathcal{S}\) for every \(a\in\mathbb{R}\), and therefore \(f\) is an \(\mathcal{S}\)-measurable function by 2.52.

Suppose \(f\) is an \(\mathcal{S}\)-measurable function.

Define \(g,h:X\times\mathbb{R}\to\mathbb{R}\) by

\begin{equation*} g(x,y)=f(x)\qquad\text{and}\qquad h(x,y)=y . \end{equation*}

Both are \(\mathcal{S}\otimes\mathcal{B}\)-measurable. Indeed, if \(B\subseteq\mathbb{R}\) is a Borel set, then

\begin{equation*} g^{-1}(B)=f^{-1}(B)\times\mathbb{R} \qquad\text{and}\qquad h^{-1}(B)=X\times B , \end{equation*}

and \(f^{-1}(B)\in\mathcal{S}\) because \(f\) is \(\mathcal{S}\)-measurable, so both of these sets are measurable rectangles in \(\mathcal{S}\otimes\mathcal{B}\).

By 2.46(a), the function \(g-h\) is \(\mathcal{S}\otimes\mathcal{B}\)-measurable, where

\begin{equation*} (g-h)(x,y)=f(x)-y . \end{equation*}

Because \(\{0\}\) is a Borel subset of \(\mathbb{R}\), we conclude that

\begin{equation*} \operatorname{graph}(f)=\{(x,y)\in X\times\mathbb{R}:f(x)-y=0\}=(g-h)^{-1}\bigl(\{0\}\bigr)\in\mathcal{S}\otimes\mathcal{B}, \end{equation*}

as desired.

For a proof that avoids citing 2.46 and instead exhibits \(\operatorname{graph}(f)\) explicitly from measurable rectangles, note that

\begin{equation*} \operatorname{graph}(f)=\bigcap_{k=1}^{\infty}\ \bigcup_{j\in\mathbb{Z}} \Bigl(f^{-1}\bigl([\tfrac{j}{k},\tfrac{j+1}{k})\bigr)\times[\tfrac{j}{k},\tfrac{j+1}{k})\Bigr). \end{equation*}

To verify the inclusion \(\subseteq\), suppose \(y=f(x)\) and let \(k\in\mathbb{Z}^+\). Take \(j\) to be the unique integer with \(j\le kf(x)<j+1\); then \(f(x)\in[\tfrac{j}{k},\tfrac{j+1}{k})\), so \(x\in f^{-1}([\tfrac{j}{k},\tfrac{j+1}{k}))\) and also \(y=f(x)\in[\tfrac{j}{k},\tfrac{j+1}{k})\). Hence \((x,y)\) lies in the \(k^{\text{th}}\) union, for every \(k\).

To verify the inclusion \(\supseteq\), suppose \((x,y)\) lies in the set on the right. For each \(k\in\mathbb{Z}^+\) there exists \(j_k\in\mathbb{Z}\) with \(f(x)\in[\tfrac{j_k}{k},\tfrac{j_k+1}{k})\) and \(y\in[\tfrac{j_k}{k},\tfrac{j_k+1}{k})\). Both numbers lie in an interval of length \(\tfrac{1}{k}\), so \(|y-f(x)|<\tfrac{1}{k}\). As this holds for every \(k\in\mathbb{Z}^+\), we get \(y=f(x)\), so \((x,y)\in\operatorname{graph}(f)\).

Finally, each interval \([\tfrac{j}{k},\tfrac{j+1}{k})\) is a Borel subset of \(\mathbb{R}\), so \(f^{-1}([\tfrac{j}{k},\tfrac{j+1}{k}))\in\mathcal{S}\) and each set \(f^{-1}([\tfrac{j}{k},\tfrac{j+1}{k}))\times[\tfrac{j}{k},\tfrac{j+1}{k})\) is a measurable rectangle in \(\mathcal{S}\otimes\mathcal{B}\). The union over \(j\in\mathbb{Z}\) is a countable union and the intersection over \(k\in\mathbb{Z}^+\) is a countable intersection, and a \(\sigma\)-algebra is closed under both operations. Hence \(\operatorname{graph}(f)\in\mathcal{S}\otimes\mathcal{B}\).

Recall the definitions used in this section: \(\|(x_1,\ldots,x_n)\|_\infty = \max\{|x_1|,\ldots,|x_n|\}\), the open cube \(B(x,\delta) = \{ y \in \mathbb{R}^n : \|y - x\|_\infty < \delta \}\), and \(G \subseteq \mathbb{R}^n\) is open means that for every \(x \in G\) there exists \(\delta > 0\) with \(B(x, \delta) \subseteq G\).

Write \(\|x\|_2 = \sqrt{x_1^2 + \cdots + x_n^2}\) for the Euclidean norm, and for \(b \in \mathbb{R}^n\) and \(r>0\) let

\begin{equation*} D(b, r) = \{ a \in \mathbb{R}^n : \|a - b\|_2 < r \} \end{equation*}

denote the Euclidean open ball; this is the set displayed in the statement of the exercise. So we must show that \(G\) is open if and only if every \(b \in G\) has some \(r > 0\) with \(D(b,r) \subseteq G\).

The proof rests on the elementary comparison of the two norms:

\begin{equation*} \|x\|_\infty \le \|x\|_2 \le \sqrt{n}\, \|x\|_\infty \quad \text{for all } x \in \mathbb{R}^n. \end{equation*}

To verify the first inequality, choose \(j\) with \(|x_j| = \|x\|_\infty\); then \(\|x\|_\infty^2 = x_j^2 \le x_1^2 + \cdots + x_n^2 = \|x\|_2^2\), and taking square roots gives \(\|x\|_\infty \le \|x\|_2\). For the second inequality, \(x_k^2 \le \|x\|_\infty^2\) for each \(k\), so \(\|x\|_2^2 = x_1^2 + \cdots + x_n^2 \le n \|x\|_\infty^2\), and taking square roots gives \(\|x\|_2 \le \sqrt{n}\, \|x\|_\infty\).

These two inequalities say exactly that

\begin{equation*} D(b, \delta) \subseteq B(b, \delta) \quad \text{and} \quad B(b, r/\sqrt{n}\,) \subseteq D(b, r) \end{equation*}

for all \(b \in \mathbb{R}^n\), \(\delta > 0\), and \(r > 0\). Indeed, if \(\|a - b\|_2 < \delta\) then \(\|a-b\|_\infty \le \|a-b\|_2 < \delta\), which gives the first inclusion; and if \(\|a - b\|_\infty < r/\sqrt{n}\) then \(\|a - b\|_2 \le \sqrt{n}\,\|a-b\|_\infty < r\), which gives the second.

Now suppose \(G\) is open and \(b = (b_1,\ldots,b_n) \in G\). Then there exists \(\delta > 0\) with \(B(b, \delta) \subseteq G\). Taking \(r = \delta\), the first inclusion above gives

\begin{equation*} D(b, r) = D(b,\delta) \subseteq B(b, \delta) \subseteq G, \end{equation*}

which is the desired conclusion.

Conversely, suppose that for each \(b \in G\) there exists \(r > 0\) with \(D(b, r) \subseteq G\). Let \(b \in G\) and choose such an \(r\). Put \(\delta = r/\sqrt{n} > 0\). Then the second inclusion above gives

\begin{equation*} B(b, \delta) = B(b, r/\sqrt{n}\,) \subseteq D(b, r) \subseteq G. \end{equation*}

Thus every point of \(G\) is the center of an open cube contained in \(G\), which means that \(G\) is open.

Hence the two conditions are equivalent. (In other words, the topology on \(\mathbb{R}^n\) generated by cubes, which is the one used throughout this section, is the same as the usual topology generated by Euclidean balls; equivalently, \(\|\cdot\|_\infty\) and \(\|\cdot\|_2\) are equivalent norms on \(\mathbb{R}^n\).)

Recall from 5.4 that \([E]_a = \{ y \in \mathbb{R} : (a,y) \in E \}\) and \([E]^a = \{ x \in \mathbb{R} : (x,a) \in E \}\).

The strategy is a counting argument: we exhibit a family of \(2^{\mathfrak{c}}\) distinct subsets of \(\mathbb{R}^2\) all of whose cross sections are open (here \(\mathfrak{c} = 2^{\aleph_0}\) is the cardinality of \(\mathbb{R}\)), and we show that \(\mathcal{B}_2\) has only \(\mathfrak{c}\) elements. Since \(2^{\mathfrak{c}} > \mathfrak{c}\) by Cantor’s Theorem, some member of the family is not a Borel set.

Step 1: a large family of sets with open cross sections.

For each \(A \subseteq (0, \infty)\) define \(f_A : \mathbb{R} \to \mathbb{R}\) by

\begin{equation*} f_A(x) = \begin{cases} -x & \text{if } |x| \in A, \\ x & \text{if } |x| \notin A. \end{cases} \end{equation*}

Because \(|f_A(x)| = |x|\) for every \(x \in \mathbb{R}\), we have \(f_A\bigl(f_A(x)\bigr) = x\) for every \(x\); thus \(f_A\) is a bijection of \(\mathbb{R}\) onto \(\mathbb{R}\) that is its own inverse. Let

\begin{equation*} \Gamma_A = \{ (x, f_A(x)) : x \in \mathbb{R} \} \quad \text{and} \quad E_A = \mathbb{R}^2 \setminus \Gamma_A . \end{equation*}

Fix \(a \in \mathbb{R}\). Then \((a, y) \in \Gamma_A\) if and only if \(y = f_A(a)\), so

\begin{equation*} [E_A]_a = \mathbb{R} \setminus \{ f_A(a) \}, \end{equation*}

which is an open subset of \(\mathbb{R}\). Also \((x, a) \in \Gamma_A\) if and only if \(f_A(x) = a\), which (because \(f_A\) is a bijection equal to its own inverse) happens if and only if \(x = f_A(a)\). Hence

\begin{equation*} [E_A]^a = \mathbb{R} \setminus \{ f_A(a) \}, \end{equation*}

which is again an open subset of \(\mathbb{R}\). So every set \(E_A\) has all its cross sections open.

The map \(A \mapsto E_A\) is injective. Indeed, \(E_A\) determines \(\Gamma_A = \mathbb{R}^2 \setminus E_A\), and for \(t > 0\) we have \(f_A(t) = -t\) if \(t \in A\) and \(f_A(t) = t\) if \(t \notin A\); because \(-t \ne t\) for \(t > 0\), this gives

\begin{equation*} A = \{ t > 0 : (t, -t) \in \Gamma_A \}, \end{equation*}

so \(A\) is recovered from \(E_A\). Since \(|(0,\infty)| = \mathfrak{c}\), the collection \(\{ E_A : A \subseteq (0,\infty) \}\) consists of \(2^{\mathfrak{c}}\) distinct subsets of \(\mathbb{R}^2\), each of which has all cross sections open.

Step 2: \(\mathcal{B}_2\) has cardinality at most \(\mathfrak{c}\).

Let \(\Sigma_0\) denote the collection of open subsets of \(\mathbb{R}^2\). By the proof of 5.38(a), every open subset of \(\mathbb{R}^2\) is the union of a subcollection of the countable collection \(\mathcal{Q}\) of open cubes whose center has rational coordinates and whose side length is rational. The map sending a subcollection of \(\mathcal{Q}\) to its union therefore maps the power set of \(\mathcal{Q}\) onto \(\Sigma_0\), so \(|\Sigma_0| \le 2^{\aleph_0} = \mathfrak{c}\).

Let \(\omega_1\) denote the first uncountable ordinal. For each ordinal \(\alpha\) with \(0 < \alpha < \omega_1\) define

\begin{equation*} \Sigma_\alpha = \Bigl\{ \bigcup_{k=1}^{\infty} F_k : \text{for each } k, \ F_k \in \textstyle\bigcup_{\beta < \alpha} \Sigma_\beta \text{ or } \mathbb{R}^2 \setminus F_k \in \textstyle\bigcup_{\beta < \alpha} \Sigma_\beta \Bigr\}, \end{equation*}

and let \(\Sigma = \bigcup_{\alpha < \omega_1} \Sigma_\alpha\). Taking \(F_1 = F_2 = \cdots = E\) shows that \(\Sigma_\beta \subseteq \Sigma_\alpha\) whenever \(\beta < \alpha < \omega_1\).

The collection \(\Sigma\) is a \(\sigma\)-algebra on \(\mathbb{R}^2\). It is closed under complementation: if \(E \in \Sigma_\alpha\), then taking \(F_1 = F_2 = \cdots = \mathbb{R}^2 \setminus E\) (whose complement \(E\) lies in \(\Sigma_\alpha\)) shows \(\mathbb{R}^2 \setminus E \in \Sigma_{\alpha + 1} \subseteq \Sigma\). It is closed under countable unions: if \(E_1, E_2, \ldots \in \Sigma\), choose \(\alpha_k < \omega_1\) with \(E_k \in \Sigma_{\alpha_k}\); the supremum \(\alpha\) of the countable set \(\{\alpha_k\}\) is a countable ordinal, so \(\alpha < \omega_1\), and then every \(E_k\) lies in \(\Sigma_\alpha\), whence \(\bigcup_k E_k \in \Sigma_{\alpha+1} \subseteq \Sigma\). Finally \(\Sigma_0 \subseteq \Sigma\), so \(\Sigma\) is a \(\sigma\)-algebra containing every open subset of \(\mathbb{R}^2\); therefore \(\mathcal{B}_2 \subseteq \Sigma\).

Now we bound \(|\Sigma|\) by transfinite induction. We have \(|\Sigma_0| \le \mathfrak{c}\). Suppose \(0 < \alpha < \omega_1\) and \(|\Sigma_\beta| \le \mathfrak{c}\) for every \(\beta < \alpha\). Because \(\alpha\) is a countable ordinal, \(\bigl|\bigcup_{\beta<\alpha}\Sigma_\beta\bigr| \le \aleph_0 \cdot \mathfrak{c} = \mathfrak{c}\); hence the collection \(\mathcal{G}\) of sets \(F\) with \(F\) or \(\mathbb{R}^2 \setminus F\) in \(\bigcup_{\beta<\alpha}\Sigma_\beta\) satisfies \(|\mathcal{G}| \le \mathfrak{c}\). Every element of \(\Sigma_\alpha\) is the union of the terms of a sequence in \(\mathcal{G}\), so

\begin{equation*} |\Sigma_\alpha| \le \mathfrak{c}^{\aleph_0} = \bigl(2^{\aleph_0}\bigr)^{\aleph_0} = 2^{\aleph_0} = \mathfrak{c}. \end{equation*}

Thus \(|\Sigma_\alpha| \le \mathfrak{c}\) for every \(\alpha < \omega_1\), and therefore

\begin{equation*} |\mathcal{B}_2| \le |\Sigma| \le \aleph_1 \cdot \mathfrak{c} = \mathfrak{c}, \end{equation*}

where we used \(\aleph_1 \le \mathfrak{c}\).

Step 3: conclusion.

By Cantor’s Theorem \(2^{\mathfrak{c}} > \mathfrak{c} \ge |\mathcal{B}_2|\). Since \(A \mapsto E_A\) is injective on the \(2^{\mathfrak{c}}\) subsets \(A\) of \((0,\infty)\), the sets \(E_A\) cannot all belong to \(\mathcal{B}_2\). Choose \(A \subseteq (0,\infty)\) with \(E_A \notin \mathcal{B}_2\) and set \(E = E_A\). By Step 1, \([E]_a\) and \([E]^a\) are open subsets of \(\mathbb{R}\) for every \(a \in \mathbb{R}\), and \(E \notin \mathcal{B}_2\), as desired.

Two remarks. First, this shows that the converse of 5.6 fails badly: measurability of all cross sections, even openness of all cross sections, does not imply measurability of the set. Second, the argument is a pure counting argument and produces no explicit \(E\); it uses the axiom of choice, both for the cardinal arithmetic in Step 2 and for the comparison \(\aleph_1 \le \mathfrak{c}\).

Throughout, we identify \(\bigl((x,y),z\bigr)\) and \(\bigl(x,(y,z)\bigr)\) with \((x,y,z)\); under this identification, \((A \times B) \times C\), \(A \times (B \times C)\), and \(A \times B \times C\) are the same subset of \(X \times Y \times Z\), and all three \(\sigma\)-algebras in the statement are collections of subsets of \(X \times Y \times Z\). Write \(\mathcal{P} = \mathcal{S} \otimes \mathcal{T} \otimes \mathcal{U}\), the smallest \(\sigma\)-algebra on \(X \times Y \times Z\) containing every set \(A \times B \times C\) with \(A \in \mathcal{S}\), \(B \in \mathcal{T}\), \(C \in \mathcal{U}\).

First we prove \(\mathcal{P} = (\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}\).

The inclusion \(\mathcal{P} \subseteq (\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}\) is easy. If \(A \in \mathcal{S}\), \(B \in \mathcal{T}\), and \(C \in \mathcal{U}\), then \(A \times B\) is a measurable rectangle in \(\mathcal{S} \otimes \mathcal{T}\), so \(A \times B \in \mathcal{S} \otimes \mathcal{T}\); hence \((A \times B) \times C\) is a measurable rectangle in \((\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}\) and thus lies in \((\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}\). Under our identification \((A \times B) \times C = A \times B \times C\). Thus \((\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}\) is a \(\sigma\)-algebra on \(X \times Y \times Z\) containing all the generators of \(\mathcal{P}\), and therefore contains \(\mathcal{P}\), because \(\mathcal{P}\) is the smallest such \(\sigma\)-algebra.

For the reverse inclusion, it suffices to show that \(D \times C \in \mathcal{P}\) for all \(D \in \mathcal{S} \otimes \mathcal{T}\) and all \(C \in \mathcal{U}\); these sets are the measurable rectangles generating \((\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}\), so once they are known to lie in the \(\sigma\)-algebra \(\mathcal{P}\) we get \((\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U} \subseteq \mathcal{P}\).

So temporarily fix \(C \in \mathcal{U}\) and let

\begin{equation*} \mathcal{E} = \{ D \subseteq X \times Y : D \times C \in \mathcal{P} \}. \end{equation*}

We claim \(\mathcal{E}\) is a \(\sigma\)-algebra on \(X \times Y\).

Since \(X \in \mathcal{S}\) and \(Y \in \mathcal{T}\), the set \((X \times Y) \times C = X \times Y \times C\) is one of the generators of \(\mathcal{P}\) and hence lies in \(\mathcal{P}\); thus \(X \times Y \in \mathcal{E}\).

If \(D_1, D_2, \ldots \in \mathcal{E}\), then

\begin{equation*} \Bigl( \bigcup_{k=1}^{\infty} D_k \Bigr) \times C = \bigcup_{k=1}^{\infty} (D_k \times C) \in \mathcal{P}, \end{equation*}

because \(\mathcal{P}\) is closed under countable unions. Hence \(\mathcal{E}\) is closed under countable unions.

If \(D \in \mathcal{E}\), then

\begin{equation*} \bigl( (X \times Y) \setminus D \bigr) \times C = \bigl( (X \times Y \times Z) \setminus (D \times C) \bigr) \cap \bigl( X \times Y \times C \bigr). \end{equation*}

To verify this set equality, note that a point \((x,y,z)\) belongs to the left side if and only if \((x,y) \notin D\) and \(z \in C\); and it belongs to the right side if and only if \(z \in C\) and it is false that both \((x,y) \in D\) and \(z \in C\), which (given \(z \in C\)) says exactly that \((x,y) \notin D\). Since \(D \times C \in \mathcal{P}\) and \(X \times Y \times C \in \mathcal{P}\), and \(\mathcal{P}\) is closed under complementation and finite intersections, the right side lies in \(\mathcal{P}\). Hence \((X \times Y) \setminus D \in \mathcal{E}\), so \(\mathcal{E}\) is closed under complementation.

Thus \(\mathcal{E}\) is a \(\sigma\)-algebra on \(X \times Y\). Moreover if \(A \in \mathcal{S}\) and \(B \in \mathcal{T}\), then \((A \times B) \times C = A \times B \times C \in \mathcal{P}\), so \(A \times B \in \mathcal{E}\). Hence \(\mathcal{E}\) is a \(\sigma\)-algebra containing all the measurable rectangles in \(\mathcal{S} \otimes \mathcal{T}\), which gives \(\mathcal{S} \otimes \mathcal{T} \subseteq \mathcal{E}\). In other words, \(D \times C \in \mathcal{P}\) for every \(D \in \mathcal{S} \otimes \mathcal{T}\). Because \(C \in \mathcal{U}\) was arbitrary, we conclude \((\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U} \subseteq \mathcal{P}\), completing the proof that

\begin{equation*} \mathcal{P} = (\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U}. \end{equation*}

The proof that \(\mathcal{P} = \mathcal{S} \otimes (\mathcal{T} \otimes \mathcal{U})\) is symmetric. The inclusion \(\mathcal{P} \subseteq \mathcal{S} \otimes (\mathcal{T} \otimes \mathcal{U})\) holds because \(A \times B \times C = A \times (B \times C)\) is a measurable rectangle in \(\mathcal{S} \otimes (\mathcal{T} \otimes \mathcal{U})\) whenever \(A \in \mathcal{S}\), \(B \in \mathcal{T}\), \(C \in \mathcal{U}\). For the reverse inclusion, temporarily fix \(A \in \mathcal{S}\) and let

\begin{equation*} \mathcal{F} = \{ F \subseteq Y \times Z : A \times F \in \mathcal{P} \}. \end{equation*}

Then \(Y \times Z \in \mathcal{F}\) because \(A \times Y \times Z \in \mathcal{P}\); \(\mathcal{F}\) is closed under countable unions because \(A \times \bigcup_k F_k = \bigcup_k (A \times F_k)\); and \(\mathcal{F}\) is closed under complementation because

\begin{equation*} A \times \bigl( (Y \times Z) \setminus F \bigr) = \bigl( (X \times Y \times Z) \setminus (A \times F) \bigr) \cap \bigl( A \times Y \times Z \bigr), \end{equation*}

which is verified exactly as above: \((x,y,z)\) lies in the left side if and only if \(x \in A\) and \((y,z) \notin F\), and this is precisely the condition described by the right side. So \(\mathcal{F}\) is a \(\sigma\)-algebra on \(Y \times Z\), and it contains every measurable rectangle \(B \times C\) with \(B \in \mathcal{T}\), \(C \in \mathcal{U}\) (because \(A \times (B \times C) = A \times B \times C \in \mathcal{P}\)). Hence \(\mathcal{T} \otimes \mathcal{U} \subseteq \mathcal{F}\), which says that \(A \times F \in \mathcal{P}\) for all \(A \in \mathcal{S}\) and \(F \in \mathcal{T} \otimes \mathcal{U}\). These sets generate \(\mathcal{S} \otimes (\mathcal{T} \otimes \mathcal{U})\), so \(\mathcal{S} \otimes (\mathcal{T} \otimes \mathcal{U}) \subseteq \mathcal{P}\).

Combining the two displayed conclusions gives

\begin{equation*} \mathcal{S} \otimes \mathcal{T} \otimes \mathcal{U} = (\mathcal{S} \otimes \mathcal{T}) \otimes \mathcal{U} = \mathcal{S} \otimes (\mathcal{T} \otimes \mathcal{U}), \end{equation*}

as desired. In particular, parentheses may be dispensed with when forming products of three \(\sigma\)-algebras, which is the general version of the associativity noted after 5.39 for the Borel \(\sigma\)-algebras.

Fix \(a \in \mathbb{R}^n\).

Step 1: \(a + E \in \mathcal{B}_n\) for every \(E \in \mathcal{B}_n\).

Let \(\mathcal{E} = \{ E \in \mathcal{B}_n : a + E \in \mathcal{B}_n \}\). If \(G \subseteq \mathbb{R}^n\) is open, then \(a + G\) is open, because \(a + B(x, \delta) = B(a + x, \delta)\) for all \(x \in \mathbb{R}^n\) and \(\delta > 0\) (translation preserves the \(\|\cdot\|_\infty\) distance), so an open cube centered at \(x\) and contained in \(G\) translates to an open cube centered at \(a + x\) and contained in \(a + G\). Hence \(\mathcal{E}\) contains every open subset of \(\mathbb{R}^n\). Because \(x \mapsto a + x\) is a bijection of \(\mathbb{R}^n\) onto \(\mathbb{R}^n\),

\begin{equation*} a + (\mathbb{R}^n \setminus E) = \mathbb{R}^n \setminus (a + E) \quad \text{and} \quad a + \bigcup_{k=1}^{\infty} E_k = \bigcup_{k=1}^{\infty} (a + E_k), \end{equation*}

so \(\mathcal{E}\) is closed under complementation and countable unions. Thus \(\mathcal{E}\) is a \(\sigma\)-algebra on \(\mathbb{R}^n\) containing all open sets, which gives \(\mathcal{E} = \mathcal{B}_n\). In other words, \(a + E \in \mathcal{B}_n\) whenever \(E \in \mathcal{B}_n\).

Step 2: the case \(n = 1\).

Lebesgue measure \(\lambda_1\) is the restriction of outer measure to the Borel subsets of \(\mathbb{R}\), and outer measure is translation invariant by 2.7. Hence \(\lambda_1(a + E) = |a + E| = |E| = \lambda_1(E)\) for every \(E \in \mathcal{B}_1\) and every \(a \in \mathbb{R}\).

Step 3: translation invariance of the integral in a dimension where translation invariance of the measure is known.

Suppose \(m \in \mathbb{Z}^+\) and we already know that \(\lambda_m(b + F) = \lambda_m(F)\) for all \(F \in \mathcal{B}_m\) and all \(b \in \mathbb{R}^m\). We claim that if \(h : \mathbb{R}^m \to [0, \infty]\) is \(\mathcal{B}_m\)-measurable and \(b \in \mathbb{R}^m\), then the function \(x \mapsto h(x - b)\) is \(\mathcal{B}_m\)-measurable and

\begin{equation*} \int_{\mathbb{R}^m} h(x - b) \, d\lambda_m(x) = \int_{\mathbb{R}^m} h \, d\lambda_m . \end{equation*}

Measurability is immediate from Step 1: if \(B\) is a Borel subset of \([0,\infty]\), then \(\{ x : h(x-b) \in B \} = b + h^{-1}(B) \in \mathcal{B}_m\).

If \(h = \chi_F\) with \(F \in \mathcal{B}_m\), then \(\chi_F(x - b) = 1\) exactly when \(x \in b + F\), so \(x \mapsto h(x-b)\) equals \(\chi_{b+F}\) and

\begin{equation*} \int_{\mathbb{R}^m} \chi_F(x-b) \, d\lambda_m(x) = \lambda_m(b + F) = \lambda_m(F) = \int_{\mathbb{R}^m} \chi_F \, d\lambda_m . \end{equation*}

If \(h = c_1 \chi_{F_1} + \cdots + c_j \chi_{F_j}\) is a nonnegative simple \(\mathcal{B}_m\)-measurable function, then \(h(x - b) = c_1 \chi_{b + F_1}(x) + \cdots + c_j \chi_{b + F_j}(x)\), so the claim follows from the previous sentence together with additivity and homogeneity of the integral of nonnegative functions (3.16 and 3.20).

For general \(\mathcal{B}_m\)-measurable \(h : \mathbb{R}^m \to [0,\infty]\), use 2.89 to choose an increasing sequence \(s_1 \le s_2 \le \cdots\) of nonnegative simple \(\mathcal{B}_m\)-measurable functions with \(\lim_{k \to \infty} s_k(x) = h(x)\) for every \(x\). Then \(k \mapsto s_k(x - b)\) is increasing with limit \(h(x-b)\) for every \(x\). Two applications of the Monotone Convergence Theorem (3.11) give

\begin{equation*} \int_{\mathbb{R}^m} h(x-b) \, d\lambda_m(x) = \lim_{k \to \infty} \int_{\mathbb{R}^m} s_k(x-b) \, d\lambda_m(x) = \lim_{k \to \infty} \int_{\mathbb{R}^m} s_k \, d\lambda_m = \int_{\mathbb{R}^m} h \, d\lambda_m , \end{equation*}

proving the claim.

Step 4: induction on \(n\).

Step 2 gives the result for \(n = 1\). Suppose \(n > 1\) and \(\lambda_{n-1}(b + F) = \lambda_{n-1}(F)\) for all \(F \in \mathcal{B}_{n-1}\) and all \(b \in \mathbb{R}^{n-1}\).

Let \(E \in \mathcal{B}_n\) and write \(a = (a^{\prime}, a_n)\) with \(a^{\prime} \in \mathbb{R}^{n-1}\) and \(a_n \in \mathbb{R}\), thinking of \(\mathbb{R}^n\) as \(\mathbb{R}^{n-1} \times \mathbb{R}\) and of \(\lambda_n\) as \(\lambda_{n-1} \times \lambda_1\) (recall \(\mathcal{B}_n = \mathcal{B}_{n-1} \otimes \mathcal{B}_1\) by 5.39). Every \(\lambda_k\) is a \(\sigma\)-finite measure, because \(\mathbb{R}^k = \bigcup_{m=1}^{\infty} C_m\) where \(C_m\) is the open cube centered at the origin with side length \(m\), and \(\lambda_k(C_m) = m^k < \infty\); so the hypotheses of 5.20 and 5.25 are met throughout what follows. By Step 1, \(a + E \in \mathcal{B}_n\). For \(x \in \mathbb{R}^{n-1}\),

\begin{equation*} [a + E]_x = \{ y \in \mathbb{R} : (x, y) \in a + E \} = \{ y \in \mathbb{R} : (x - a^{\prime}, y - a_n) \in E \} = a_n + [E]_{x - a^{\prime}}, \end{equation*}

where the middle equality holds because \((x,y) \in a + E\) if and only if \((x,y) - a = (x - a^{\prime}, y - a_n) \in E\). Hence, by Step 2,

\begin{equation*} \lambda_1\bigl( [a+E]_x \bigr) = \lambda_1\bigl( a_n + [E]_{x-a^{\prime}} \bigr) = \lambda_1\bigl( [E]_{x - a^{\prime}} \bigr). \end{equation*}

Define \(g : \mathbb{R}^{n-1} \to [0, \infty]\) by \(g(x) = \lambda_1([E]_x)\); this is a \(\mathcal{B}_{n-1}\)-measurable function by 5.20(a), and the equation displayed just above says that \(x \mapsto \lambda_1([a+E]_x)\) is the function \(x \mapsto g(x - a^{\prime})\). Therefore, using the formula for \(\lambda_n\) coming from 5.25,

\begin{equation*} \begin{aligned} \lambda_n(a + E) &= \int_{\mathbb{R}^{n-1}} \lambda_1\bigl( [a+E]_x \bigr) \, d\lambda_{n-1}(x) \\ &= \int_{\mathbb{R}^{n-1}} g(x - a^{\prime}) \, d\lambda_{n-1}(x) \\ &= \int_{\mathbb{R}^{n-1}} g \, d\lambda_{n-1} \\ &= \int_{\mathbb{R}^{n-1}} \lambda_1\bigl( [E]_x \bigr) \, d\lambda_{n-1}(x) \\ &= \lambda_n(E), \end{aligned} \end{equation*}

where the third equality is Step 3 applied with \(m = n - 1\) and \(b = a^{\prime}\) (legitimate by the induction hypothesis).

This completes the induction, so \(\lambda_n(a + E) = \lambda_n(E)\) for every \(n \in \mathbb{Z}^+\), every \(E \in \mathcal{B}_n\), and every \(a \in \mathbb{R}^n\). Together with Step 1, Lebesgue measure on \(\mathbb{R}^n\) is translation invariant.

We first extend 5.41, which is stated for \(t > 0\), to all nonzero \(t\).

Lemma. If \(t \in \mathbb{R} \setminus \{0\}\) and \(E \in \mathcal{B}_n\), then \(tE \in \mathcal{B}_n\) and \(\lambda_n(tE) = |t|^n \lambda_n(E)\).

Proof of the Lemma. For \(t > 0\) this is exactly 5.41. So it suffices to treat the map \(x \mapsto -x\), since for \(t < 0\) we have \(tE = -\bigl(|t| E\bigr)\).

Write \(-F = \{ -x : x \in F \}\). The collection \(\{ F \in \mathcal{B}_n : -F \in \mathcal{B}_n \}\) contains every open set (if \(G\) is open then so is \(-G\), because \(-B(x,\delta) = B(-x,\delta)\)) and is closed under complementation and countable unions (because \(x \mapsto -x\) is a bijection of \(\mathbb{R}^n\), so \(-(\mathbb{R}^n \setminus F) = \mathbb{R}^n \setminus (-F)\) and \(-\bigcup_k F_k = \bigcup_k (-F_k)\)). Hence it is a \(\sigma\)-algebra containing the open sets, so it equals \(\mathcal{B}_n\): we get \(-F \in \mathcal{B}_n\) for all \(F \in \mathcal{B}_n\).

Next, \(\lambda_n(-F) = \lambda_n(F)\) for all \(F \in \mathcal{B}_n\), by induction on \(n\). For \(n = 1\): \(\lambda_1\) is the restriction of outer measure, and if \(I_1, I_2, \ldots\) is a sequence of open intervals whose union contains \(F\), then \(-I_1, -I_2, \ldots\) is a sequence of open intervals of the same lengths whose union contains \(-F\) (and conversely); hence \(|-F| = |F|\), so \(\lambda_1(-F) = \lambda_1(F)\).

Now let \(n > 1\) and assume \(\lambda_{n-1}(-F) = \lambda_{n-1}(F)\) for all \(F \in \mathcal{B}_{n-1}\). Exactly as in Step 3 of the solution to Exercise 4 in this section (first characteristic functions, then nonnegative simple functions using 3.16 and 3.20, then general nonnegative measurable functions using 2.89 and the Monotone Convergence Theorem 3.11), the induction hypothesis implies that

\begin{equation*} \int_{\mathbb{R}^{n-1}} h(-x) \, d\lambda_{n-1}(x) = \int_{\mathbb{R}^{n-1}} h \, d\lambda_{n-1} \end{equation*}

for every \(\mathcal{B}_{n-1}\)-measurable \(h : \mathbb{R}^{n-1} \to [0, \infty]\); note that \(x \mapsto h(-x)\) is \(\mathcal{B}_{n-1}\)-measurable because \(\{ x : h(-x) \in B \} = -h^{-1}(B) \in \mathcal{B}_{n-1}\), and that \(\chi_F(-x) = \chi_{-F}(x)\).

Let \(F \in \mathcal{B}_n\) and view \(\mathbb{R}^n = \mathbb{R}^{n-1} \times \mathbb{R}\), so \(\lambda_n = \lambda_{n-1} \times \lambda_1\) and \(\mathcal{B}_n = \mathcal{B}_{n-1} \otimes \mathcal{B}_1\) (5.39). Each \(\lambda_k\) is \(\sigma\)-finite, since \(\mathbb{R}^k\) is the union of the increasing sequence of open cubes centered at the origin with side lengths \(1, 2, 3, \ldots\), each of finite measure; so 5.20 and 5.25 apply. For \(x \in \mathbb{R}^{n-1}\),

\begin{equation*} [-F]_x = \{ y \in \mathbb{R} : (x,y) \in -F \} = \{ y \in \mathbb{R} : (-x, -y) \in F \} = -[F]_{-x} . \end{equation*}

Hence \(\lambda_1([-F]_x) = \lambda_1([F]_{-x}) = g(-x)\), where \(g(x) = \lambda_1([F]_x)\) is \(\mathcal{B}_{n-1}\)-measurable by 5.20(a). Therefore, using 5.25,

\begin{equation*} \lambda_n(-F) = \int_{\mathbb{R}^{n-1}} g(-x) \, d\lambda_{n-1}(x) = \int_{\mathbb{R}^{n-1}} g \, d\lambda_{n-1} = \lambda_n(F). \end{equation*}

This completes the induction.

Finally, for \(t < 0\) and \(E \in \mathcal{B}_n\) we have \(|t| E \in \mathcal{B}_n\) by 5.41, hence \(tE = -(|t|E) \in \mathcal{B}_n\), and

\begin{equation*} \lambda_n(tE) = \lambda_n\bigl( -(|t|E) \bigr) = \lambda_n\bigl( |t| E \bigr) = |t|^n \lambda_n(E), \end{equation*}

where the last equality is 5.41. This proves the Lemma.

(a) Let \(B \subseteq \mathbb{R}\) be a Borel set. Since \(t \ne 0\),

\begin{equation*} f_t^{-1}(B) = \{ x \in \mathbb{R}^n : f(tx) \in B \} = \{ x \in \mathbb{R}^n : tx \in f^{-1}(B) \} = \tfrac{1}{t} f^{-1}(B). \end{equation*}

Because \(f\) is \(\mathcal{B}_n\)-measurable, \(f^{-1}(B) \in \mathcal{B}_n\), and then the Lemma (applied with the nonzero number \(1/t\)) gives \(\tfrac{1}{t} f^{-1}(B) \in \mathcal{B}_n\). Thus \(f_t^{-1}(B) \in \mathcal{B}_n\) for every Borel \(B \subseteq \mathbb{R}\), which by 2.35 says that \(f_t\) is \(\mathcal{B}_n\)-measurable.

(b) We build up in the usual way. Throughout, for a function \(h\) on \(\mathbb{R}^n\) we write \(h_t(x) = h(tx)\); part (a) and its proof show that \(h_t\) is \(\mathcal{B}_n\)-measurable whenever \(h\) is (the argument in (a) never used that \(f\) is real valued, so it applies to \(\mathcal{B}_n\)-measurable functions with values in \([0,\infty]\) as well).

Characteristic functions. Let \(E \in \mathcal{B}_n\) and \(h = \chi_E\). Then \(h_t(x) = \chi_E(tx)\), which equals \(1\) exactly when \(tx \in E\), that is, exactly when \(x \in \tfrac{1}{t}E\). Thus \(h_t = \chi_{(1/t)E}\) and, by the Lemma,

\begin{equation*} \int_{\mathbb{R}^n} h_t \, d\lambda_n = \lambda_n\Bigl( \tfrac{1}{t} E \Bigr) = \Bigl| \tfrac{1}{t} \Bigr|^{\,n} \lambda_n(E) = \frac{1}{|t|^n} \int_{\mathbb{R}^n} h \, d\lambda_n . \end{equation*}

Nonnegative simple functions. If \(h = c_1 \chi_{E_1} + \cdots + c_j \chi_{E_j}\) with \(c_1, \ldots, c_j \in [0,\infty)\) and \(E_1, \ldots, E_j \in \mathcal{B}_n\), then \(h_t = c_1 (\chi_{E_1})_t + \cdots + c_j (\chi_{E_j})_t\), so additivity and homogeneity of the integral of nonnegative functions (3.16 and 3.20) together with the previous paragraph give \(\int h_t \, d\lambda_n = |t|^{-n} \int h \, d\lambda_n\).

Nonnegative measurable functions. Let \(h : \mathbb{R}^n \to [0,\infty]\) be \(\mathcal{B}_n\)-measurable. By 2.89 there is an increasing sequence \(s_1 \le s_2 \le \cdots\) of nonnegative simple \(\mathcal{B}_n\)-measurable functions converging pointwise to \(h\). Then \((s_1)_t \le (s_2)_t \le \cdots\) converges pointwise to \(h_t\) (evaluate at \(tx\)). Applying the Monotone Convergence Theorem (3.11) twice,

\begin{equation*} \int_{\mathbb{R}^n} h_t \, d\lambda_n = \lim_{k \to \infty} \int_{\mathbb{R}^n} (s_k)_t \, d\lambda_n = \lim_{k \to \infty} \frac{1}{|t|^n} \int_{\mathbb{R}^n} s_k \, d\lambda_n = \frac{1}{|t|^n} \int_{\mathbb{R}^n} h \, d\lambda_n . \end{equation*}

The general case. Suppose \(\int_{\mathbb{R}^n} f \, d\lambda_n\) is defined; by 3.18 this means that \(f\) is \(\mathcal{B}_n\)-measurable and at least one of \(\int f^+ \, d\lambda_n\) and \(\int f^- \, d\lambda_n\) is finite. For every \(x \in \mathbb{R}^n\),

\begin{equation*} (f_t)^+(x) = \max\{ f(tx), 0 \} = f^+(tx) = (f^+)_t(x), \qquad (f_t)^-(x) = (f^-)_t(x), \end{equation*}

so \((f_t)^{\pm} = (f^{\pm})_t\). By the previous paragraph,

\begin{equation*} \int_{\mathbb{R}^n} (f_t)^+ \, d\lambda_n = \frac{1}{|t|^n} \int_{\mathbb{R}^n} f^+ \, d\lambda_n, \qquad \int_{\mathbb{R}^n} (f_t)^- \, d\lambda_n = \frac{1}{|t|^n} \int_{\mathbb{R}^n} f^- \, d\lambda_n . \end{equation*}

Because \(0 < |t|^{-n} < \infty\), at least one of the two integrals on the left is finite; hence \(\int_{\mathbb{R}^n} f_t \, d\lambda_n\) is defined and

\begin{equation*} \int_{\mathbb{R}^n} f_t \, d\lambda_n = \int_{\mathbb{R}^n} (f_t)^+ \, d\lambda_n - \int_{\mathbb{R}^n} (f_t)^- \, d\lambda_n = \frac{1}{|t|^n} \Bigl( \int_{\mathbb{R}^n} f^+ \, d\lambda_n - \int_{\mathbb{R}^n} f^- \, d\lambda_n \Bigr) = \frac{1}{|t|^n} \int_{\mathbb{R}^n} f \, d\lambda_n , \end{equation*}

where the subtraction is legitimate because not both terms are \(\infty\). This completes the proof of (b).

First note that \(\lambda\) on \((\mathbb{R}, \mathcal{L})\) is a \(\sigma\)-finite measure, because \(\mathbb{R} = \bigcup_{k=1}^{\infty} (-k, k)\) and \(\lambda\bigl( (-k,k) \bigr) = 2k < \infty\) for each \(k \in \mathbb{Z}^+\). Hence the product measure \(\lambda \times \lambda\) on \((\mathbb{R}^2, \mathcal{L} \otimes \mathcal{L})\) is defined (5.25) and is a measure (5.27).

There exists a set \(V \subseteq \mathbb{R}\) that is not Lebesgue measurable. Indeed, 2.18 produces disjoint sets \(A, B \subseteq \mathbb{R}\) with \(|A \cup B| \ne |A| + |B|\), whereas outer measure is a measure on \((\mathbb{R}, \mathcal{L})\) by 2.72(b) and hence is finitely additive on disjoint sets in \(\mathcal{L}\); so \(A\) and \(B\) cannot both lie in \(\mathcal{L}\). (This is the observation made in the discussion following 2.72.) Fix such a \(V\), so \(V \notin \mathcal{L}\).

Define

\begin{equation*} F = \{0\} \times \mathbb{R} \qquad \text{and} \qquad E = \{0\} \times V . \end{equation*}

Clearly \(E \subseteq F\).

\(F\) is a measurable rectangle: \(\{0\} \in \mathcal{L}\) (indeed \(\{0\}\) is a Borel set, and every Borel set is Lebesgue measurable) and \(\mathbb{R} \in \mathcal{L}\). Hence \(F \in \mathcal{L} \otimes \mathcal{L}\).

Next we compute \((\lambda \times \lambda)(F)\). For \(x \in \mathbb{R}\) the cross section is

\begin{equation*} [F]_x = \begin{cases} \mathbb{R} & \text{if } x = 0, \\ \emptyset & \text{if } x \ne 0, \end{cases} \end{equation*}

so \(\lambda([F]_x) = \infty \cdot \chi_{\{0\}}(x)\). By 5.25,

\begin{equation*} (\lambda \times \lambda)(F) = \int_{\mathbb{R}} \int_{\mathbb{R}} \chi_F(x,y) \, d\lambda(y) \, d\lambda(x) = \int_{\mathbb{R}} \lambda\bigl( [F]_x \bigr) \, d\lambda(x) = \int_{\mathbb{R}} \infty \cdot \chi_{\{0\}} \, d\lambda = 0, \end{equation*}

the last equality holding because the integrand vanishes outside the set \(\{0\}\), which has \(\lambda\)-measure \(0\): for every \(c \in [0,\infty)\) we have \(\int c \chi_{\{0\}} \, d\lambda = c \lambda(\{0\}) = 0\), and \(\infty \cdot \chi_{\{0\}}\) is the increasing pointwise limit of \(k \chi_{\{0\}}\) as \(k \to \infty\), so the Monotone Convergence Theorem (3.11) gives \(\int \infty \cdot \chi_{\{0\}} \, d\lambda = \lim_{k \to \infty} 0 = 0\). (Equivalently, use 5.26: \((\lambda \times \lambda)(F) = \lambda(\{0\}) \cdot \lambda(\mathbb{R}) = 0 \cdot \infty = 0\).) Thus \(F\) satisfies the first bullet point.

Finally, suppose toward a contradiction that \(E \in \mathcal{L} \otimes \mathcal{L}\). By 5.6, every cross section of a set in \(\mathcal{L} \otimes \mathcal{L}\) is measurable; in particular \([E]_0 \in \mathcal{L}\). But

\begin{equation*} [E]_0 = \{ y \in \mathbb{R} : (0, y) \in E \} = V, \end{equation*}

so \(V \in \mathcal{L}\), contradicting the choice of \(V\). Hence \(E \notin \mathcal{L} \otimes \mathcal{L}\), and \(E\) satisfies the second bullet point.

Thus \(E \subseteq F\), \(F \in \mathcal{L} \otimes \mathcal{L}\) with \((\lambda \times \lambda)(F) = 0\), and yet \(E \notin \mathcal{L} \otimes \mathcal{L}\). In the terminology of the bracketed remark, the measure space \((\mathbb{R}^2, \mathcal{L} \otimes \mathcal{L}, \lambda \times \lambda)\) is not complete, even though \((\mathbb{R}, \mathcal{L}, \lambda)\) is.

Recall the setting of the proof of 5.41: a number \(t > 0\) is fixed, \(C_m\) denotes the open cube in \(\mathbb{R}^n\) centered at the origin with side length \(m\), and

\begin{equation*} \mathcal{E}_m = \{ E \in \mathcal{B}_n : E \subseteq C_m \text{ and } \lambda_n(tE) = t^n \lambda_n(E) \}. \end{equation*}

Recall also from the first paragraph of that proof that \(tE \in \mathcal{B}_n\) for every \(E \in \mathcal{B}_n\), so the condition \(\lambda_n(tE) = t^n \lambda_n(E)\) makes sense. By 5.15, we must show that \(\mathcal{E}_m\) is closed under countable increasing unions and under countable decreasing intersections.

Two elementary observations will be used repeatedly. First, because \(t > 0\), the map \(x \mapsto tx\) is a bijection of \(\mathbb{R}^n\) onto \(\mathbb{R}^n\), and therefore it carries unions to unions and intersections to intersections:

\begin{equation*} t \Bigl( \bigcup_{k=1}^{\infty} E_k \Bigr) = \bigcup_{k=1}^{\infty} (t E_k) \qquad \text{and} \qquad t \Bigl( \bigcap_{k=1}^{\infty} E_k \Bigr) = \bigcap_{k=1}^{\infty} (t E_k). \end{equation*}

(The inclusion \(\subseteq\) in the second equation holds for any map; the reverse inclusion uses injectivity of \(x \mapsto tx\).) Also, \(E \subseteq F\) implies \(tE \subseteq tF\).

Second, \(\lambda_n(C_m) = m^n < \infty\). Indeed \(C_m = (-\tfrac{m}{2}, \tfrac{m}{2})^n\), and by induction on \(n\), using \(\lambda_n = \lambda_{n-1} \times \lambda_1\) and 5.26,

\begin{equation*} \lambda_n(C_m) = \lambda_{n-1}\bigl( (-\tfrac{m}{2}, \tfrac{m}{2})^{n-1} \bigr) \cdot \lambda_1\bigl( (-\tfrac{m}{2}, \tfrac{m}{2}) \bigr) = m^{n-1} \cdot m = m^n , \end{equation*}

with the case \(n = 1\) holding because \(\lambda_1\) of an open interval is its length. Similarly \(tC_m\) is the open cube centered at the origin with side length \(tm\), so \(\lambda_n(t C_m) = ( tm)^n < \infty\).

Countable increasing unions. Suppose \(E_1 \subseteq E_2 \subseteq \cdots\) is an increasing sequence of sets in \(\mathcal{E}_m\), and let \(E = \bigcup_{k=1}^{\infty} E_k\). Then \(E \in \mathcal{B}_n\) (as \(\mathcal{B}_n\) is a \(\sigma\)-algebra) and \(E \subseteq C_m\) (as each \(E_k \subseteq C_m\)). Moreover \(tE_1 \subseteq tE_2 \subseteq \cdots\) is an increasing sequence in \(\mathcal{B}_n\) with union \(tE\). Hence, using 2.59 twice,

\begin{equation*} \lambda_n(tE) = \lim_{k \to \infty} \lambda_n(t E_k) = \lim_{k \to \infty} t^n \lambda_n(E_k) = t^n \lim_{k \to \infty} \lambda_n(E_k) = t^n \lambda_n(E), \end{equation*}

where the second equality holds because each \(E_k \in \mathcal{E}_m\). Therefore \(E \in \mathcal{E}_m\).

Countable decreasing intersections. Suppose \(E_1 \supseteq E_2 \supseteq \cdots\) is a decreasing sequence of sets in \(\mathcal{E}_m\), and let \(E = \bigcap_{k=1}^{\infty} E_k\). Then \(E \in \mathcal{B}_n\) and \(E \subseteq C_m\). Also \(tE_1 \supseteq tE_2 \supseteq \cdots\) is a decreasing sequence in \(\mathcal{B}_n\) with intersection \(tE\).

The finiteness hypothesis needed for 2.60 holds in both of the sequences we use: since \(E_1 \subseteq C_m\), we have \(\lambda_n(E_1) \le \lambda_n(C_m) = m^n < \infty\) by 2.57(a); and \(tE_1 \subseteq tC_m\), so \(\lambda_n(tE_1) \le \lambda_n(tC_m) = ( tm)^n < \infty\). This is exactly why the sets in \(\mathcal{E}_m\) are required to sit inside the bounded cube \(C_m\).

Hence, applying 2.60 to \(tE_1 \supseteq tE_2 \supseteq \cdots\) and then to \(E_1 \supseteq E_2 \supseteq \cdots\),

\begin{equation*} \lambda_n(tE) = \lim_{k \to \infty} \lambda_n(t E_k) = \lim_{k \to \infty} t^n \lambda_n(E_k) = t^n \lim_{k \to \infty} \lambda_n(E_k) = t^n \lambda_n(E), \end{equation*}

where again the second equality uses \(E_k \in \mathcal{E}_m\). Therefore \(E \in \mathcal{E}_m\).

We have shown that \(\mathcal{E}_m\) is closed under countable increasing unions and under countable decreasing intersections, so \(\mathcal{E}_m\) is a monotone class on \(C_m\), as claimed in the proof of 5.41.

Recall (5.43) that the open unit ball is

\begin{equation*} B_n = \{(x_1,\dots,x_n) \in \mathbb{R}^n : x_1{}^2 + \cdots + x_n{}^2 < 1\}, \end{equation*}

and that a subset \(G\) of \(\mathbb{R}^n\) is open if for every \(x \in G\) there exists \(\delta > 0\) with \(B(x,\delta) \subseteq G\), where \(B(x,\delta) = \{y \in \mathbb{R}^n : \|y - x\|_\infty < \delta\}\) and \(\|(y_1,\dots,y_n)\|_\infty = \max\{|y_1|,\dots,|y_n|\}\).

Write \(\|y\|_2 = (y_1{}^2 + \cdots + y_n{}^2)^{1/2}\), so that \(B_n = \{y \in \mathbb{R}^n : \|y\|_2 < 1\}\). Two standard facts about \(\|\cdot\|_2\) will be used.

First, the triangle inequality \(\|u + v\|_2 \le \|u\|_2 + \|v\|_2\). This follows from the Cauchy–Schwarz inequality \(\sum_{j=1}^n u_j v_j \le \|u\|_2 \|v\|_2\), because

\begin{equation*} \|u+v\|_2{}^2 = \|u\|_2{}^2 + 2\sum_{j=1}^n u_j v_j + \|v\|_2{}^2 \le \|u\|_2{}^2 + 2\|u\|_2\|v\|_2 + \|v\|_2{}^2 = (\|u\|_2 + \|v\|_2)^2 . \end{equation*}

Second, the comparison

\begin{equation*} \|y\|_2 = \Bigl(\sum_{j=1}^n y_j{}^2\Bigr)^{1/2} \le \Bigl(\sum_{j=1}^n \|y\|_\infty{}^2\Bigr)^{1/2} = \sqrt{n}\,\|y\|_\infty , \end{equation*}

which holds because \(|y_j| \le \|y\|_\infty\) for each \(j\).

Now suppose \(x \in B_n\). Then \(\|x\|_2 < 1\), so

\begin{equation*} r = 1 - \|x\|_2 > 0 . \end{equation*}

Set \(\delta = r/\sqrt{n} > 0\). Suppose \(y \in B(x,\delta)\), meaning \(\|y - x\|_\infty < \delta\). Then

\begin{equation*} \|y - x\|_2 \le \sqrt{n}\,\|y - x\|_\infty < \sqrt{n}\,\delta = r , \end{equation*}

and hence, by the triangle inequality applied to \(y = x + (y - x)\),

\begin{equation*} \|y\|_2 \le \|x\|_2 + \|y - x\|_2 < \|x\|_2 + r = 1 . \end{equation*}

Thus \(y \in B_n\). Since \(y\) was an arbitrary element of \(B(x,\delta)\), we have shown \(B(x,\delta) \subseteq B_n\).

Every point of \(B_n\) therefore lies in an open cube contained in \(B_n\), which means \(B_n\) is an open subset of \(\mathbb{R}^n\). In particular \(B_n\) is a Borel subset of \(\mathbb{R}^n\), as asserted after 5.43.

Throughout we use the identification of \(\mathbb{R}^m \times \mathbb{R}^n\) with \(\mathbb{R}^{m+n}\) and the identity

\begin{equation*} B(x,\delta) \times B(y,\delta) = B\bigl((x,y),\delta\bigr) \end{equation*}

for \(x \in \mathbb{R}^m\), \(y \in \mathbb{R}^n\), \(\delta > 0\), both recorded in the review at the start of Section 5C. This identity holds because \(\|(u,v)\|_\infty = \max\{\|u\|_\infty, \|v\|_\infty\}\) under that identification.

First suppose \(G_1\) is open in \(\mathbb{R}^m\) and \(G_2\) is open in \(\mathbb{R}^n\). Then \(G_1 \times G_2\) is open in \(\mathbb{R}^{m+n}\) by 5.36. (This direction does not use the hypothesis that \(G_1\) and \(G_2\) are nonempty.)

Now suppose \(G_1 \times G_2\) is an open subset of \(\mathbb{R}^{m+n}\). Because \(G_2 \neq \varnothing\), we may fix some \(y_0 \in G_2\).

To prove \(G_1\) is open, let \(x \in G_1\). Then \((x,y_0) \in G_1 \times G_2\), so openness of \(G_1 \times G_2\) gives \(\delta > 0\) with

\begin{equation*} B\bigl((x,y_0),\delta\bigr) \subseteq G_1 \times G_2 . \end{equation*}

Let \(u \in B(x,\delta)\). Since \(y_0 \in B(y_0,\delta)\), we get

\begin{equation*} (u,y_0) \in B(x,\delta) \times B(y_0,\delta) = B\bigl((x,y_0),\delta\bigr) \subseteq G_1 \times G_2 , \end{equation*}

and therefore \(u \in G_1\). Thus \(B(x,\delta) \subseteq G_1\). As \(x \in G_1\) was arbitrary, \(G_1\) is an open subset of \(\mathbb{R}^m\).

The proof that \(G_2\) is open is symmetric: because \(G_1 \neq \varnothing\) we may fix \(x_0 \in G_1\); given \(y \in G_2\), choose \(\delta > 0\) with \(B((x_0,y),\delta) \subseteq G_1 \times G_2\), and then for each \(v \in B(y,\delta)\) we have \((x_0,v) \in B(x_0,\delta) \times B(y,\delta) = B((x_0,y),\delta) \subseteq G_1 \times G_2\), so \(v \in G_2\). Hence \(B(y,\delta) \subseteq G_2\) and \(G_2\) is open in \(\mathbb{R}^n\).

The hypothesis that both sets are nonempty is genuinely needed for the second direction: if \(G_2 = \varnothing\) then \(G_1 \times G_2 = \varnothing\) is open no matter what \(G_1\) is, so no conclusion about \(G_1\) could be drawn.

As in the previous exercise we identify \(\mathbb{R}^m \times \mathbb{R}^n\) with \(\mathbb{R}^{m+n}\) and use \(B(x,\delta) \times B(y,\delta) = B((x,y),\delta)\). Recall that a subset of \(\mathbb{R}^k\) is closed exactly when its complement is open.

First suppose \(F_1\) is closed in \(\mathbb{R}^m\) and \(F_2\) is closed in \(\mathbb{R}^n\). A point \((x,y)\) fails to lie in \(F_1 \times F_2\) precisely when \(x \notin F_1\) or \(y \notin F_2\). Hence

\begin{equation*} \mathbb{R}^{m+n} \setminus (F_1 \times F_2) = \bigl[(\mathbb{R}^m \setminus F_1) \times \mathbb{R}^n\bigr] \cup \bigl[\mathbb{R}^m \times (\mathbb{R}^n \setminus F_2)\bigr]. \end{equation*}

The sets \(\mathbb{R}^m \setminus F_1\) and \(\mathbb{R}^n \setminus F_2\) are open by hypothesis, and \(\mathbb{R}^m\) and \(\mathbb{R}^n\) are open in themselves. Thus each of the two sets on the right side above is a product of an open subset of \(\mathbb{R}^m\) with an open subset of \(\mathbb{R}^n\), hence is open in \(\mathbb{R}^{m+n}\) by 5.36. A union of two open sets is open. Therefore \(\mathbb{R}^{m+n} \setminus (F_1 \times F_2)\) is open, which means \(F_1 \times F_2\) is closed. (This direction does not use nonemptiness.)

Now suppose \(F_1 \times F_2\) is a closed subset of \(\mathbb{R}^{m+n}\); thus \(\mathbb{R}^{m+n} \setminus (F_1 \times F_2)\) is open. Because \(F_2 \neq \varnothing\), fix \(y_0 \in F_2\).

To prove \(F_1\) is closed we show \(\mathbb{R}^m \setminus F_1\) is open. Let \(x \in \mathbb{R}^m \setminus F_1\). Then \((x,y_0) \notin F_1 \times F_2\), so there exists \(\delta > 0\) with

\begin{equation*} B\bigl((x,y_0),\delta\bigr) \subseteq \mathbb{R}^{m+n} \setminus (F_1 \times F_2). \end{equation*}

Let \(u \in B(x,\delta)\). Then \((u,y_0) \in B(x,\delta) \times B(y_0,\delta) = B((x,y_0),\delta)\), so \((u,y_0) \notin F_1 \times F_2\). Since \(y_0 \in F_2\), this forces \(u \notin F_1\). Hence \(B(x,\delta) \subseteq \mathbb{R}^m \setminus F_1\). As \(x\) was an arbitrary point of \(\mathbb{R}^m \setminus F_1\), that complement is open and so \(F_1\) is closed in \(\mathbb{R}^m\).

The argument that \(F_2\) is closed is symmetric, using a fixed \(x_0 \in F_1\) (available since \(F_1 \neq \varnothing\)): given \(y \in \mathbb{R}^n \setminus F_2\), choose \(\delta > 0\) with \(B((x_0,y),\delta) \cap (F_1 \times F_2) = \varnothing\); for \(v \in B(y,\delta)\) we have \((x_0,v) \in B((x_0,y),\delta)\), so \((x_0,v) \notin F_1 \times F_2\), and since \(x_0 \in F_1\) this gives \(v \notin F_2\). Hence \(B(y,\delta) \subseteq \mathbb{R}^n \setminus F_2\), proving \(F_2\) closed.

Again nonemptiness is needed only for the second direction: \(\varnothing \times F_2 = \varnothing\) is closed regardless of \(F_2\).

The set \(A\) is the projection of \(E\) onto the first factor. As before we identify \(\mathbb{R}^m \times \mathbb{R}^n\) with \(\mathbb{R}^{m+n}\) and use \(B(x,\delta) \times B(y,\delta) = B((x,y),\delta)\).

(a) Suppose \(E\) is open and let \(x \in A\). By the definition of \(A\) there exists \(y \in \mathbb{R}^n\) with \((x,y) \in E\). Because \(E\) is open, there exists \(\delta > 0\) with

\begin{equation*} B\bigl((x,y),\delta\bigr) = B(x,\delta) \times B(y,\delta) \subseteq E . \end{equation*}

Now let \(u \in B(x,\delta)\). Since \(y \in B(y,\delta)\), we have \((u,y) \in B(x,\delta) \times B(y,\delta) \subseteq E\). So \(y\) witnesses that \(u \in A\). Hence \(B(x,\delta) \subseteq A\).

Every point of \(A\) is thus the center of an open cube contained in \(A\), so \(A\) is an open subset of \(\mathbb{R}^m\).

(b) The statement is false; here is a counterexample. Take \(m = n = 1\) and

\begin{equation*} E = \{(x,y) \in \mathbb{R}^2 : xy = 1\}, \end{equation*}

the hyperbola.

\(E\) is closed. To see this directly, suppose \((x_0,y_0) \notin E\), so \(x_0 y_0 \neq 1\). Put \(\varepsilon = |x_0 y_0 - 1| > 0\) and choose \(\delta > 0\) small enough that \(\delta \le 1\) and \(\delta(|x_0| + |y_0| + 1) < \varepsilon\). If \(\|(x,y) - (x_0,y_0)\|_\infty < \delta\), then \(|x - x_0| < \delta\), \(|y - y_0| < \delta\), and \(|y| < |y_0| + \delta \le |y_0| + 1\), so

\begin{equation*} |xy - x_0 y_0| \le |y|\,|x - x_0| + |x_0|\,|y - y_0| < (|y_0| + 1)\delta + |x_0|\delta = \delta(|x_0| + |y_0| + 1) < \varepsilon . \end{equation*}

Hence \(xy \neq 1\), i.e. \((x,y) \notin E\). Thus \(B((x_0,y_0),\delta) \subseteq \mathbb{R}^2 \setminus E\), so \(\mathbb{R}^2 \setminus E\) is open and \(E\) is closed. (Equivalently: \(E\) is the inverse image of the closed set \(\{1\}\) under the continuous map \((x,y) \mapsto xy\).)

Now compute \(A\). If \(x \neq 0\) then \((x, 1/x) \in E\), so \(x \in A\). If \(x = 0\) then \(xy = 0 \neq 1\) for every \(y\), so \(0 \notin A\). Therefore

\begin{equation*} A = \mathbb{R} \setminus \{0\}, \end{equation*}

whose complement \(\{0\}\) is not open in \(\mathbb{R}\) (every open cube centered at \(0\) contains nonzero points). Hence \(A\) is not closed.

The same phenomenon occurs for arbitrary \(m\) and \(n\): take

\begin{equation*} E = \{(x,y) \in \mathbb{R}^m \times \mathbb{R}^n : x_1 y_1 = 1,\ x_2 = \cdots = x_m = 0,\ y_2 = \cdots = y_n = 0\}, \end{equation*}

which is an intersection of finitely many closed sets and hence closed. Indeed \(\{(x,y) : x_1 y_1 = 1\}\) is closed in \(\mathbb{R}^{m+n}\) by exactly the estimate above, because \(\|(x,y) - (x^0,y^0)\|_\infty < \delta\) forces both \(|x_1 - x_1{}^0| < \delta\) and \(|y_1 - y_1{}^0| < \delta\); each set \(\{(x,y) : x_j = 0\}\) is closed because a point with \(x_j \neq 0\) has the open cube \(B((x,y), |x_j|)\) inside the complement, and similarly for each \(\{(x,y) : y_k = 0\}\); and a finite intersection of closed sets is closed because its complement is a finite union of open sets. Then

\begin{equation*} A = \{x \in \mathbb{R}^m : x_1 \neq 0,\ x_2 = \cdots = x_m = 0\}, \end{equation*}

which is not closed: the origin lies in \(\mathbb{R}^m \setminus A\), yet every open cube \(B(0,\delta)\) contains the point \((\delta/2, 0,\dots,0) \in A\), so \(\mathbb{R}^m \setminus A\) is not open.

The key tool is the recursion established in the course of proving 5.44: for every integer \(n > 2\),

\begin{equation*} \lambda_n(B_n) = \frac{2\pi}{n}\,\lambda_{n-2}(B_{n-2}). \end{equation*}

We also use the values supplied by 5.44:

\begin{equation*} \lambda_1(B_1) = 2,\quad \lambda_2(B_2) = \pi,\quad \lambda_3(B_3) = \frac{4\pi}{3},\quad \lambda_4(B_4) = \frac{\pi^2}{2},\quad \lambda_5(B_5) = \frac{8\pi^2}{15},\quad \lambda_6(B_6) = \frac{\pi^3}{6}. \end{equation*}

(For example \(\lambda_5(B_5) = 2^{3}\pi^{2}/(1 \cdot 3 \cdot 5) = 8\pi^2/15\) from the odd case of 5.44, and \(\lambda_6(B_6) = \pi^{3}/3! = \pi^3/6\) from the even case.)

(a) Since \(\pi < 3.5\) we have \(2\pi < 7\), so

\begin{equation*} r := \frac{2\pi}{7} < 1 . \end{equation*}

If \(n \ge 5\), then \(n + 2 \ge 7\), so the recursion gives

\begin{equation*} \lambda_{n+2}(B_{n+2}) = \frac{2\pi}{n+2}\,\lambda_n(B_n) \le \frac{2\pi}{7}\,\lambda_n(B_n) = r\,\lambda_n(B_n). \end{equation*}

Iterating this inequality starting from \(n = 5\) and from \(n = 6\) yields, for every integer \(j \ge 0\),

\begin{equation*} \lambda_{5+2j}(B_{5+2j}) \le r^{\,j}\,\lambda_5(B_5) \qquad\text{and}\qquad \lambda_{6+2j}(B_{6+2j}) \le r^{\,j}\,\lambda_6(B_6). \end{equation*}

Let \(M = \max\{\lambda_5(B_5), \lambda_6(B_6)\}\). Every integer \(n \ge 5\) equals \(5 + 2j\) or \(6 + 2j\) for the integer \(j = \lfloor (n-5)/2 \rfloor\), and \(j \ge (n-6)/2\). Hence for all \(n \ge 5\),

\begin{equation*} 0 \le \lambda_n(B_n) \le M\,r^{\,(n-6)/2}. \end{equation*}

Because \(0 < r < 1\), we have \(r^{(n-6)/2} \to 0\) as \(n \to \infty\). The squeeze theorem now gives

\begin{equation*} \lim_{n \to \infty} \lambda_n(B_n) = 0 . \end{equation*}

(b) The answer is \(n = 5\), where \(\lambda_5(B_5) = 8\pi^2/15 \approx 5.26\).

First, \(\lambda_n(B_n)\) is strictly decreasing for \(n \ge 5\) within each parity class: if \(n \ge 5\), then \(n + 2 \ge 7 > 2\pi\), so

\begin{equation*} \lambda_{n+2}(B_{n+2}) = \frac{2\pi}{n+2}\,\lambda_n(B_n) < \lambda_n(B_n), \end{equation*}

using \(\lambda_n(B_n) > 0\). Consequently

\begin{equation*} \lambda_5(B_5) > \lambda_7(B_7) > \lambda_9(B_9) > \cdots \qquad\text{and}\qquad \lambda_6(B_6) > \lambda_8(B_8) > \lambda_{10}(B_{10}) > \cdots . \end{equation*}

So \(\sup_n \lambda_n(B_n) = \max\{\lambda_1(B_1), \lambda_2(B_2), \lambda_3(B_3), \lambda_4(B_4), \lambda_5(B_5), \lambda_6(B_6)\}\), and it suffices to compare those six numbers.

Using \(3 < \pi < 3.2\):

\(\lambda_1(B_1) = 2 < \pi = \lambda_2(B_2)\).

\(\lambda_2(B_2) = \pi < \tfrac{4}{3}\pi = \lambda_3(B_3)\).

\(\lambda_3(B_3) < \lambda_4(B_4)\) because \(\frac{\pi^2/2}{4\pi/3} = \frac{3\pi}{8} > \frac{3 \cdot 3}{8} > 1\).

\(\lambda_4(B_4) < \lambda_5(B_5)\) because \(\frac{8\pi^2/15}{\pi^2/2} = \frac{16}{15} > 1\).

\(\lambda_6(B_6) < \lambda_5(B_5)\) because \(\frac{8\pi^2/15}{\pi^3/6} = \frac{48}{15\pi} = \frac{16}{5\pi}\), and \(5\pi < 16\) since \(\pi < 3.2\).

Thus the largest of the six is \(\lambda_5(B_5)\), and therefore \(\lambda_n(B_n)\) attains its maximum over all positive integers \(n\) uniquely at \(n = 5\), with value \(8\pi^2/15\).

It is worth noting the contrast with the smallest cube containing \(B_n\), which has \(n\)-dimensional Lebesgue measure \(2^n \to \infty\). So the ratio \(\lambda_n(B_n)/2^n\) tends to \(0\) very fast: in high dimensions essentially all of the cube lies outside the inscribed ball.

Recall the two standard properties of the gamma function \(\Gamma(x) = \int_0^\infty t^{x-1}e^{-t}\,dt\) for \(x > 0\) that we need:

\begin{equation*} \Gamma(x+1) = x\,\Gamma(x) \quad \text{for } x > 0, \qquad \Gamma(1) = 1, \qquad \Gamma\bigl(\tfrac12\bigr) = \sqrt{\pi}. \end{equation*}

We argue by induction on \(n\), in steps of \(2\), using the recursion derived in the proof of 5.44: for every integer \(n > 2\),

\begin{equation*} \lambda_n(B_n) = \frac{2\pi}{n}\,\lambda_{n-2}(B_{n-2}). \end{equation*}

Base case \(n = 1\). Here \(\lambda_1(B_1) = 2\), since \(B_1 = (-1,1)\). On the other side,

\begin{equation*} \Gamma\bigl(\tfrac12 + 1\bigr) = \tfrac12\,\Gamma\bigl(\tfrac12\bigr) = \frac{\sqrt{\pi}}{2}, \end{equation*}

so

\begin{equation*} \frac{\pi^{1/2}}{\Gamma\bigl(\frac12 + 1\bigr)} = \frac{\sqrt{\pi}}{\sqrt{\pi}/2} = 2 = \lambda_1(B_1). \end{equation*}

Base case \(n = 2\). Here \(\lambda_2(B_2) = \pi\), and

\begin{equation*} \frac{\pi^{2/2}}{\Gamma\bigl(\frac{2}{2} + 1\bigr)} = \frac{\pi}{\Gamma(2)} = \frac{\pi}{1 \cdot \Gamma(1)} = \pi = \lambda_2(B_2). \end{equation*}

Induction step. Let \(n > 2\) be an integer and suppose the formula holds for \(n - 2\) (which is a positive integer, so the induction hypothesis applies, and every positive integer is reached from \(n = 1\) or \(n = 2\) by repeatedly adding \(2\)). Then, by the recursion and the induction hypothesis,

\begin{equation*} \lambda_n(B_n) = \frac{2\pi}{n} \cdot \frac{\pi^{(n-2)/2}}{\Gamma\bigl(\frac{n-2}{2} + 1\bigr)} = \frac{2\pi \cdot \pi^{(n-2)/2}}{n\,\Gamma\bigl(\frac{n}{2}\bigr)} = \frac{2\,\pi^{n/2}}{n\,\Gamma\bigl(\frac{n}{2}\bigr)} . \end{equation*}

Now apply the functional equation with \(x = n/2 > 0\):

\begin{equation*} \Gamma\Bigl(\frac{n}{2} + 1\Bigr) = \frac{n}{2}\,\Gamma\Bigl(\frac{n}{2}\Bigr), \qquad\text{so}\qquad n\,\Gamma\Bigl(\frac{n}{2}\Bigr) = 2\,\Gamma\Bigl(\frac{n}{2} + 1\Bigr). \end{equation*}

Substituting gives

\begin{equation*} \lambda_n(B_n) = \frac{2\,\pi^{n/2}}{2\,\Gamma\bigl(\frac{n}{2} + 1\bigr)} = \frac{\pi^{n/2}}{\Gamma\bigl(\frac{n}{2} + 1\bigr)}, \end{equation*}

completing the induction.

Hence \(\lambda_n(B_n) = \pi^{n/2}/\Gamma(\frac{n}{2}+1)\) for every positive integer \(n\).

As a consistency check, this single formula reproduces both cases of 5.44. If \(n\) is even, then \(\frac{n}{2}\) is a positive integer and repeated use of \(\Gamma(x+1) = x\Gamma(x)\) together with \(\Gamma(1) = 1\) gives \(\Gamma(\frac{n}{2}+1) = (\frac{n}{2})!\), so \(\lambda_n(B_n) = \pi^{n/2}/(n/2)!\). If \(n\) is odd, write \(n = 2k+1\); then

\begin{equation*} \Gamma\Bigl(\frac{n}{2}+1\Bigr) = \frac{2k+1}{2}\cdot\frac{2k-1}{2}\cdots\frac{3}{2}\cdot\frac{1}{2}\cdot\Gamma\Bigl(\frac12\Bigr) = \frac{1 \cdot 3 \cdot 5 \cdots n}{2^{(n+1)/2}}\sqrt{\pi}, \end{equation*}

so

\begin{equation*} \lambda_n(B_n) = \frac{\pi^{n/2}\,2^{(n+1)/2}}{1 \cdot 3 \cdot 5 \cdots n \cdot \sqrt{\pi}} = \frac{2^{(n+1)/2}\pi^{(n-1)/2}}{1 \cdot 3 \cdot 5 \cdots n}, \end{equation*}

matching the odd case of 5.44.

Write \(G = \mathbb{R}^2 \setminus \{(0,0)\}\), an open subset of \(\mathbb{R}^2\). On \(G\) we have

\begin{equation*} f(x,y) = \frac{x^3 y - x y^3}{x^2 + y^2}, \end{equation*}

a quotient of polynomials whose denominator does not vanish. Hence \(f\) is infinitely differentiable on \(G\), and in particular all the partial derivatives appearing below exist and are continuous at every point of \(G\).

Step 1: the first-order partial derivatives on all of \(\mathbb{R}^2\).

On \(G\), the quotient rule gives

\begin{equation*} (D_1 f)(x,y) = \frac{(3x^2y - y^3)(x^2+y^2) - (x^3y - xy^3)\cdot 2x}{(x^2+y^2)^2} = \frac{x^4 y + 4x^2 y^3 - y^5}{(x^2+y^2)^2} \end{equation*}

and

\begin{equation*} (D_2 f)(x,y) = \frac{(x^3 - 3xy^2)(x^2+y^2) - (x^3y - xy^3)\cdot 2y}{(x^2+y^2)^2} = \frac{x^5 - 4x^3 y^2 - x y^4}{(x^2+y^2)^2}. \end{equation*}

(Expanding the numerators: \((3x^2y-y^3)(x^2+y^2) = 3x^4y + 2x^2y^3 - y^5\) and \(2x(x^3y-xy^3) = 2x^4y - 2x^2y^3\), whose difference is \(x^4y + 4x^2y^3 - y^5\); similarly \((x^3-3xy^2)(x^2+y^2) = x^5 - 2x^3y^2 - 3xy^4\) and \(2y(x^3y-xy^3) = 2x^3y^2 - 2xy^4\), whose difference is \(x^5 - 4x^3y^2 - xy^4\).)

At the origin, note that \(f(t,0) = 0\) and \(f(0,t) = 0\) for every \(t \in \mathbb{R}\) (this is immediate from the formula when \(t \neq 0\), and from the definition when \(t = 0\)). Hence

\begin{equation*} (D_1 f)(0,0) = \lim_{t \to 0} \frac{f(t,0) - f(0,0)}{t} = 0, \qquad (D_2 f)(0,0) = \lim_{t \to 0} \frac{f(0,t) - f(0,0)}{t} = 0 . \end{equation*}

So \(D_1 f\) and \(D_2 f\) exist at every point of \(\mathbb{R}^2\).

Step 2: the restrictions of \(D_1 f\) and \(D_2 f\) to the axes.

For \(y \neq 0\) the formula for \(D_1 f\) gives

\begin{equation*} (D_1 f)(0,y) = \frac{0 + 0 - y^5}{(y^2)^2} = -y, \end{equation*}

and \((D_1 f)(0,0) = 0 = -0\). Hence

\begin{equation*} (D_1 f)(0,y) = -y \quad \text{for all } y \in \mathbb{R}. \end{equation*}

Likewise, for \(x \neq 0\),

\begin{equation*} (D_2 f)(x,0) = \frac{x^5 - 0 - 0}{(x^2)^2} = x, \end{equation*}

and \((D_2 f)(0,0) = 0\), so

\begin{equation*} (D_2 f)(x,0) = x \quad \text{for all } x \in \mathbb{R}. \end{equation*}

(a) Existence of the two mixed partial derivatives everywhere.

On \(G\) this is immediate: \(D_1 f\) and \(D_2 f\) agree there with the rational functions displayed in Step 1, which are infinitely differentiable on \(G\); hence \(D_1(D_2 f)\) and \(D_2(D_1 f)\) exist at every point of \(G\). (Explicitly, differentiating the formula for \(D_2 f\) with respect to \(x\) and simplifying gives, for \((x,y) \in G\),

\begin{equation*} \bigl(D_1(D_2 f)\bigr)(x,y) = \frac{(5x^4 - 12x^2y^2 - y^4)(x^2+y^2) - 4x^2(x^4 - 4x^2y^2 - y^4)}{(x^2+y^2)^3} = \frac{x^6 + 9x^4 y^2 - 9x^2 y^4 - y^6}{(x^2+y^2)^3}, \end{equation*}

and the same expression is obtained for \(D_2(D_1 f)\) on \(G\), as it must be by 5.48 applied on the open set \(G\), where \(f\) is smooth.)

At the origin, by definition

\begin{equation*} \bigl(D_1(D_2 f)\bigr)(0,0) = \lim_{t \to 0} \frac{(D_2 f)(t,0) - (D_2 f)(0,0)}{t}, \end{equation*}

which makes sense because \(D_2 f\) exists everywhere by Step 1. By Step 2, \((D_2 f)(t,0) = t\), so this limit equals \(\lim_{t \to 0} t/t = 1\); in particular it exists. Similarly

\begin{equation*} \bigl(D_2(D_1 f)\bigr)(0,0) = \lim_{t \to 0} \frac{(D_1 f)(0,t) - (D_1 f)(0,0)}{t} = \lim_{t \to 0} \frac{-t}{t} = -1, \end{equation*}

which also exists. Thus \(D_1(D_2 f)\) and \(D_2(D_1 f)\) exist at every point of \(\mathbb{R}^2\).

(b) From the computation just performed,

\begin{equation*} \bigl(D_1(D_2 f)\bigr)(0,0) = 1 \neq -1 = \bigl(D_2(D_1 f)\bigr)(0,0). \end{equation*}

(c) Result 5.48 requires more than the mere existence of \(D_1 f\), \(D_2 f\), \(D_1(D_2 f)\), and \(D_2(D_1 f)\) on the open set \(G\) in its statement: it requires all four of these functions to be continuous on that open set. For our \(f\), at least one of \(D_1(D_2 f)\), \(D_2(D_1 f)\) fails to be continuous at the origin, so 5.48 simply does not apply on any open set containing \((0,0)\).

Here is a proof of that failure that uses no computation beyond part (b). On the open set \(G\) the function \(f\) is infinitely differentiable, so all four functions \(D_1 f\), \(D_2 f\), \(D_1(D_2 f)\), \(D_2(D_1 f)\) are continuous on \(G\), and 5.48 applies there to give

\begin{equation*} D_1(D_2 f) = D_2(D_1 f) \quad \text{on } G . \end{equation*}

If both mixed partials were also continuous at \((0,0)\), then taking the limit as \((x,y) \to (0,0)\) within \(G\) of the two equal functions would give \(\bigl(D_1(D_2 f)\bigr)(0,0) = \bigl(D_2(D_1 f)\bigr)(0,0)\), contradicting (b). Hence continuity of the mixed partials at the origin fails, and the hypotheses of 5.48 are not satisfied on any open set containing the origin.

One can also see the discontinuity explicitly from the formula in (a). For \(x \neq 0\),

\begin{equation*} \bigl(D_1(D_2 f)\bigr)(x,0) = \frac{x^6}{x^6} = 1, \qquad\text{while for } y \neq 0, \quad \bigl(D_1(D_2 f)\bigr)(0,y) = \frac{-y^6}{y^6} = -1 . \end{equation*}

So \(D_1(D_2 f)\) takes the value \(1\) at points arbitrarily close to the origin along the horizontal axis and the value \(-1\) at points arbitrarily close to the origin along the vertical axis; since \(\bigl(D_1(D_2 f)\bigr)(0,0) = 1\), no limit at the origin exists and \(D_1(D_2 f)\) is discontinuous there. The same computation shows \(D_2(D_1 f)\), which equals the same rational function on \(G\) but has value \(-1\) at the origin, is discontinuous there as well.

This example is exactly the one referred to in the marginal comment preceding 5.48: the continuity hypotheses in 5.48 can be weakened somewhat, but they cannot be eliminated.

Recall (6.1) that a function \(d \colon V \times V \to [0,\infty)\) is a metric provided that \(d(f,f) = 0\) for all \(f\), that \(d(f,g) = 0\) implies \(f = g\), that \(d(f,g) = d(g,f)\), and that \(d(f,h) \le d(f,g) + d(g,h)\). We check these four conditions for each of the five examples in 6.2. In every case the first three conditions are nearly immediate; the content is in checking that \(d\) really takes values in \([0,\infty)\) (finiteness) and in the triangle inequality.

The discrete metric. Suppose \(V\) is a nonempty set and \(d(f,g) = 1\) if \(f \ne g\) and \(d(f,g) = 0\) if \(f = g\). Then \(d\) takes values in \(\{0,1\} \subseteq [0,\infty)\). By definition \(d(f,f) = 0\), and if \(d(f,g) = 0\) then we cannot have \(f \ne g\) (which would force \(d(f,g) = 1\)), so \(f = g\). Because the relation \(f \ne g\) is symmetric in \(f\) and \(g\), we have \(d(f,g) = d(g,f)\). For the triangle inequality, suppose \(f, g, h \in V\). If \(f = h\) then \(d(f,h) = 0 \le d(f,g) + d(g,h)\) because \(d\) is nonnegative. If \(f \ne h\), then \(g\) cannot equal both \(f\) and \(h\); hence at least one of \(d(f,g)\), \(d(g,h)\) equals \(1\), and thus

\begin{equation*} d(f,h) = 1 \le d(f,g) + d(g,h). \end{equation*}

The absolute-value metric on \(\mathbb{R}\). Here \(d(x,y) = |x-y| \in [0,\infty)\). We have \(d(x,x) = |x - x| = 0\); if \(|x - y| = 0\) then \(x - y = 0\), so \(x = y\); and \(|x-y| = |-(x-y)| = |y-x|\). For \(x,y,z \in \mathbb{R}\), the triangle inequality for absolute value gives

\begin{equation*} d(x,z) = |x - z| = |(x-y) + (y-z)| \le |x-y| + |y-z| = d(x,y) + d(y,z). \end{equation*}

The max metric on \(\mathbb{R}^n\). For \(x = (x_1,\dots,x_n)\) and \(y = (y_1,\dots,y_n)\) in \(\mathbb{R}^n\), set \(d(x,y) = \max\{|x_1-y_1|,\dots,|x_n-y_n|\}\). This is a maximum of finitely many nonnegative real numbers, hence is a well-defined element of \([0,\infty)\). Clearly \(d(x,x) = \max\{0,\dots,0\} = 0\). If \(d(x,y) = 0\) then \(|x_j - y_j| \le 0\) for each \(j \in \{1,\dots,n\}\), so \(x_j = y_j\) for each \(j\) and thus \(x = y\). Symmetry holds because \(|x_j - y_j| = |y_j - x_j|\) for each \(j\). For the triangle inequality, let \(x, y, z \in \mathbb{R}^n\) and let \(j \in \{1,\dots,n\}\). Then

\begin{equation*} |x_j - z_j| \le |x_j - y_j| + |y_j - z_j| \le d(x,y) + d(y,z), \end{equation*}

where the second inequality holds because each summand is at most the corresponding maximum. The right side does not depend on \(j\), so taking the maximum over \(j \in \{1,\dots,n\}\) on the left gives \(d(x,z) \le d(x,y) + d(y,z)\).

The sup metric on \(C([0,1])\). Here \(C([0,1])\) is the set of continuous real-valued functions on \([0,1]\) and \(d(f,g) = \sup\{|f(t) - g(t)| : t \in [0,1]\}\). First we check that \(d(f,g) \in [0,\infty)\). If \(f, g \in C([0,1])\) then \(|f - g|\) is a continuous real-valued function on the closed bounded interval \([0,1]\) and hence is bounded; thus the set \(\{|f(t)-g(t)| : t \in [0,1]\}\) is a nonempty bounded set of nonnegative numbers and its supremum is a nonnegative real number. Next, \(d(f,f) = \sup\{0\} = 0\). If \(d(f,g) = 0\) then \(|f(t) - g(t)| \le 0\) for every \(t \in [0,1]\), so \(f(t) = g(t)\) for every \(t \in [0,1]\), which means \(f = g\). Symmetry holds because \(|f(t) - g(t)| = |g(t) - f(t)|\) for each \(t\). For the triangle inequality, suppose \(f, g, h \in C([0,1])\) and \(t \in [0,1]\). Then

\begin{equation*} |f(t) - h(t)| \le |f(t) - g(t)| + |g(t) - h(t)| \le d(f,g) + d(g,h). \end{equation*}

Thus \(d(f,g) + d(g,h)\) is an upper bound for \(\{|f(t)-h(t)| : t \in [0,1]\}\), and therefore \(d(f,h) \le d(f,g) + d(g,h)\).

The \(\ell^1\) metric. Here \(\ell^1\) is the set of sequences \((a_1,a_2,\dots)\) of real numbers with \(\sum_{k=1}^{\infty} |a_k| < \infty\), and

\begin{equation*} d\bigl((a_1,a_2,\dots),(b_1,b_2,\dots)\bigr) = \sum_{k=1}^{\infty} |a_k - b_k|. \end{equation*}

To see that \(d\) takes values in \([0,\infty)\), note that all terms are nonnegative and that \(|a_k - b_k| \le |a_k| + |b_k|\) for each \(k\), so

\begin{equation*} \sum_{k=1}^{\infty} |a_k - b_k| \le \sum_{k=1}^{\infty} |a_k| + \sum_{k=1}^{\infty} |b_k| < \infty \end{equation*}

whenever \((a_k)\) and \((b_k)\) are in \(\ell^1\). Clearly \(d(a,a) = 0\). If \(d(a,b) = 0\) then, because every term of the series is nonnegative, \(|a_k - b_k| = 0\) for every \(k\); hence \(a_k = b_k\) for every \(k\) and the two sequences are equal. Symmetry is immediate from \(|a_k - b_k| = |b_k - a_k|\). Finally, suppose \(a, b, c \in \ell^1\). For every \(k\) we have \(|a_k - c_k| \le |a_k - b_k| + |b_k - c_k|\), so for every \(n \in \mathbb{Z}^+\) the partial sums satisfy

\begin{equation*} \sum_{k=1}^{n} |a_k - c_k| \le \sum_{k=1}^{n} |a_k - b_k| + \sum_{k=1}^{n} |b_k - c_k| \le d(a,b) + d(b,c), \end{equation*}

where the last inequality holds because each of the two series has nonnegative terms and hence has partial sums bounded by its sum. Letting \(n \to \infty\) gives

\begin{equation*} d(a,c) = \sum_{k=1}^{\infty} |a_k - c_k| \le d(a,b) + d(b,c). \end{equation*}

Thus each of the five functions in 6.2 is a metric on the indicated set.

Suppose \((V,d)\) is a metric space and \(E \subseteq V\) is finite. By 6.6 we must show that \(V \setminus E\) is an open subset of \(V\).

First suppose \(E = \emptyset\). Then \(V \setminus E = V\), and \(V\) is open because \(B(f,1) \subseteq V\) for every \(f \in V\). Hence \(\emptyset\) is closed.

Now suppose \(E \ne \emptyset\), say \(E = \{f_1,\dots,f_n\}\) with \(n \in \mathbb{Z}^+\). Suppose \(g \in V \setminus E\). Then \(g \ne f_j\) for each \(j \in \{1,\dots,n\}\). The second bullet point in 6.1 (in contrapositive form) shows that \(d(g,f_j) \ne 0\) for each such \(j\); because \(d\) takes values in \([0,\infty)\), this means \(d(g,f_j) > 0\) for each \(j \in \{1,\dots,n\}\). Let

\begin{equation*} r = \min\{d(g,f_1),\dots,d(g,f_n)\}. \end{equation*}

As the minimum of a finite nonempty collection of positive numbers, \(r > 0\).

We claim \(B(g,r) \subseteq V \setminus E\). Indeed, if some \(h \in B(g,r)\) were an element of \(E\), then \(h = f_j\) for some \(j \in \{1,\dots,n\}\), and hence

\begin{equation*} d(g,f_j) = d(g,h) < r \le d(g,f_j), \end{equation*}

a contradiction. Thus \(B(g,r) \cap E = \emptyset\), which is to say \(B(g,r) \subseteq V \setminus E\).

Because every \(g \in V \setminus E\) is the center of an open ball contained in \(V \setminus E\), the set \(V \setminus E\) is open (6.4). Hence \(E\) is closed.

Suppose \((V,d)\) is a metric space, \(f \in V\), and \(r > 0\). Recall from 6.3 that

\begin{equation*} \overline{B}(f,r) = \{g \in V : d(f,g) \le r\}. \end{equation*}

By 6.6 we must show that \(V \setminus \overline{B}(f,r)\) is open.

Suppose \(h \in V \setminus \overline{B}(f,r)\), which means \(d(f,h) > r\). Let

\begin{equation*} s = d(f,h) - r, \end{equation*}

so \(s > 0\). We claim that \(B(h,s) \subseteq V \setminus \overline{B}(f,r)\).

To see this, suppose \(g \in B(h,s)\), so \(d(h,g) < s\). The triangle inequality (the last bullet point of 6.1), together with the symmetry of \(d\), gives

\begin{equation*} d(f,h) \le d(f,g) + d(g,h) < d(f,g) + s. \end{equation*}

Hence

\begin{equation*} d(f,g) > d(f,h) - s = d(f,h) - \bigl(d(f,h) - r\bigr) = r, \end{equation*}

so \(g \notin \overline{B}(f,r)\). This proves the claim.

Thus every element of \(V \setminus \overline{B}(f,r)\) is the center of an open ball contained in \(V \setminus \overline{B}(f,r)\), so \(V \setminus \overline{B}(f,r)\) is open by 6.4. Therefore \(\overline{B}(f,r)\) is a closed subset of \(V\).

Throughout, \(d\) denotes the metric on \(V\), and we use the definition of open set given by 6.4: a subset \(G\) of \(V\) is open if for every \(f \in G\) there exists \(r > 0\) with \(B(f,r) \subseteq G\).

(a) Suppose \(\mathcal{A}\) is a collection of open subsets of \(V\) and let

\begin{equation*} G = \bigcup_{A \in \mathcal{A}} A . \end{equation*}

Suppose \(f \in G\). Then there exists \(A \in \mathcal{A}\) with \(f \in A\). Because \(A\) is open, there exists \(r > 0\) such that \(B(f,r) \subseteq A\). Since \(A \subseteq G\), we conclude that \(B(f,r) \subseteq G\). As \(f\) was an arbitrary element of \(G\), this shows that \(G\) is open.

(Note that if \(\mathcal{A} = \emptyset\) then \(G = \emptyset\) and the condition in 6.4 holds vacuously, so the empty set is open; the argument above covers this case automatically.)

(b) Suppose \(G_1,\dots,G_n\) are open subsets of \(V\), where \(n \in \mathbb{Z}^+\), and let

\begin{equation*} G = G_1 \cap \cdots \cap G_n . \end{equation*}

Suppose \(f \in G\). Then \(f \in G_j\) for each \(j \in \{1,\dots,n\}\), so for each such \(j\) the openness of \(G_j\) provides \(r_j > 0\) with \(B(f,r_j) \subseteq G_j\). Let

\begin{equation*} r = \min\{r_1,\dots,r_n\}, \end{equation*}

which is positive because it is the minimum of finitely many positive numbers. If \(g \in B(f,r)\) and \(j \in \{1,\dots,n\}\), then \(d(f,g) < r \le r_j\), so \(g \in B(f,r_j) \subseteq G_j\). Hence \(B(f,r) \subseteq G_j\) for every \(j\), and therefore \(B(f,r) \subseteq G\). As \(f\) was an arbitrary element of \(G\), this shows that \(G\) is open.

(If one adopts the convention that the intersection of the empty collection of subsets of \(V\) equals \(V\), then that case is also fine: \(V\) is open because \(B(f,1) \subseteq V\) for every \(f \in V\).)

The hypothesis that the collection in (b) is finite cannot be dropped: in \(\mathbb{R}\) with the metric \(d(x,y) = |x-y|\), each set \(\bigl(-\frac{1}{k},\frac{1}{k}\bigr)\) is open, but \(\bigcap_{k=1}^{\infty}\bigl(-\frac{1}{k},\frac{1}{k}\bigr) = \{0\}\), which is not open.

By 6.6, a subset of \(V\) is closed exactly when its complement in \(V\) is open. Thus both parts follow from Exercise 6A.4 by taking complements and using De Morgan’s laws.

(a) Suppose \(\mathcal{F}\) is a nonempty collection of closed subsets of \(V\) and let

\begin{equation*} F = \bigcap_{A \in \mathcal{F}} A . \end{equation*}

De Morgan’s laws give

\begin{equation*} V \setminus F = V \setminus \bigcap_{A \in \mathcal{F}} A = \bigcup_{A \in \mathcal{F}} (V \setminus A). \end{equation*}

Indeed, \(f \in V \setminus F\) if and only if \(f \in V\) and \(f\) fails to belong to at least one \(A \in \mathcal{F}\), which happens if and only if \(f \in V \setminus A\) for some \(A \in \mathcal{F}\).

Because each \(A \in \mathcal{F}\) is closed, each set \(V \setminus A\) is open. Hence \(V \setminus F\) is a union of a collection of open subsets of \(V\), so \(V \setminus F\) is open by Exercise 6A.4(a). Therefore \(F\) is closed.

(If \(\mathcal{F} = \emptyset\) and one interprets the intersection as \(V\), the conclusion still holds: \(V \setminus V = \emptyset\) is open, so \(V\) is closed.)

(b) Suppose \(F_1,\dots,F_n\) are closed subsets of \(V\), where \(n \in \mathbb{Z}^+\), and let

\begin{equation*} F = F_1 \cup \cdots \cup F_n . \end{equation*}

De Morgan’s laws give

\begin{equation*} V \setminus F = V \setminus \bigcup_{j=1}^{n} F_j = \bigcap_{j=1}^{n} (V \setminus F_j), \end{equation*}

since an element of \(V\) lies outside \(F\) if and only if it lies outside every \(F_j\).

Because each \(F_j\) is closed, each set \(V \setminus F_j\) is open. Hence \(V \setminus F\) is the intersection of a finite collection of open subsets of \(V\), so \(V \setminus F\) is open by Exercise 6A.4(b). Therefore \(F\) is closed.

(The union of the empty collection is \(\emptyset\), whose complement \(V\) is open, so \(\emptyset\) is closed as well.)

Again finiteness in (b) is essential: in \(\mathbb{R}\) with \(d(x,y) = |x-y|\), each singleton \(\{1/k\}\) is closed by Exercise 6A.2, but \(\bigcup_{k=1}^{\infty}\{1/k\}\) is not closed, since \(0\) lies in its closure (it is the limit of the sequence \(1, \frac12, \frac13, \dots\)) but not in the set; apply 6.9(e).

Here \(B(f,r) = \{g \in V : d(f,g) < r\}\) is the open ball and \(\overline{B}(f,r) = \{g \in V : d(f,g) \le r\}\) is the closed ball (6.3), while \(\overline{B(f,r)}\) denotes the closure of the open ball in the sense of 6.7.

(a) Suppose \(g \in \overline{B(f,r)}\). By 6.7 this means

\begin{equation*} B(g,\varepsilon) \cap B(f,r) \ne \emptyset \quad \text{for every } \varepsilon > 0 . \end{equation*}

Let \(\varepsilon > 0\) and choose \(h \in B(g,\varepsilon) \cap B(f,r)\). Then \(d(g,h) < \varepsilon\) and \(d(f,h) < r\), so the triangle inequality and the symmetry of \(d\) (6.1) give

\begin{equation*} d(f,g) \le d(f,h) + d(h,g) < r + \varepsilon . \end{equation*}

Thus \(d(f,g) < r + \varepsilon\) for every \(\varepsilon > 0\), which forces \(d(f,g) \le r\). Hence \(g \in \overline{B}(f,r)\), completing the proof that \(\overline{B(f,r)} \subseteq \overline{B}(f,r)\).

Alternatively, one can argue as follows: \(\overline{B}(f,r)\) is a closed subset of \(V\) by Exercise 6A.3, and it contains \(B(f,r)\) because \(d(f,g) < r\) implies \(d(f,g) \le r\); since by 6.9(b) the closure \(\overline{B(f,r)}\) is the intersection of all closed subsets of \(V\) containing \(B(f,r)\), we get \(\overline{B(f,r)} \subseteq \overline{B}(f,r)\).

(b) Let \(V = \{0,1\}\) with the discrete metric from the first bullet point of 6.2, so \(d(x,y) = 1\) if \(x \ne y\) and \(d(x,y) = 0\) if \(x = y\). Take \(f = 0\) and \(r = 1\).

Then

\begin{equation*} B(0,1) = \{g \in V : d(0,g) < 1\} = \{0\}, \qquad \overline{B}(0,1) = \{g \in V : d(0,g) \le 1\} = \{0,1\} = V, \end{equation*}

because \(d(0,1) = 1\), which is not less than \(1\) but is at most \(1\).

Now \(\overline{B(0,1)} = \overline{\{0\}} = \{0\}\). To see this directly from 6.7, note that \(0 \in \overline{\{0\}}\) trivially, while \(1 \notin \overline{\{0\}}\) because \(B(1,\tfrac12) = \{1\}\) (as \(d(1,0) = 1 > \tfrac12\)), so \(B(1,\tfrac12) \cap \{0\} = \emptyset\). (One can also just invoke Exercise 6A.2 and 6.9(d): the finite set \(\{0\}\) is closed, hence equals its own closure.)

Therefore

\begin{equation*} \overline{B(0,1)} = \{0\} \ne \{0,1\} = \overline{B}(0,1). \end{equation*}

The same phenomenon occurs, for example, in \(V = \mathbb{Z}\) with the metric \(d(m,n) = |m-n|\) inherited from \(\mathbb{R}\): there \(B(0,1) = \{0\}\), whose closure is \(\{0\}\), while \(\overline{B}(0,1) = \{-1,0,1\}\).

Suppose \((V,d)\) is a metric space, \(f_1, f_2, \dots\) is a sequence in \(V\), and \(f, g \in V\) satisfy

\begin{equation*} \lim_{k \to \infty} f_k = f \quad \text{and} \quad \lim_{k \to \infty} f_k = g . \end{equation*}

We show that \(f = g\); this proves that a sequence cannot have two different limits, which is exactly the assertion that it has at most one limit. (Of course a sequence may have no limit at all.)

Let \(\varepsilon > 0\). By 6.8, \(\lim_{k \to \infty} d(f_k,f) = 0\), so there exists \(n_1 \in \mathbb{Z}^+\) such that

\begin{equation*} d(f_k,f) < \tfrac{\varepsilon}{2} \quad \text{for all integers } k \ge n_1 . \end{equation*}

Similarly there exists \(n_2 \in \mathbb{Z}^+\) such that

\begin{equation*} d(f_k,g) < \tfrac{\varepsilon}{2} \quad \text{for all integers } k \ge n_2 . \end{equation*}

Let \(n = \max\{n_1,n_2\}\). Then both inequalities hold for \(k = n\), and hence the triangle inequality together with the symmetry of \(d\) (6.1) gives

\begin{equation*} d(f,g) \le d(f,f_n) + d(f_n,g) < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

Thus \(0 \le d(f,g) < \varepsilon\) for every \(\varepsilon > 0\). If we had \(d(f,g) > 0\), then taking \(\varepsilon = d(f,g)\) would yield the contradiction \(d(f,g) < d(f,g)\). Hence \(d(f,g) = 0\), and the second bullet point of 6.1 implies \(f = g\).

Therefore each sequence in a metric space has at most one limit, which justifies the notation \(\lim_{k \to \infty} f_k\) for the limit of a convergent sequence.

Throughout, \(d\) denotes the metric on \(V\). Recall the definition 6.7 of the closure:

\begin{equation*} \overline{E} = \{g \in V : B(g,\varepsilon) \cap E \neq \emptyset \text{ for every } \varepsilon > 0\}. \end{equation*}

Preliminary observation: \(E \subseteq \overline{E}\). Indeed, if \(f \in E\) and \(\varepsilon > 0\), then \(d(f,f) = 0 < \varepsilon\), so \(f \in B(f,\varepsilon) \cap E\), showing that this intersection is nonempty; hence \(f \in \overline{E}\).

Proof of (a). Let \(A\) denote the set on the right side of (a).

Suppose \(g \in \overline{E}\). For each \(k \in \mathbb{Z}^+\), the set \(B(g, 1/k) \cap E\) is nonempty; choose \(f_k\) in it. Then \(f_k \in E\) and \(d(f_k, g) < 1/k\) for each \(k\), so \(\lim_{k \to \infty} d(f_k, g) = 0\), which by 6.8 means \(\lim_{k \to \infty} f_k = g\). Thus \(g \in A\).

Conversely, suppose \(g \in A\), so there exist \(f_1, f_2, \ldots\) in \(E\) with \(\lim_{k \to \infty} f_k = g\). Let \(\varepsilon > 0\). Because \(\lim_{k \to \infty} d(f_k, g) = 0\), there exists \(k \in \mathbb{Z}^+\) with \(d(f_k, g) < \varepsilon\). Then \(f_k \in B(g,\varepsilon) \cap E\), so this intersection is nonempty. As \(\varepsilon > 0\) was arbitrary, \(g \in \overline{E}\).

Hence \(\overline{E} = A\).

Proof of (c). We show that \(V \setminus \overline{E}\) is open; then \(\overline{E}\) is closed by 6.6. Suppose \(g \in V \setminus \overline{E}\). By the definition of closure, there exists \(\varepsilon > 0\) such that \(B(g,\varepsilon) \cap E = \emptyset\). We claim \(B(g,\varepsilon) \subseteq V \setminus \overline{E}\). To see this, let \(h \in B(g,\varepsilon)\) and set \(\delta = \varepsilon - d(g,h) > 0\). If \(u \in B(h,\delta)\), then

\begin{equation*} d(g,u) \leq d(g,h) + d(h,u) < d(g,h) + \delta = \varepsilon, \end{equation*}

so \(B(h,\delta) \subseteq B(g,\varepsilon)\) and therefore \(B(h,\delta) \cap E = \emptyset\). Thus \(h \notin \overline{E}\), proving the claim. Hence every element of \(V \setminus \overline{E}\) is the center of an open ball contained in \(V \setminus \overline{E}\), so \(V \setminus \overline{E}\) is open and \(\overline{E}\) is closed.

Proof of (b). First we record the following statement, labelled \((\ast)\): if \(F\) is a closed subset of \(V\) with \(E \subseteq F\), then \(\overline{E} \subseteq F\).

To prove \((\ast)\), suppose \(g \in V \setminus F\). Because \(V \setminus F\) is open, there exists \(\varepsilon > 0\) with \(B(g,\varepsilon) \subseteq V \setminus F \subseteq V \setminus E\), and hence \(B(g,\varepsilon) \cap E = \emptyset\), which gives \(g \notin \overline{E}\). Taking complements yields \(\overline{E} \subseteq F\).

Now let \(C\) be the intersection of all closed subsets of \(V\) that contain \(E\). By \((\ast)\), the set \(\overline{E}\) is contained in each such closed set, so \(\overline{E} \subseteq C\). On the other hand, by the preliminary observation and (c), the set \(\overline{E}\) is itself a closed subset of \(V\) containing \(E\); hence \(\overline{E}\) is one of the sets being intersected, and therefore \(C \subseteq \overline{E}\). Thus \(C = \overline{E}\), as claimed.

Proof of (d). If \(E = \overline{E}\), then \(E\) is closed by (c). Conversely, suppose \(E\) is closed. Then \(E\) is a closed set containing \(E\), so \((\ast)\) gives \(\overline{E} \subseteq E\); combined with the preliminary observation \(E \subseteq \overline{E}\), this gives \(E = \overline{E}\).

Proof of (e). Suppose \(E\) is closed. If \(f_1, f_2, \ldots\) is a sequence in \(E\) converging to some \(g \in V\), then \(g \in \overline{E}\) by (a), and \(\overline{E} = E\) by (d); hence \(g \in E\). Thus \(E\) contains the limit of every convergent sequence of its elements.

Conversely, suppose \(E\) contains the limit of every convergent sequence of elements of \(E\). If \(g \in \overline{E}\), then by (a) there is a sequence of elements of \(E\) converging to \(g\), so \(g \in E\) by hypothesis. Thus \(\overline{E} \subseteq E\), and with the preliminary observation we get \(E = \overline{E}\), so \(E\) is closed by (d).

Let \(d\) be the metric on \(V\) and let \(G\) be an open subset of \(V\). For each \(k \in \mathbb{Z}^+\), define

\begin{equation*} F_k = V \setminus \bigcup_{h \in V \setminus G} B\!\left(h, \tfrac{1}{k}\right). \end{equation*}

(If \(G = V\), the union is over the empty index set and is thus \(\emptyset\), so \(F_k = V\) in that case.) Explicitly,

\begin{equation*} F_k = \left\{ f \in V : d(f,h) \geq \tfrac{1}{k} \text{ for every } h \in V \setminus G \right\}. \end{equation*}

Each \(F_k\) is closed. Indeed, each ball \(B(h, 1/k)\) is an open subset of \(V\) by 6.5, and a union of open sets is open: if \(f\) belongs to a union \(\bigcup_{\alpha} G_\alpha\) of open sets, then \(f \in G_\alpha\) for some \(\alpha\), so there exists \(r > 0\) with \(B(f,r) \subseteq G_\alpha \subseteq \bigcup_\alpha G_\alpha\). Hence \(V \setminus F_k\) is open, which by 6.6 means precisely that \(F_k\) is closed.

Next, \(F_k \subseteq G\) for every \(k\). Indeed, if \(f \in V \setminus G\), then \(d(f,f) = 0 < 1/k\), so \(f \in B(f, 1/k)\) and therefore \(f \notin F_k\). Taking contrapositives, \(f \in F_k\) implies \(f \in G\). Consequently

\begin{equation*} \bigcup_{k=1}^{\infty} F_k \subseteq G. \end{equation*}

Finally, \(G \subseteq \bigcup_{k=1}^{\infty} F_k\). Suppose \(f \in G\). Because \(G\) is open, there exists \(r > 0\) with \(B(f,r) \subseteq G\). Choose \(k \in \mathbb{Z}^+\) with \(1/k \leq r\). If we had \(f \in B(h, 1/k)\) for some \(h \in V \setminus G\), then \(d(f,h) < 1/k \leq r\), so \(h \in B(f,r) \subseteq G\), contradicting \(h \notin G\). Hence \(f\) lies in no such ball, which means \(f \in F_k\).

Combining the last two displays, \(G = \bigcup_{k=1}^{\infty} F_k\), exhibiting the open set \(G\) as the union of the sequence \(F_1, F_2, \ldots\) of closed subsets of \(V\).

The statement is true; we prove it.

Let \(d\) be the metric on \(V\). Recall from 6.7 that for \(E \subseteq V\),

\begin{equation*} \overline{E} = \{g \in V : B(g,\varepsilon) \cap E \neq \emptyset \text{ for every } \varepsilon > 0\}. \end{equation*}

First note that the closure operation is order preserving: if \(E_1 \subseteq E_2 \subseteq V\) and \(g \in \overline{E_1}\), then for every \(\varepsilon > 0\) we have \(\emptyset \neq B(g,\varepsilon) \cap E_1 \subseteq B(g,\varepsilon) \cap E_2\), so \(g \in \overline{E_2}\). Thus \(E_1 \subseteq E_2\) implies \(\overline{E_1} \subseteq \overline{E_2}\).

Applying this to \(U \subseteq U \cup W\) and to \(W \subseteq U \cup W\) gives \(\overline{U} \subseteq \overline{U \cup W}\) and \(\overline{W} \subseteq \overline{U \cup W}\), hence

\begin{equation*} \overline{U} \cup \overline{W} \subseteq \overline{U \cup W}. \end{equation*}

For the reverse inclusion, suppose \(g \in V\) and \(g \notin \overline{U} \cup \overline{W}\); we show \(g \notin \overline{U \cup W}\). Because \(g \notin \overline{U}\), there exists \(\varepsilon_1 > 0\) with \(B(g,\varepsilon_1) \cap U = \emptyset\). Because \(g \notin \overline{W}\), there exists \(\varepsilon_2 > 0\) with \(B(g,\varepsilon_2) \cap W = \emptyset\). Let \(\varepsilon = \min\{\varepsilon_1, \varepsilon_2\} > 0\). Then \(B(g,\varepsilon) \subseteq B(g,\varepsilon_1)\) and \(B(g,\varepsilon) \subseteq B(g,\varepsilon_2)\), so

\begin{equation*} B(g,\varepsilon) \cap (U \cup W) = \bigl(B(g,\varepsilon) \cap U\bigr) \cup \bigl(B(g,\varepsilon) \cap W\bigr) = \emptyset . \end{equation*}

Hence \(g \notin \overline{U \cup W}\). Taking contrapositives gives \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W}\).

The two inclusions give \(\overline{U} \cup \overline{W} = \overline{U \cup W}\).

Alternative proof of the inclusion \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W}\): by 6.9(c) each of \(\overline{U}\) and \(\overline{W}\) is closed, so their union is closed (a union of two closed sets is closed, because the complement of the union is the intersection of the two open complements, which is open). Since \(U \cup W \subseteq \overline{U} \cup \overline{W}\) by the preliminary observation \(E \subseteq \overline{E}\), and since by 6.9(b) the set \(\overline{U \cup W}\) is the intersection of all closed sets containing \(U \cup W\), we conclude \(\overline{U \cup W} \subseteq \overline{U} \cup \overline{W}\).

Note that the corresponding statement for infinite unions is false: in \(\mathbb{R}\) with its usual metric, writing \(\mathbb{Q} = \{r_1, r_2, \ldots\}\), each singleton \(\{r_k\}\) is closed and equals its own closure, but \(\overline{\bigcup_k \{r_k\}} = \overline{\mathbb{Q}} = \mathbb{R} \neq \mathbb{Q} = \bigcup_k \overline{\{r_k\}}\).

The statement is false. Here is a counterexample.

Let \(V = \mathbb{R}\) with the usual metric \(d(x,y) = |x - y|\) (see the second bullet point of 6.2), and let

\begin{equation*} U = (0,1), \qquad W = (1,2). \end{equation*}

First, \(U \cap W = \emptyset\), and \(\overline{\emptyset} = \emptyset\): for any \(g \in \mathbb{R}\) and any \(\varepsilon > 0\) we have \(B(g,\varepsilon) \cap \emptyset = \emptyset\), so no \(g\) satisfies the defining condition 6.7. Thus

\begin{equation*} \overline{U \cap W} = \emptyset . \end{equation*}

Second, \(1 \in \overline{U} \cap \overline{W}\). To see that \(1 \in \overline{U}\), let \(\varepsilon > 0\) and put

\begin{equation*} t = \max\left\{\tfrac{1}{2},\, 1 - \tfrac{\varepsilon}{2}\right\}. \end{equation*}

Then \(t \geq 1/2 > 0\) and \(t < 1\) (since both \(1/2 < 1\) and \(1 - \varepsilon/2 < 1\)), so \(t \in (0,1) = U\); moreover \(1 - t \leq \varepsilon/2 < \varepsilon\), so \(t \in B(1,\varepsilon)\). Hence \(B(1,\varepsilon) \cap U \neq \emptyset\) for every \(\varepsilon > 0\), so \(1 \in \overline{U}\).

Symmetrically, to see that \(1 \in \overline{W}\), let \(\varepsilon > 0\) and put \(s = \min\{3/2,\, 1 + \varepsilon/2\}\). Then \(s > 1\) and \(s \leq 3/2 < 2\), so \(s \in (1,2) = W\), while \(s - 1 \leq \varepsilon/2 < \varepsilon\), so \(s \in B(1,\varepsilon)\). Hence \(1 \in \overline{W}\).

Therefore \(1 \in \overline{U} \cap \overline{W}\) but \(1 \notin \overline{U \cap W} = \emptyset\), so

\begin{equation*} \overline{U} \cap \overline{W} \neq \overline{U \cap W}, \end{equation*}

and the proposed statement fails.

An even more dramatic counterexample in \(V = \mathbb{R}\): take \(U = \mathbb{Q}\) and \(W = \mathbb{R} \setminus \mathbb{Q}\). Every open interval contains both a rational and an irrational number, so \(\overline{U} = \overline{W} = \mathbb{R}\) and hence \(\overline{U} \cap \overline{W} = \mathbb{R}\), while \(U \cap W = \emptyset\) gives \(\overline{U \cap W} = \emptyset\).

One inclusion does always hold. The closure operation is order preserving: if \(E_1 \subseteq E_2 \subseteq V\) and \(g \in \overline{E_1}\), then for every \(\varepsilon > 0\) we have \(\emptyset \neq B(g,\varepsilon) \cap E_1 \subseteq B(g,\varepsilon) \cap E_2\), so \(g \in \overline{E_2}\). Since \(U \cap W \subseteq U\) and \(U \cap W \subseteq W\), this gives \(\overline{U \cap W} \subseteq \overline{U}\) and \(\overline{U \cap W} \subseteq \overline{W}\), hence

\begin{equation*} \overline{U \cap W} \subseteq \overline{U} \cap \overline{W}. \end{equation*}

The counterexamples above show that this inclusion can be strict.

(a) We use definition 6.10 directly. Fix \(f \in U\) and let \(\varepsilon > 0\).

Because \(S\) is continuous, \(S\) is continuous at the element \(T(f) \in V\), so there exists \(\gamma > 0\) such that

\begin{equation*} d_W\bigl(S(T(f)), S(v)\bigr) < \varepsilon \quad \text{for all } v \in V \text{ with } d_V\bigl(T(f), v\bigr) < \gamma . \end{equation*}

Because \(T\) is continuous, \(T\) is continuous at \(f\), so with this \(\gamma\) in the role of the tolerance there exists \(\delta > 0\) such that

\begin{equation*} d_V\bigl(T(f), T(g)\bigr) < \gamma \quad \text{for all } g \in U \text{ with } d_U(f,g) < \delta . \end{equation*}

Now suppose \(g \in U\) satisfies \(d_U(f,g) < \delta\). Then \(d_V(T(f), T(g)) < \gamma\), and applying the first display with \(v = T(g)\) gives

\begin{equation*} d_W\bigl((S \circ T)(f), (S \circ T)(g)\bigr) = d_W\bigl(S(T(f)), S(T(g))\bigr) < \varepsilon . \end{equation*}

Thus \(S \circ T\) is continuous at \(f\). Since \(f \in U\) was arbitrary, \(S \circ T\) is continuous.

(b) We verify condition 6.11(b) for \(S \circ T\). Suppose \(f_1, f_2, \ldots\) is a sequence in \(U\) with \(\lim_{k \to \infty} f_k = f\) for some \(f \in U\).

Because \(T\) is continuous, the implication 6.11(a) \(\Rightarrow\) 6.11(b) applied to \(T\) gives

\begin{equation*} \lim_{k \to \infty} T(f_k) = T(f) \quad \text{in } V . \end{equation*}

Because \(S\) is continuous, the implication 6.11(a) \(\Rightarrow\) 6.11(b) applied to \(S\), with the convergent sequence \(T(f_1), T(f_2), \ldots\) in \(V\), gives

\begin{equation*} \lim_{k \to \infty} S\bigl(T(f_k)\bigr) = S\bigl(T(f)\bigr) \quad \text{in } W , \end{equation*}

that is, \(\lim_{k \to \infty} (S \circ T)(f_k) = (S \circ T)(f)\).

Thus \(S \circ T\) satisfies 6.11(b), and therefore the implication 6.11(b) \(\Rightarrow\) 6.11(a) shows that \(S \circ T\) is continuous.

(c) We verify condition 6.11(c) for \(S \circ T\). First note that for every set \(G \subseteq W\),

\begin{equation*} (S \circ T)^{-1}(G) = T^{-1}\bigl(S^{-1}(G)\bigr), \end{equation*}

because for \(f \in U\) we have the chain of equivalences: \(f \in (S \circ T)^{-1}(G)\) if and only if \(S(T(f)) \in G\), if and only if \(T(f) \in S^{-1}(G)\), if and only if \(f \in T^{-1}(S^{-1}(G))\).

Now suppose \(G\) is an open subset of \(W\). Because \(S\) is continuous, the implication 6.11(a) \(\Rightarrow\) 6.11(c) applied to \(S\) shows that \(S^{-1}(G)\) is an open subset of \(V\). Because \(T\) is continuous, the same implication applied to \(T\), with the open set \(S^{-1}(G) \subseteq V\), shows that \(T^{-1}(S^{-1}(G))\) is an open subset of \(U\). By the displayed identity, \((S \circ T)^{-1}(G)\) is an open subset of \(U\).

Thus \(S \circ T\) satisfies 6.11(c), and therefore the implication 6.11(c) \(\Rightarrow\) 6.11(a) shows that \(S \circ T\) is continuous.

Write \(d_V\) and \(d_W\) for the metrics on \(V\) and \(W\). The text proves the implication (b) \(\Rightarrow\) (d) and the equivalence (c) \(\Leftrightarrow\) (d). To complete the proof that all four conditions are equivalent, it suffices to prove (a) \(\Rightarrow\) (b) and (c) \(\Rightarrow\) (a). Indeed, these together with the text’s results give the cycle

\begin{equation*} \text{(a)} \Rightarrow \text{(b)} \Rightarrow \text{(d)} \Rightarrow \text{( c)} \Rightarrow \text{(a)}, \end{equation*}

from which every one of the four conditions implies every other.

Proof that (a) implies (b). Suppose \(T\) is continuous. Suppose \(f_1, f_2, \ldots\) is a sequence in \(V\) with \(\lim_{k \to \infty} f_k = f\) for some \(f \in V\). Let \(\varepsilon > 0\). Because \(T\) is continuous at \(f\), definition 6.10 provides \(\delta > 0\) such that

\begin{equation*} d_W\bigl(T(f), T(g)\bigr) < \varepsilon \quad \text{for all } g \in V \text{ with } d_V(f,g) < \delta . \end{equation*}

Because \(\lim_{k \to \infty} f_k = f\) means \(\lim_{k \to \infty} d_V(f_k, f) = 0\) (definition 6.8), there exists \(n \in \mathbb{Z}^+\) such that \(d_V(f_k, f) < \delta\) for all integers \(k \geq n\). For such \(k\) the display above (with \(g = f_k\)) gives \(d_W(T(f), T(f_k)) < \varepsilon\).

Thus for every \(\varepsilon > 0\) there exists \(n \in \mathbb{Z}^+\) with \(d_W(T(f_k), T(f)) < \varepsilon\) for all \(k \geq n\); that is, \(\lim_{k \to \infty} d_W(T(f_k), T(f)) = 0\), which means \(\lim_{k \to \infty} T(f_k) = T(f)\). Hence (b) holds.

Proof that (c) implies (a). Suppose \(T^{-1}(G)\) is an open subset of \(V\) for every open set \(G \subseteq W\). Let \(f \in V\) and let \(\varepsilon > 0\).

The open ball \(B\bigl(T(f), \varepsilon\bigr)\) is an open subset of \(W\) by 6.5. Hence

\begin{equation*} G := T^{-1}\Bigl(B\bigl(T(f), \varepsilon\bigr)\Bigr) \end{equation*}

is an open subset of \(V\). Since \(d_W(T(f), T(f)) = 0 < \varepsilon\), we have \(T(f) \in B(T(f),\varepsilon)\) and therefore \(f \in G\). Because \(G\) is open, definition 6.4 provides \(\delta > 0\) with \(B(f,\delta) \subseteq G\).

Now suppose \(g \in V\) satisfies \(d_V(f,g) < \delta\). Then \(g \in B(f,\delta) \subseteq G\), so \(T(g) \in B(T(f),\varepsilon)\), which means

\begin{equation*} d_W\bigl(T(f), T(g)\bigr) < \varepsilon . \end{equation*}

Thus \(T\) is continuous at \(f\). Since \(f \in V\) was arbitrary, \(T\) is continuous, so (a) holds.

For completeness, here is a direct proof of the implication (a) \(\Rightarrow\) (c), which shows the equivalence without routing through (b) and (d). Suppose \(T\) is continuous and \(G \subseteq W\) is open. Let \(f \in T^{-1}(G)\). Then \(T(f) \in G\), so there exists \(\varepsilon > 0\) with \(B(T(f),\varepsilon) \subseteq G\). By continuity of \(T\) at \(f\), there exists \(\delta > 0\) such that \(d_W(T(f), T(g)) < \varepsilon\) whenever \(d_V(f,g) < \delta\). Hence \(g \in B(f,\delta)\) implies \(T(g) \in B(T(f),\varepsilon) \subseteq G\), that is, \(B(f,\delta) \subseteq T^{-1}(G)\). Therefore \(T^{-1}(G)\) is open.

Suppose \((V,d)\) is a metric space and \(f_1, f_2, \ldots\) is a Cauchy sequence in \(V\). Suppose \(k_1 < k_2 < \cdots\) are positive integers and \(f \in V\) is such that

\begin{equation*} \lim_{m \to \infty} f_{k_m} = f . \end{equation*}

We show that \(\lim_{k \to \infty} f_k = f\).

First we record that \(k_m \geq m\) for every \(m \in \mathbb{Z}^+\). This holds by induction: \(k_1 \geq 1\) because \(k_1\) is a positive integer, and if \(k_m \geq m\) then \(k_{m+1} > k_m \geq m\), so \(k_{m+1} \geq m+1\) because \(k_{m+1}\) is an integer.

Now let \(\varepsilon > 0\). Because \(f_1, f_2, \ldots\) is a Cauchy sequence (definition 6.12), there exists \(n \in \mathbb{Z}^+\) such that

\begin{equation*} d(f_j, f_k) < \frac{\varepsilon}{2} \quad \text{for all integers } j \geq n \text{ and } k \geq n . \end{equation*}

Because \(\lim_{m \to \infty} f_{k_m} = f\), which by 6.8 means \(\lim_{m \to \infty} d(f_{k_m}, f) = 0\), there exists \(M \in \mathbb{Z}^+\) such that \(d(f_{k_m}, f) < \varepsilon/2\) for all \(m \geq M\). Choose an integer \(m \geq \max\{M, n\}\) and set \(j = k_m\). Then

\begin{equation*} d(f_j, f) = d(f_{k_m}, f) < \frac{\varepsilon}{2}, \end{equation*}

and also \(j = k_m \geq m \geq n\) by the observation of the previous paragraph.

Now suppose \(k \in \mathbb{Z}^+\) with \(k \geq n\). Since \(j \geq n\) as well, the Cauchy estimate gives \(d(f_k, f_j) < \varepsilon/2\), and the triangle inequality then gives

\begin{equation*} d(f_k, f) \leq d(f_k, f_j) + d(f_j, f) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon . \end{equation*}

Thus for every \(\varepsilon > 0\) there exists \(n \in \mathbb{Z}^+\) such that \(d(f_k, f) < \varepsilon\) for all integers \(k \geq n\). In other words \(\lim_{k \to \infty} d(f_k, f) = 0\), which by 6.8 means \(\lim_{k \to \infty} f_k = f\). Hence the Cauchy sequence converges (to the same limit as its convergent subsequence).

Recall the five metric spaces of 6.2: (i) an arbitrary nonempty set \(V\) with \(d(f,g) = 1\) if \(f \neq g\) and \(d(f,f) = 0\); (ii) \(\mathbb{R}\) with \(d(x,y) = |x-y|\); (iii) \(\mathbb{R}^n\) with \(d\big((x_1,\dots,x_n),(y_1,\dots,y_n)\big) = \max\{|x_1-y_1|,\dots,|x_n-y_n|\}\); (iv) \(C([0,1])\), the continuous real-valued functions on \([0,1]\), with \(d(f,g) = \sup\{|f(t)-g(t)| : t \in [0,1]\}\); (v) \(\ell^1\), the set of real sequences \((a_1,a_2,\dots)\) with \(\sum_{k=1}^{\infty}|a_k| < \infty\), with \(d\big((a_1,a_2,\dots),(b_1,b_2,\dots)\big) = \sum_{k=1}^{\infty}|a_k-b_k|\).

(i) The discrete metric. Suppose \(f_1, f_2, \dots\) is a Cauchy sequence in \(V\). Applying the definition 6.12 with \(\varepsilon = 1\), there exists \(n \in \mathbb{Z}^+\) such that \(d(f_j,f_k) < 1\) for all \(j,k \geq n\). Because \(d\) takes only the values \(0\) and \(1\), this forces \(d(f_j,f_k) = 0\), and hence \(f_j = f_k\), for all \(j,k \geq n\). Thus \(f_k = f_n\) for every \(k \geq n\), so \(d(f_k, f_n) = 0 < \varepsilon\) for every \(k \geq n\) and every \(\varepsilon > 0\). Hence \(\lim_{k \to \infty} f_k = f_n \in V\), and \(V\) is complete.

(ii) \(\mathbb{R}\) with the absolute-value metric. Suppose \(x_1, x_2, \dots\) is a Cauchy sequence of real numbers. First, the sequence is bounded: taking \(\varepsilon = 1\) in 6.12 gives \(n \in \mathbb{Z}^+\) with \(|x_j - x_k| < 1\) for all \(j,k \geq n\); hence \(|x_k| \leq |x_n| + 1\) for all \(k \geq n\), and therefore

\begin{equation*} |x_k| \leq \max\{|x_1|, \dots, |x_{n-1}|, |x_n| + 1\} \end{equation*}

for all \(k \in \mathbb{Z}^+\). Let \(M\) denote the right side above. For each \(m \in \mathbb{Z}^+\) the set \(\{x_k : k \geq m\}\) is nonempty and bounded above by \(M\), so by the completeness (least upper bound) property of \(\mathbb{R}\) the number

\begin{equation*} s_m = \sup\{x_k : k \geq m\} \end{equation*}

exists; furthermore \(-M \leq s_m\) and \(s_1 \geq s_2 \geq \cdots\). Hence \(\{s_m : m \in \mathbb{Z}^+\}\) is a nonempty set of real numbers bounded below by \(-M\), so \(L = \inf\{s_m : m \in \mathbb{Z}^+\}\) exists in \(\mathbb{R}\).

We claim \(\lim_{k \to \infty} x_k = L\). Suppose \(\varepsilon > 0\). There exists \(n \in \mathbb{Z}^+\) such that \(|x_j - x_k| < \varepsilon/2\) for all \(j,k \geq n\). For every \(k \geq n\) we have \(x_k \leq x_n + \varepsilon/2\); taking the supremum over \(k \geq n\) gives \(s_n \leq x_n + \varepsilon/2\), and thus \(L \leq s_n \leq x_n + \varepsilon/2\). On the other hand, for every \(m \geq n\) we have \(s_m \geq x_m > x_n - \varepsilon/2\); because \(s_1 \geq s_2 \geq \cdots\) we have \(L = \inf\{s_m : m \geq n\} \geq x_n - \varepsilon/2\). Combining the two estimates gives \(|x_n - L| \leq \varepsilon/2\). Hence for every \(j \geq n\),

\begin{equation*} |x_j - L| \leq |x_j - x_n| + |x_n - L| < \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} = \varepsilon . \end{equation*}

Thus \(x_k \to L \in \mathbb{R}\), and \(\mathbb{R}\) is complete.

(iii) \(\mathbb{R}^n\) with the maximum metric. Suppose \(x^{(1)}, x^{(2)}, \dots\) is a Cauchy sequence in \(\mathbb{R}^n\), where \(x^{(k)} = (x^{(k)}_1, \dots, x^{(k)}_n)\). For each \(i \in \{1,\dots,n\}\) and all \(j,k\) we have

\begin{equation*} |x^{(j)}_i - x^{(k)}_i| \leq \max_{1 \leq p \leq n} |x^{(j)}_p - x^{(k)}_p| = d\big(x^{(j)}, x^{(k)}\big), \end{equation*}

so the real sequence \(x^{(1)}_i, x^{(2)}_i, \dots\) is Cauchy. By (ii) it converges to some \(y_i \in \mathbb{R}\). Let \(y = (y_1, \dots, y_n)\). Suppose \(\varepsilon > 0\). For each \(i \in \{1,\dots,n\}\) there exists \(m_i \in \mathbb{Z}^+\) such that \(|x^{(k)}_i - y_i| < \varepsilon\) for all \(k \geq m_i\). Because there are only finitely many coordinates, \(N = \max\{m_1, \dots, m_n\}\) is a positive integer, and for \(k \geq N\) we have

\begin{equation*} d\big(x^{(k)}, y\big) = \max_{1 \leq i \leq n} |x^{(k)}_i - y_i| < \varepsilon . \end{equation*}

Thus \(x^{(k)} \to y \in \mathbb{R}^n\), and \(\mathbb{R}^n\) is complete. (Finiteness of the number of coordinates is essential here; it is what allows the single \(N\).)

(iv) \(C([0,1])\) with the supremum metric. Suppose \(f_1, f_2, \dots\) is a Cauchy sequence in \(C([0,1])\). For each \(t \in [0,1]\) and all \(j,k\) we have \(|f_j(t) - f_k(t)| \leq d(f_j,f_k)\), so \(f_1(t), f_2(t), \dots\) is a Cauchy sequence of real numbers; by (ii) it converges. Define \(f : [0,1] \to \mathbb{R}\) by

\begin{equation*} f(t) = \lim_{k \to \infty} f_k(t). \end{equation*}

We first show that \(\sup\{|f(t) - f_k(t)| : t \in [0,1]\} \to 0\). Suppose \(\varepsilon > 0\). There exists \(n \in \mathbb{Z}^+\) such that \(d(f_j,f_k) < \varepsilon/3\) for all \(j,k \geq n\). Fix \(k \geq n\) and \(t \in [0,1]\). Then \(|f_j(t) - f_k(t)| < \varepsilon/3\) for all \(j \geq n\); letting \(j \to \infty\) and using the continuity of the absolute value on \(\mathbb{R}\) gives

\begin{equation*} |f(t) - f_k(t)| \leq \tfrac{\varepsilon}{3} \end{equation*}

for all \(t \in [0,1]\) and all \(k \geq n\). Hence \(\sup\{|f(t) - f_k(t)| : t \in [0,1]\} \leq \varepsilon/3\) for \(k \geq n\).

Next we show \(f\) is continuous. Suppose \(t \in [0,1]\) and \(\varepsilon > 0\); take \(n\) as above and set \(k = n\). Because \(f_n\) is continuous at \(t\), there exists \(\delta > 0\) such that \(|f_n(t) - f_n(s)| < \varepsilon/3\) for all \(s \in [0,1]\) with \(|s - t| < \delta\). For such \(s\),

\begin{equation*} |f(t) - f(s)| \leq |f(t) - f_n(t)| + |f_n(t) - f_n(s)| + |f_n(s) - f(s)| < \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} = \varepsilon . \end{equation*}

Thus \(f\) is continuous at each \(t \in [0,1]\), so \(f \in C([0,1])\). Now \(d(f_k, f) = \sup\{|f(t)-f_k(t)| : t \in [0,1]\} \leq \varepsilon/3 < \varepsilon\) for all \(k \geq n\), so \(\lim_{k \to \infty} f_k = f\) in \(C([0,1])\). Hence \(C([0,1])\) is complete.

(v) \(\ell^1\). Suppose \(a^{(1)}, a^{(2)}, \dots\) is a Cauchy sequence in \(\ell^1\), where \(a^{(m)} = (a^{(m)}_1, a^{(m)}_2, \dots)\). For each \(k \in \mathbb{Z}^+\) and all \(j,m\),

\begin{equation*} |a^{(j)}_k - a^{(m)}_k| \leq \sum_{p=1}^{\infty} |a^{(j)}_p - a^{(m)}_p| = d\big(a^{(j)}, a^{(m)}\big), \end{equation*}

so the real sequence \(a^{(1)}_k, a^{(2)}_k, \dots\) is Cauchy; by (ii) it converges to some \(b_k \in \mathbb{R}\). Let \(b = (b_1, b_2, \dots)\).

Suppose \(\varepsilon > 0\). There exists \(n \in \mathbb{Z}^+\) such that \(d(a^{(j)}, a^{(m)}) < \varepsilon/2\) for all \(j,m \geq n\). Fix \(m \geq n\) and \(N \in \mathbb{Z}^+\). For every \(j \geq n\),

\begin{equation*} \sum_{k=1}^{N} |a^{(j)}_k - a^{(m)}_k| \leq d\big(a^{(j)}, a^{(m)}\big) \leq \tfrac{\varepsilon}{2}. \end{equation*}

The left side is a finite sum, so letting \(j \to \infty\) (each of the \(N\) terms converges) gives

\begin{equation*} \sum_{k=1}^{N} |b_k - a^{(m)}_k| \leq \tfrac{\varepsilon}{2}. \end{equation*}

This holds for every \(N \in \mathbb{Z}^+\); the partial sums of the nonnegative series \(\sum_k |b_k - a^{(m)}_k|\) are therefore bounded above by \(\varepsilon/2\), and hence

\begin{equation*} \sum_{k=1}^{\infty} |b_k - a^{(m)}_k| \leq \tfrac{\varepsilon}{2} < \varepsilon \end{equation*}

for all \(m \geq n\).

Taking \(m = n\), the sequence \(b - a^{(n)}\) lies in \(\ell^1\). Because \(a^{(n)} \in \ell^1\) and

\begin{equation*} \sum_{k=1}^{\infty} |b_k| \leq \sum_{k=1}^{\infty} |b_k - a^{(n)}_k| + \sum_{k=1}^{\infty} |a^{(n)}_k| < \infty, \end{equation*}

we conclude \(b \in \ell^1\). The displayed estimate now reads \(d(a^{(m)}, b) < \varepsilon\) for all \(m \geq n\), so \(\lim_{m \to \infty} a^{(m)} = b\) in \(\ell^1\). Hence \(\ell^1\) is complete.

Throughout we use freely the fact that \(\mathbb{R}\) with the metric \(|x-y|\) is complete (Example 6.15, verified in Exercise 15 of this section).

(a) Reflexivity. If \((f_1,f_2,\dots) \in W\), then \(d(f_k,f_k) = 0\) for every \(k\) by the first bullet point of 6.1, so \(\lim_{k \to \infty} d(f_k,f_k) = 0\) and \((f_1,f_2,\dots) \equiv (f_1,f_2,\dots)\).

Symmetry. Suppose \((f_1,f_2,\dots) \equiv (g_1,g_2,\dots)\). Because \(d(g_k,f_k) = d(f_k,g_k)\) for every \(k\) (third bullet point of 6.1), we get \(\lim_{k \to \infty} d(g_k,f_k) = 0\), so \((g_1,g_2,\dots) \equiv (f_1,f_2,\dots)\).

Transitivity. Suppose \((f_1,f_2,\dots) \equiv (g_1,g_2,\dots)\) and \((g_1,g_2,\dots) \equiv (h_1,h_2,\dots)\). The triangle inequality (fourth bullet point of 6.1) gives

\begin{equation*} 0 \leq d(f_k,h_k) \leq d(f_k,g_k) + d(g_k,h_k) \end{equation*}

for every \(k \in \mathbb{Z}^+\). Both terms on the right tend to \(0\), so \(\lim_{k \to \infty} d(f_k,h_k) = 0\) and \((f_1,f_2,\dots) \equiv (h_1,h_2,\dots)\).

Thus \(\equiv\) is an equivalence relation on \(W\).

(b) Two things must be checked: that the limit defining \(d_V\) exists, and that its value does not depend on the representatives chosen from the two equivalence classes.

The key inequality. Suppose \(f, f^{\prime}, g, g^{\prime} \in U\). Two applications of the triangle inequality give

\begin{equation*} d(f,g) \leq d(f,f^{\prime}) + d(f^{\prime},g^{\prime}) + d(g^{\prime},g), \end{equation*}

so \(d(f,g) - d(f^{\prime},g^{\prime}) \leq d(f,f^{\prime}) + d(g,g^{\prime})\). Interchanging the roles of the primed and unprimed elements gives the reverse inequality, and hence

\begin{equation*} \big| d(f,g) - d(f^{\prime},g^{\prime}) \big| \leq d(f,f^{\prime}) + d(g,g^{\prime}) . \end{equation*}

Existence of the limit. Suppose \((f_1,f_2,\dots)\) and \((g_1,g_2,\dots)\) are in \(W\). Applying the key inequality with \(f = f_j\), \(f^{\prime} = f_k\), \(g = g_j\), \(g^{\prime} = g_k\) gives

\begin{equation*} \big| d(f_j,g_j) - d(f_k,g_k) \big| \leq d(f_j,f_k) + d(g_j,g_k) . \end{equation*}

Suppose \(\varepsilon > 0\). Because both sequences are Cauchy, there exists \(n \in \mathbb{Z}^+\) such that \(d(f_j,f_k) < \varepsilon/2\) and \(d(g_j,g_k) < \varepsilon/2\) for all \(j,k \geq n\); the displayed inequality then gives \(|d(f_j,g_j) - d(f_k,g_k)| < \varepsilon\) for all \(j,k \geq n\). Thus \(d(f_1,g_1), d(f_2,g_2), \dots\) is a Cauchy sequence of real numbers, so it converges to some real number. Each term is nonnegative, so the limit lies in \([0,\infty)\), as required by the stated codomain of \(d_V\).

Independence of representatives. Suppose \((f_1,f_2,\dots) \equiv (f^{\prime}_1,f^{\prime}_2,\dots)\) and \((g_1,g_2,\dots) \equiv (g^{\prime}_1,g^{\prime}_2,\dots)\), all four sequences being in \(W\). Applying the key inequality with \(f = f_k\), \(f^{\prime} = f^{\prime}_k\), \(g = g_k\), \(g^{\prime} = g^{\prime}_k\) gives

\begin{equation*} \big| d(f_k,g_k) - d(f^{\prime}_k,g^{\prime}_k) \big| \leq d(f_k,f^{\prime}_k) + d(g_k,g^{\prime}_k) \to 0 \end{equation*}

as \(k \to \infty\). Because both limits exist by the previous paragraph, we conclude

\begin{equation*} \lim_{k \to \infty} d(f_k,g_k) = \lim_{k \to \infty} d(f^{\prime}_k,g^{\prime}_k) . \end{equation*}

Hence the value assigned by \(d_V\) depends only on the two equivalence classes, and \(d_V : V \times V \to [0,\infty)\) is a well-defined function.

\(d_V\) is a metric. Write \(F = \widehat{(f_1,f_2,\dots)}\), \(G = \widehat{(g_1,g_2,\dots)}\), \(H = \widehat{(h_1,h_2,\dots)}\) for arbitrary elements of \(V\).

If \(F \in V\), then \(d_V(F,F) = \lim_{k \to \infty} d(f_k,f_k) = 0\).

If \(d_V(F,G) = 0\), then \(\lim_{k \to \infty} d(f_k,g_k) = 0\), which says exactly that \((f_1,f_2,\dots) \equiv (g_1,g_2,\dots)\); hence these two sequences lie in the same equivalence class, which means \(F = G\).

Symmetry: \(d_V(F,G) = \lim_{k \to \infty} d(f_k,g_k) = \lim_{k \to \infty} d(g_k,f_k) = d_V(G,F)\).

Triangle inequality: for every \(k\) we have \(d(f_k,h_k) \leq d(f_k,g_k) + d(g_k,h_k)\). All three limits exist by the first part of (b), so taking limits as \(k \to \infty\) preserves the inequality:

\begin{equation*} d_V(F,H) \leq d_V(F,G) + d_V(G,H) . \end{equation*}

Thus \(d_V\) satisfies all four bullet points of 6.1, and \((V, d_V)\) is a metric space. (Note that \(V\) is nonempty because \(U\) is nonempty: any constant sequence is Cauchy.)

(c) We begin by recording the constant-sequence embedding, which is also the content of (d) and which will be used several times below. Suppose \(f \in U\). The constant sequence \((f,f,f,\dots)\) satisfies \(d(f,f) = 0 < \varepsilon\) for every \(\varepsilon > 0\), so it is a Cauchy sequence and thus lies in \(W\). Define \(\iota : U \to V\) by

\begin{equation*} \iota(f) = \widehat{(f,f,f,\dots)} . \end{equation*}

For all \(f,g \in U\) the sequence whose \(k\)-th term is \(d(f,g)\) is constant, so

\begin{equation*} d_V\big(\iota(f), \iota(g)\big) = \lim_{k \to \infty} d(f,g) = d(f,g) . \end{equation*}

Now suppose \(F^{(1)}, F^{(2)}, \dots\) is a Cauchy sequence in \((V, d_V)\). For each \(m \in \mathbb{Z}^+\) choose a representative, so that \(F^{(m)} = \widehat{(f^{(m)}_1, f^{(m)}_2, \dots)}\) with \((f^{(m)}_1, f^{(m)}_2, \dots) \in W\). Because this sequence is Cauchy in \(U\), there exists \(k_m \in \mathbb{Z}^+\) such that

\begin{equation*} d\big(f^{(m)}_j, f^{(m)}_k\big) < \tfrac{1}{m} \quad \text{for all } j, k \geq k_m . \end{equation*}

Set \(h_m = f^{(m)}_{k_m}\). Then for every \(k \geq k_m\) we have \(d(f^{(m)}_k, h_m) < 1/m\), and therefore

\begin{equation*} d_V\big(F^{(m)}, \iota(h_m)\big) = \lim_{k \to \infty} d\big(f^{(m)}_k, h_m\big) \leq \tfrac{1}{m} . \end{equation*}

The sequence \(h_1, h_2, \dots\) is Cauchy in \(U\). Using the displayed identity \(d(h_j,h_m) = d_V(\iota(h_j), \iota(h_m))\) and then the triangle inequality in \(V\),

\begin{equation*} \begin{aligned} d(h_j, h_m) &= d_V\big(\iota(h_j), \iota(h_m)\big) \\ &\leq d_V\big(\iota(h_j), F^{(j)}\big) + d_V\big(F^{(j)}, F^{(m)}\big) + d_V\big(F^{(m)}, \iota(h_m)\big) \\ &\leq \tfrac{1}{j} + d_V\big(F^{(j)}, F^{(m)}\big) + \tfrac{1}{m}, \end{aligned} \end{equation*}

where the last line uses the displayed bound \(d_V(F^{(m)}, \iota(h_m)) \leq 1/m\). Suppose \(\varepsilon > 0\). Because \(F^{(1)}, F^{(2)}, \dots\) is Cauchy in \(V\), there exists \(n_0 \in \mathbb{Z}^+\) such that \(d_V(F^{(j)}, F^{(m)}) < \varepsilon/3\) for all \(j,m \geq n_0\). Let \(n\) be an integer with \(n \geq n_0\) and \(n > 3/\varepsilon\). Then for all \(j,m \geq n\),

\begin{equation*} d(h_j,h_m) < \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} = \varepsilon . \end{equation*}

Thus \((h_1,h_2,\dots) \in W\); let \(H = \widehat{(h_1,h_2,\dots)} \in V\).

\(F^{(m)} \to H\). First note that for each \(m\),

\begin{equation*} d_V\big(\iota(h_m), H\big) = \lim_{k \to \infty} d(h_m, h_k). \end{equation*}

Suppose \(\varepsilon > 0\). Because \(h_1, h_2, \dots\) is Cauchy, there exists \(n_1 \in \mathbb{Z}^+\) such that \(d(h_m, h_k) < \varepsilon/3\) for all \(m,k \geq n_1\); letting \(k \to \infty\) gives \(d_V(\iota(h_m), H) \leq \varepsilon/3\) for all \(m \geq n_1\). Now let \(n\) be an integer with \(n \geq n_1\) and \(n > 3/\varepsilon\). For all \(m \geq n\), using \(d_V(F^{(m)}, \iota(h_m)) \leq 1/m\) and the triangle inequality in \(V\),

\begin{equation*} d_V\big(F^{(m)}, H\big) \leq d_V\big(F^{(m)}, \iota(h_m)\big) + d_V\big(\iota(h_m), H\big) \leq \tfrac{1}{m} + \tfrac{\varepsilon}{3} < \tfrac{\varepsilon}{3} + \tfrac{\varepsilon}{3} < \varepsilon . \end{equation*}

Hence \(\lim_{m \to \infty} F^{(m)} = H\) in \((V, d_V)\). Every Cauchy sequence in \(V\) converges to an element of \(V\), so \((V,d_V)\) is a complete metric space by 6.14.

(d) This is the identity recorded at the start of (c), which we now restate. For \(f \in U\) the constant sequence \((f,f,f,\dots)\) is Cauchy and hence belongs to \(W\), so the map \(\iota : U \to V\) given by \(\iota(f) = \widehat{(f,f,f,\dots)}\) is defined. If \(f, g \in U\), then \(d(f_k, g_k) = d(f,g)\) for every \(k\) when \(f_k = f\) and \(g_k = g\) for all \(k\); this constant sequence of real numbers has limit \(d(f,g)\), and therefore

\begin{equation*} d_V\big(\widehat{(f,f,f,\dots)}, \widehat{(g,g,g,\dots)}\big) = \lim_{k \to \infty} d(f,g) = d(f,g) . \end{equation*}

Thus \(\iota\) preserves distances. (This computation uses only the definition of \(d_V\), which part (b) showed to be independent of the representatives chosen, so no circularity arises from having invoked it inside (c).)

(e) By (d), the map \(\iota : U \to V\) preserves distances. In particular \(\iota\) is injective: if \(\iota(f) = \iota(g)\), then \(d(f,g) = d_V(\iota(f), \iota(g)) = d_V(\iota(f), \iota(f)) = 0\), and hence \(f = g\) by the second bullet point of 6.1. Thus \(\iota\) is a bijection from \(U\) onto the subset \(\iota(U)\) of \(V\), and it carries the metric \(d\) to the metric that \(\iota(U)\) inherits from \((V,d_V)\): for \(f,g \in U\),

\begin{equation*} d(f,g) = d_V\big(\iota(f), \iota(g)\big). \end{equation*}

So \((U,d)\) and \((\iota(U), d_V|_{\iota(U) \times \iota(U)})\) are indistinguishable as metric spaces; every metric statement about one is the same statement about the other, with \(f\) replaced by \(\iota(f)\). Identifying each \(f \in U\) with \(\iota(f)\), we may regard \(U\) itself as a subset of \(V\), carrying its original metric. Since \((V,d_V)\) is complete by (c), every metric space is (after this identification) a subset of a complete metric space.

Stated without the identification: every metric space is isometric to a subset of a complete metric space. One can say more, though it is not asked here: \(\iota(U)\) is dense in \(V\), since for \(H = \widehat{(h_1,h_2,\dots)} \in V\) the estimate in the second half of (c) shows \(d_V(\iota(h_m), H) \to 0\). This makes \((V, d_V)\) the completion of \((U,d)\).

Because \(a + bi \neq 0\), we do not have \(a = b = 0\); hence \(a^2 + b^2 > 0\) and the right side of the displayed equation makes sense.

Set

\begin{equation*} w = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}\, i . \end{equation*}

To prove that \(w = \frac{1}{a+bi}\) it suffices, by the definition of the multiplicative inverse in the field \(\mathbb{C}\), to show that \((a + bi)w = 1\).

Using the definition of complex multiplication (6.17), which states that \((a + bi)(c + di) = (ac - bd) + (ad + bc)i\), with \(c = \frac{a}{a^2+b^2}\) and \(d = -\frac{b}{a^2+b^2}\), we get

\begin{equation*} \begin{aligned} (a + bi)w &= \left( a \cdot \frac{a}{a^2 + b^2} - b \cdot \left( -\frac{b}{a^2 + b^2} \right) \right)

• \left( a \cdot \left( -\frac{b}{a^2 + b^2} \right) + b \cdot \frac{a}{a^2 + b^2} \right) i \\ &= \frac{a^2 + b^2}{a^2 + b^2} + \frac{-ab + ab}{a^2 + b^2}\, i \\ &= 1 + 0i \\ &= 1 . \end{aligned} \end{equation*}

Because \(\mathbb{C}\) is a field, the multiplicative inverse of the nonzero element \(a + bi\) is unique, and thus

\begin{equation*} \frac{1}{a + bi} = w = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}\, i , \end{equation*}

as desired.

Equivalently, one can obtain the formula rather than merely verify it: by 6.24 we have \(z\bar{z} = |z|^2\) for every \(z \in \mathbb{C}\). Taking \(z = a + bi\), so that \(\bar{z} = a - bi\) and \(|z|^2 = a^2 + b^2 \neq 0\), dividing the equation \(z \bar z = |z|^2\) by \(z|z|^2\) gives

\begin{equation*} \frac{1}{a + bi} = \frac{\overline{a + bi}}{|a + bi|^2} = \frac{a - bi}{a^2 + b^2} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}\, i . \end{equation*}

Write \(z = a + bi\) with \(a, b \in \mathbb{R}\), so that \(\operatorname{Re} z = a\), \(\operatorname{Im} z = b\), and \(|z| = \sqrt{a^2 + b^2}\) by 6.18. Let

\begin{equation*} M = \max\{|a|, |b|\} . \end{equation*}

First we prove the left inequality. Because \(a^2 \geq 0\) and \(b^2 \geq 0\), we have

\begin{equation*} M^2 = \max\{a^2, b^2\} \leq a^2 + b^2 = |z|^2 . \end{equation*}

Here the first equality holds because \(t \mapsto t^2\) is increasing on \([0, \infty)\), so the larger of \(|a|, |b|\) has the larger square. Because \(M \geq 0\) and \(|z| \geq 0\), taking square roots (which preserves the order on \([0,\infty)\)) gives

\begin{equation*} \max\{|\operatorname{Re} z|, |\operatorname{Im} z|\} = M \leq |z| . \end{equation*}

Now we prove the right inequality. Because \(a^2 \leq M^2\) and \(b^2 \leq M^2\), we have

\begin{equation*} |z|^2 = a^2 + b^2 \leq 2M^2 = \left( \sqrt{2}\, M \right)^2 . \end{equation*}

Both \(|z|\) and \(\sqrt{2}M\) are nonnegative, so taking square roots gives

\begin{equation*} |z| \leq \sqrt{2} \, M = \sqrt{2} \max\{|\operatorname{Re} z|, |\operatorname{Im} z|\} , \end{equation*}

completing the proof.

Both inequalities are sharp: the left inequality becomes an equality when \(z\) is real or purely imaginary (for instance \(z = 1\)), and the right inequality becomes an equality when \(|\operatorname{Re} z| = |\operatorname{Im} z|\) (for instance \(z = 1 + i\), where \(|z| = \sqrt{2}\) and \(M = 1\)).

Write \(z = a + bi\) with \(a, b \in \mathbb{R}\), so \(\operatorname{Re} z = a\), \(\operatorname{Im} z = b\), and \(|z| = \sqrt{a^2 + b^2}\) by 6.18.

For the right inequality: Because \(|a| \, |b| \geq 0\), we have

\begin{equation*} \left( |a| + |b| \right)^2 = a^2 + 2|a| \, |b| + b^2 \geq a^2 + b^2 = |z|^2 . \end{equation*}

Both \(|a| + |b|\) and \(|z|\) are nonnegative, so taking square roots gives \(|z| \leq |a| + |b|\), which is the right inequality.

For the left inequality: For all real numbers \(s, t\) we have \((|s| - |t|)^2 \geq 0\), which expands to \(2|s| \, |t| \leq s^2 + t^2\). Applying this with \(s = a\) and \(t = b\) gives

\begin{equation*} \left( |a| + |b| \right)^2 = a^2 + 2|a| \, |b| + b^2 \leq 2\left( a^2 + b^2 \right) = 2|z|^2 = \left( \sqrt{2}\, |z| \right)^2 . \end{equation*}

Again both quantities being squared are nonnegative, so taking square roots gives \(|a| + |b| \leq \sqrt{2}\,|z|\), that is,

\begin{equation*} \frac{|\operatorname{Re} z| + |\operatorname{Im} z|}{\sqrt{2}} \leq |z| , \end{equation*}

completing the proof.

Both inequalities are sharp. Equality holds on the right exactly when \(|a| \, |b| = 0\), that is, exactly when \(z\) is real or purely imaginary. Equality holds on the left exactly when \((|a| - |b|)^2 = 0\), that is, exactly when \(|\operatorname{Re} z| = |\operatorname{Im} z|\).

Write \(w = a + bi\) and \(z = c + di\), where \(a, b, c, d \in \mathbb{R}\).

To prove multiplicativity of the absolute value, recall from 6.17 that

\begin{equation*} wz = (ac - bd) + (ad + bc) i . \end{equation*}

Hence, using the formula \(|u + vi|^2 = u^2 + v^2\) from 6.18 three times,

\begin{equation*} \begin{aligned} |wz|^2 &= (ac - bd)^2 + (ad + bc)^2 \\ &= a^2 c^2 - 2abcd + b^2 d^2 + a^2 d^2 + 2abcd + b^2 c^2 \\ &= a^2 c^2 + b^2 d^2 + a^2 d^2 + b^2 c^2 \\ &= \left( a^2 + b^2 \right) \left( c^2 + d^2 \right) \\ &= |w|^2 |z|^2 = \left( |w| \, |z| \right)^2 . \end{aligned} \end{equation*}

Because \(|wz|\) and \(|w| \, |z|\) are both nonnegative, taking square roots gives \(|wz| = |w| \, |z|\).

Now we prove the triangle inequality. We first record two facts.

First, if \(u \in \mathbb{C}\) then \(\operatorname{Re} u \leq |u|\). Indeed, \(\operatorname{Re} u \leq |\operatorname{Re} u| \leq |u|\), where the last inequality is the left inequality of Exercise 2 in this section (or directly: \((\operatorname{Re} u)^2 \leq (\operatorname{Re} u)^2 + (\operatorname{Im} u)^2 = |u|^2\)).

Second, \(|\bar{u}| = |u|\) for every \(u \in \mathbb{C}\) (this is one of the items of 6.24; it is immediate from 6.18 because \(\operatorname{Re}\bar u = \operatorname{Re} u\) and \(\operatorname{Im} \bar u = -\operatorname{Im} u\)).

By 6.24 we have \(u \bar{u} = |u|^2\) for all \(u \in \mathbb{C}\), and complex conjugation is additive and multiplicative. Thus

\begin{equation*} \begin{aligned} |w + z|^2 &= (w + z)\overline{(w + z)} \\ &= (w + z)(\bar{w} + \bar{z}) \\ &= w\bar{w} + z\bar{z} + w \bar{z} + \bar{w} z \\ &= |w|^2 + |z|^2 + w\bar{z} + \overline{w \bar{z}} , \end{aligned} \end{equation*}

where the last equality uses \(\overline{w\bar z} = \bar w \bar{\bar z} = \bar w z\) (multiplicativity of conjugation together with \(\bar{\bar z} = z\), both in 6.24).

By 6.24 we have \(u + \bar{u} = 2 \operatorname{Re} u\). Applying this with \(u = w \bar{z}\), and then using the two recorded facts along with the multiplicativity of the absolute value proved above, we obtain

\begin{equation*} w \bar{z} + \overline{w \bar{z}} = 2 \operatorname{Re}(w \bar{z}) \leq 2 |w \bar{z}| = 2 |w| \, |\bar{z}| = 2|w| \, |z| . \end{equation*}

Combining the two displays gives

\begin{equation*} |w + z|^2 \leq |w|^2 + 2|w| \, |z| + |z|^2 = \left( |w| + |z| \right)^2 . \end{equation*}

Because \(|w + z|\) and \(|w| + |z|\) are nonnegative, taking square roots yields

\begin{equation*} |w + z| \leq |w| + |z| , \end{equation*}

as desired.

Throughout, \(\mathbb{C}\) is identified with \(\mathbb{R}^2\) by the bijection \(z \mapsto (\operatorname{Re} z, \operatorname{Im} z)\). Write

\begin{equation*} u = \operatorname{Re} f \quad\text{and}\quad v = \operatorname{Im} f , \end{equation*}

so that \(u, v \colon X \to \mathbb{R}\) and, under the identification, \(f(x) = \bigl( u(x), v(x) \bigr)\) for each \(x \in X\). Recall from 6.19 that \(f\) is \(\mathcal{S}\)-measurable means precisely that \(u\) and \(v\) are both \(\mathcal{S}\)-measurable real-valued functions.

We prove (a) \(\Rightarrow\) (c) \(\Rightarrow\) (b) \(\Rightarrow\) (a).

First suppose (a) holds, so \(u\) and \(v\) are \(\mathcal{S}\)-measurable. Let

\begin{equation*} \mathcal{T} = \left\{ B \subseteq \mathbb{R}^2 : f^{-1}(B) \in \mathcal{S} \right\} . \end{equation*}

We claim \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}^2\). Indeed \(f^{-1}(\varnothing) = \varnothing \in \mathcal{S}\), so \(\varnothing \in \mathcal{T}\). If \(B \in \mathcal{T}\), then by 2.33(a) we have \(f^{-1}(\mathbb{R}^2 \setminus B) = X \setminus f^{-1}(B) \in \mathcal{S}\), so \(\mathbb{R}^2 \setminus B \in \mathcal{T}\); thus \(\mathcal{T}\) is closed under complementation. If \(B_1, B_2, \ldots \in \mathcal{T}\), then by 2.33(b) we have \(f^{-1}\bigl( \bigcup_{k=1}^{\infty} B_k \bigr) = \bigcup_{k=1}^{\infty} f^{-1}(B_k) \in \mathcal{S}\), so \(\mathcal{T}\) is closed under countable unions. Hence \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}^2\).

Next we show that \(\mathcal{T}\) contains every open cube in \(\mathbb{R}^2\). Recall that for \(x \in \mathbb{R}^2\) and \(\delta > 0\) the open cube \(B(x, \delta)\) consists of the points \(y\) with \(\| y - x \|_{\infty} < \delta\); thus an open cube in \(\mathbb{R}^2\) is exactly a set of the form \(C = I_1 \times I_2\), where \(I_1\) and \(I_2\) are bounded open intervals of the same length (namely \(2\delta\)). For such \(C\),

\begin{equation*} f^{-1}( C) = \left\{ x \in X : u(x) \in I_1 \text{ and } v(x) \in I_2 \right\} = u^{-1}(I_1) \cap v^{-1}(I_2) . \end{equation*}

Because \(I_1\) and \(I_2\) are open subsets of \(\mathbb{R}\), they are Borel sets, and thus \(u^{-1}(I_1) \in \mathcal{S}\) and \(v^{-1}(I_2) \in \mathcal{S}\) by the definition (2.35) of \(\mathcal{S}\)-measurability of the real-valued functions \(u\) and \(v\). Because \(\mathcal{S}\) is a \(\sigma\)-algebra it is closed under finite intersections (2.25(b)), so \(f^{-1}( C) \in \mathcal{S}\), that is, \(C \in \mathcal{T}\).

Thus \(\mathcal{T}\) is a \(\sigma\)-algebra on \(\mathbb{R}^2\) containing all open cubes in \(\mathbb{R}^2\). By 5.38(b), \(\mathcal{B}_2\) is the smallest such \(\sigma\)-algebra, and therefore \(\mathcal{B}_2 \subseteq \mathcal{T}\). In other words, \(f^{-1}(B) \in \mathcal{S}\) for every \(B \in \mathcal{B}_2\), which is (c).

Now suppose (c) holds. By the definition of \(\mathcal{B}_2\) (see 5.37), every open subset \(G\) of \(\mathbb{R}^2\) belongs to \(\mathcal{B}_2\). Hence \(f^{-1}(G) \in \mathcal{S}\) for every open \(G \subseteq \mathbb{R}^2\), which is (b).

Finally, suppose (b) holds. Fix \(a \in \mathbb{R}\). The set \((a, \infty) \times \mathbb{R}\) is an open subset of \(\mathbb{R}^2\), and

\begin{equation*} u^{-1}\bigl( (a, \infty) \bigr) = \left\{ x \in X : u(x) > a \right\} = f^{-1}\bigl( (a, \infty) \times \mathbb{R} \bigr) , \end{equation*}

which is in \(\mathcal{S}\) by (b). Because \(a \in \mathbb{R}\) was arbitrary, 2.39 implies that \(u\) is an \(\mathcal{S}\)-measurable function. Similarly, \(\mathbb{R} \times (a, \infty)\) is open in \(\mathbb{R}^2\) and

\begin{equation*} v^{-1}\bigl( (a, \infty) \bigr) = f^{-1}\bigl( \mathbb{R} \times (a, \infty) \bigr) \in \mathcal{S} \end{equation*}

for all \(a \in \mathbb{R}\), so 2.39 implies that \(v\) is \(\mathcal{S}\)-measurable. Because \(u = \operatorname{Re} f\) and \(v = \operatorname{Im} f\) are both \(\mathcal{S}\)-measurable, 6.19 tells us that \(f\) is \(\mathcal{S}\)-measurable, which is (a).

Having proved (a) \(\Rightarrow\) (c) \(\Rightarrow\) (b) \(\Rightarrow\) (a), we conclude that the three conditions are equivalent.

Write

\begin{equation*} f = u_1 + i v_1 \quad\text{and}\quad g = u_2 + i v_2 , \end{equation*}

where \(u_1 = \operatorname{Re} f\), \(v_1 = \operatorname{Im} f\), \(u_2 = \operatorname{Re} g\), \(v_2 = \operatorname{Im} g\) are real-valued functions on \(X\). By 6.19, the hypothesis that \(f\) and \(g\) are \(\mathcal{S}\)-measurable means exactly that the four real-valued functions \(u_1, v_1, u_2, v_2\) are \(\mathcal{S}\)-measurable. Also by 6.19, to prove that a complex-valued function on \(X\) is \(\mathcal{S}\)-measurable it suffices to prove that its real and imaginary parts are \(\mathcal{S}\)-measurable real-valued functions. Throughout we use 2.46, which states that sums, differences, products, and (where the denominator never vanishes) quotients of \(\mathcal{S}\)-measurable real-valued functions are \(\mathcal{S}\)-measurable.

For (a), the definition of addition in \(\mathbb{C}\) (6.17) gives

\begin{equation*} \operatorname{Re}(f + g) = u_1 + u_2 , \qquad \operatorname{Im}(f + g) = v_1 + v_2 , \end{equation*}

where these equalities hold pointwise on \(X\). Both right sides are \(\mathcal{S}\)-measurable by 2.46(a). Hence \(f + g\) is \(\mathcal{S}\)-measurable by 6.19. In the same way,

\begin{equation*} \operatorname{Re}(f - g) = u_1 - u_2 , \qquad \operatorname{Im}(f - g) = v_1 - v_2 , \end{equation*}

which are \(\mathcal{S}\)-measurable by 2.46(a); hence \(f - g\) is \(\mathcal{S}\)-measurable.

For the product, the definition of multiplication in \(\mathbb{C}\) (6.17) gives, pointwise on \(X\),

\begin{equation*} fg = (u_1 u_2 - v_1 v_2) + (u_1 v_2 + v_1 u_2) i , \end{equation*}

so

\begin{equation*} \operatorname{Re}(fg) = u_1 u_2 - v_1 v_2 , \qquad \operatorname{Im}(fg) = u_1 v_2 + v_1 u_2 . \end{equation*}

Each of \(u_1 u_2\), \(v_1 v_2\), \(u_1 v_2\), \(v_1 u_2\) is \(\mathcal{S}\)-measurable by 2.46(a) (products), and hence so are the displayed difference and sum, again by 2.46(a). Thus \(fg\) is \(\mathcal{S}\)-measurable by 6.19, completing the proof of (a).

For (b), suppose \(g(x) \neq 0\) for all \(x \in X\). Define \(h \colon X \to \mathbb{R}\) by

\begin{equation*} h = |g|^2 = u_2^{\,2} + v_2^{\,2} . \end{equation*}

Then \(u_2^{\,2}\) and \(v_2^{\,2}\) are \(\mathcal{S}\)-measurable by 2.45, and hence \(h\) is \(\mathcal{S}\)-measurable by 2.46(a). Moreover \(h(x) > 0\) for every \(x \in X\): if \(h(x) = 0\) then \(u_2(x) = v_2(x) = 0\), so \(g(x) = 0\), contradicting the hypothesis. In particular \(h(x) \neq 0\) for all \(x \in X\).

By 6.24 we have \(g \bar{g} = |g|^2 = h\), and \(h\) is nowhere zero, so dividing by \(gh\) shows that pointwise on \(X\)

\begin{equation*} \frac{f}{g} = \frac{f \bar{g}}{g \bar{g}} = \frac{f \bar{g}}{h} . \end{equation*}

By 6.23, \(\bar{g} = u_2 - i v_2\), so by the multiplication rule in 6.17,

\begin{equation*} f \bar{g} = (u_1 + i v_1)(u_2 - i v_2) = (u_1 u_2 + v_1 v_2) + (v_1 u_2 - u_1 v_2) i . \end{equation*}

Therefore

\begin{equation*} \operatorname{Re}\frac{f}{g} = \frac{u_1 u_2 + v_1 v_2}{u_2^{\,2} + v_2^{\,2}} , \qquad \operatorname{Im}\frac{f}{g} = \frac{v_1 u_2 - u_1 v_2}{u_2^{\,2} + v_2^{\,2}} , \end{equation*}

where we have used that dividing a complex number by the nonzero real number \(h\) divides its real part and its imaginary part by \(h\).

The two numerators above are \(\mathcal{S}\)-measurable by 2.46(a), and the denominator \(h\) is an \(\mathcal{S}\)-measurable function that is never \(0\). Hence both quotients are \(\mathcal{S}\)-measurable by 2.46(b). Thus \(\operatorname{Re}\frac{f}{g}\) and \(\operatorname{Im}\frac{f}{g}\) are \(\mathcal{S}\)-measurable, and so \(\frac{f}{g}\) is an \(\mathcal{S}\)-measurable function by 6.19.

The key point is that convergence in \(\mathbb{C}\) is equivalent to convergence of the real parts together with convergence of the imaginary parts. Section 6B records this after 6.18: if \(z_1, z_2, \ldots \in \mathbb{C}\) and \(L \in \mathbb{C}\), then

\begin{equation*} \lim_{k \to \infty} z_k = L \iff \lim_{k \to \infty} \operatorname{Re} z_k = \operatorname{Re} L \ \text{ and } \ \lim_{k \to \infty} \operatorname{Im} z_k = \operatorname{Im} L . \end{equation*}

For completeness, here is the proof of the direction we need. Suppose \(\lim_{k \to \infty} z_k = L\), meaning \(\lim_{k \to \infty} |z_k - L| = 0\) (see 6.18). Because \(\operatorname{Re}\) and \(\operatorname{Im}\) are additive, \(\operatorname{Re} z_k - \operatorname{Re} L = \operatorname{Re}(z_k - L)\) and \(\operatorname{Im} z_k - \operatorname{Im} L = \operatorname{Im}(z_k - L)\). The left inequality of Exercise 2 in this section gives

\begin{equation*} \left| \operatorname{Re} z_k - \operatorname{Re} L \right| \leq |z_k - L| \quad\text{and}\quad \left| \operatorname{Im} z_k - \operatorname{Im} L \right| \leq |z_k - L| . \end{equation*}

Letting \(k \to \infty\) and using the Squeeze Theorem for real sequences, we get \(\lim_{k \to \infty} \operatorname{Re} z_k = \operatorname{Re} L\) and \(\lim_{k \to \infty} \operatorname{Im} z_k = \operatorname{Im} L\).

Now fix \(x \in X\). Applying the above with \(z_k = f_k(x)\) and \(L = f(x)\) (the limit exists by hypothesis) shows that

\begin{equation*} \lim_{k \to \infty} (\operatorname{Re} f_k)(x) = (\operatorname{Re} f)(x) \quad\text{and}\quad \lim_{k \to \infty} (\operatorname{Im} f_k)(x) = (\operatorname{Im} f)(x) . \end{equation*}

In particular, both of these limits exist for every \(x \in X\).

Each \(f_k\) is an \(\mathcal{S}\)-measurable complex-valued function, so by 6.19 the real-valued functions \(\operatorname{Re} f_k\) and \(\operatorname{Im} f_k\) are \(\mathcal{S}\)-measurable. Thus \(\operatorname{Re} f_1, \operatorname{Re} f_2, \ldots\) is a sequence of \(\mathcal{S}\)-measurable functions from \(X\) to \(\mathbb{R}\) whose pointwise limit exists at every point of \(X\) and equals \(\operatorname{Re} f\). Hence 2.48 implies that \(\operatorname{Re} f\) is an \(\mathcal{S}\)-measurable function. The same argument applied to \(\operatorname{Im} f_1, \operatorname{Im} f_2, \ldots\) shows that \(\operatorname{Im} f\) is an \(\mathcal{S}\)-measurable function.

Because \(\operatorname{Re} f\) and \(\operatorname{Im} f\) are both \(\mathcal{S}\)-measurable, 6.19 tells us that \(f\) is an \(\mathcal{S}\)-measurable function, as desired.

Write \(u = \operatorname{Re} f\) and \(v = \operatorname{Im} f\), so that \(u, v : X \to \mathbb{R}\) are \(\mathcal{S}\)-measurable (by 6.19) and \(f = u + iv\). Write \(\alpha = a + bi\) with \(a, b \in \mathbb{R}\).

First we check that every integral appearing below is defined and finite. For each \(x \in X\) we have \(|u(x)| \le |f(x)|\) and \(|v(x)| \le |f(x)|\), because \(|f(x)| = \sqrt{u(x)^2 + v(x)^2}\). The functions \(|u|, |v|, |f|\) are nonnegative and \(\mathcal{S}\)-measurable (measurability of \(|f|\) is the case \(p = 1\) of 6.20), so 3.8 gives

\begin{equation*} \int |u| \, d\mu \le \int |f| \, d\mu < \infty \quad\text{and}\quad \int |v| \, d\mu \le \int |f| \, d\mu < \infty . \end{equation*}

Now \(u^{+} \le |u|\) and \(u^{-} \le |u|\) pointwise, so 3.8 gives \(\int u^{+} \, d\mu < \infty\) and \(\int u^{-} \, d\mu < \infty\); hence \(\int u \, d\mu = \int u^{+} \, d\mu - \int u^{-} \, d\mu\) is defined and is a real number. The same argument applies to \(v\). Thus \(\int f \, d\mu\) is defined by 6.21 as

\begin{equation*} \int f \, d\mu = \int u \, d\mu + i \int v \, d\mu . \end{equation*}

Next we check that \(\int \alpha f \, d\mu\) is defined. Multiplying out with the definition 6.17 of complex multiplication,

\begin{equation*} \alpha f = (a + bi)(u + iv) = (au - bv) + i\,(av + bu), \end{equation*}

so

\begin{equation*} \operatorname{Re}(\alpha f) = au - bv \quad\text{and}\quad \operatorname{Im}(\alpha f) = av + bu . \end{equation*}

Constant functions are \(\mathcal{S}\)-measurable, so \(au\), \(bv\), \(av\), \(bu\) are \(\mathcal{S}\)-measurable by 2.46(a), and therefore \(au - bv\) and \(av + bu\) are \(\mathcal{S}\)-measurable, again by 2.46(a). Thus \(\alpha f\) is an \(\mathcal{S}\)-measurable function from \(X\) to \(\mathbb{C}\), by 6.19. Moreover \(|\alpha f(x)| = |\alpha| \, |f(x)|\) for every \(x \in X\), by the multiplicativity of the absolute value on \(\mathbb{C}\) (Exercise 4 in this section; it also follows from a direct computation with 6.17 and 6.18). Since \(|f|\) is nonnegative and \(\mathcal{S}\)-measurable, its integral is defined, so 3.20 with \(c = |\alpha|\) gives

\begin{equation*} \int |\alpha f| \, d\mu = |\alpha| \int |f| \, d\mu < \infty . \end{equation*}

Hence \(\int \alpha f \, d\mu\) is defined, and 6.21 gives

\begin{equation*} \int \alpha f \, d\mu = \int (au - bv) \, d\mu + i \int (av + bu) \, d\mu . \end{equation*}

Now we evaluate the two real integrals on the right. Each of \(au\), \(bv\), \(av\), \(bu\) has absolute value with a finite integral: for instance \(\int |au| \, d\mu = |a| \int |u| \, d\mu < \infty\) by 3.20. Hence additivity of integration (3.21) applies to each of the two sums, and homogeneity of integration (3.20) applies to each real scalar multiple. This gives

\begin{equation*} \begin{aligned} \int (au - bv) \, d\mu &= \int au \, d\mu + \int (-b)v \, d\mu = a \int u \, d\mu - b \int v \, d\mu, \\ \int (av + bu) \, d\mu &= \int av \, d\mu + \int bu \, d\mu = a \int v \, d\mu + b \int u \, d\mu . \end{aligned} \end{equation*}

Therefore

\begin{equation*} \int \alpha f \, d\mu = \left( a \int u \, d\mu - b \int v \, d\mu \right) + i \left( a \int v \, d\mu + b \int u \, d\mu \right). \end{equation*}

On the other hand, using 6.21 and the definition 6.17 of complex multiplication,

\begin{equation*} \alpha \int f \, d\mu = (a + bi)\left( \int u \, d\mu + i \int v \, d\mu \right) = \left( a \int u \, d\mu - b \int v \, d\mu \right) + i \left( a \int v \, d\mu + b \int u \, d\mu \right). \end{equation*}

The two displayed expressions are identical, so \(\int \alpha f \, d\mu = \alpha \int f \, d\mu\), as desired.

Let \(\mathcal{A}\) be a collection of subspaces of \(V\) and let

\begin{equation*} U = \bigcap_{U^{\prime} \in \mathcal{A}} U^{\prime} \end{equation*}

denote the intersection of the collection, the intersection being taken inside \(V\) (with the usual convention that the intersection of the empty collection of subsets of \(V\) is \(V\) itself, which is a subspace of \(V\); the argument below covers that degenerate case as well, since each of its three verifications is then vacuously about \(V\)).

Note first that \(U \subseteq V\): this is immediate when \(\mathcal{A} \neq \emptyset\), since \(U \subseteq U^{\prime} \subseteq V\) for any \(U^{\prime} \in \mathcal{A}\), and it holds by the stated convention when \(\mathcal{A} = \emptyset\). We now verify the three conditions of 6.31.

Additive identity. Each \(U^{\prime} \in \mathcal{A}\) is a subspace of \(V\), so the forward direction of 6.31 gives \(0 \in U^{\prime}\), where \(0\) denotes the additive identity of \(V\). Since \(0 \in U^{\prime}\) for every \(U^{\prime} \in \mathcal{A}\), we have \(0 \in U\). In particular \(U\) is nonempty.

Closed under addition. Suppose \(f, g \in U\). Let \(U^{\prime} \in \mathcal{A}\). Then \(f \in U^{\prime}\) and \(g \in U^{\prime}\) by the definition of intersection, and \(U^{\prime}\) is closed under addition by 6.31, so \(f + g \in U^{\prime}\). As \(U^{\prime} \in \mathcal{A}\) was arbitrary, \(f + g\) belongs to every member of \(\mathcal{A}\), which means \(f + g \in U\).

Closed under scalar multiplication. Suppose \(\alpha \in \mathbf{F}\) and \(f \in U\). Let \(U^{\prime} \in \mathcal{A}\). Then \(f \in U^{\prime}\), and \(U^{\prime}\) is closed under scalar multiplication by 6.31, so \(\alpha f \in U^{\prime}\). As \(U^{\prime} \in \mathcal{A}\) was arbitrary, \(\alpha f \in U\).

Thus \(U\) is a subset of \(V\) satisfying the three conditions of 6.31, and therefore the reverse direction of 6.31 shows that \(U\) is a subspace of \(V\).

Here \(V\) and \(W\) are vector spaces over the same field \(\mathbf{F}\) (recall 6.25: \(\mathbf{F}\) denotes \(\mathbb{R}\) or \(\mathbb{C}\)), which is what makes the definition of \(\alpha(f,g)\) meaningful. Throughout we use repeatedly that two ordered pairs are equal if and only if their first coordinates are equal and their second coordinates are equal.

The operations are well defined. If \((f_1, g_1), (f_2, g_2) \in V \times W\), then \(f_1 + f_2 \in V\) because addition on \(V\) maps into \(V\), and \(g_1 + g_2 \in W\) likewise; hence \((f_1 + f_2, g_1 + g_2) \in V \times W\). Similarly, if \(\alpha \in \mathbf{F}\) and \((f,g) \in V \times W\), then \(\alpha f \in V\) and \(\alpha g \in W\), so \((\alpha f, \alpha g) \in V \times W\). Thus the two formulas do define an addition and a scalar multiplication on \(V \times W\) in the sense of 6.26.

We now verify each property in the definition 6.27 of a vector space. In every case the verification is carried out coordinatewise, using the corresponding property in \(V\) for the first coordinate and in \(W\) for the second.

Commutativity. For \((f_1, g_1), (f_2, g_2) \in V \times W\),

\begin{equation*} (f_1, g_1) + (f_2, g_2) = (f_1 + f_2,\ g_1 + g_2) = (f_2 + f_1,\ g_2 + g_1) = (f_2, g_2) + (f_1, g_1), \end{equation*}

where the middle equality uses commutativity of addition in \(V\) and in \(W\).

Associativity of addition. For \((f_1, g_1), (f_2, g_2), (f_3, g_3) \in V \times W\),

\begin{equation*} \begin{aligned} \bigl( (f_1, g_1) + (f_2, g_2) \bigr) + (f_3, g_3) &= (f_1 + f_2,\ g_1 + g_2) + (f_3, g_3) \\ &= \bigl( (f_1 + f_2) + f_3,\ (g_1 + g_2) + g_3 \bigr) \\ &= \bigl( f_1 + (f_2 + f_3),\ g_1 + (g_2 + g_3) \bigr) \\ &= (f_1, g_1) + (f_2 + f_3,\ g_2 + g_3) \\ &= (f_1, g_1) + \bigl( (f_2, g_2) + (f_3, g_3) \bigr), \end{aligned} \end{equation*}

where the third equality uses associativity of addition in \(V\) and in \(W\).

Associativity of scalar multiplication. For \(\alpha, \beta \in \mathbf{F}\) and \((f,g) \in V \times W\),

\begin{equation*} (\alpha\beta)(f, g) = \bigl( (\alpha\beta) f,\ (\alpha\beta) g \bigr) = \bigl( \alpha(\beta f),\ \alpha(\beta g) \bigr) = \alpha(\beta f,\ \beta g) = \alpha \bigl( \beta (f,g) \bigr), \end{equation*}

where the second equality uses the corresponding associativity property in \(V\) and in \(W\).

Additive identity. Let \(0_V\) and \(0_W\) denote the additive identities of \(V\) and \(W\), and set \(0 = (0_V, 0_W) \in V \times W\). For every \((f,g) \in V \times W\),

\begin{equation*} (f, g) + (0_V, 0_W) = (f + 0_V,\ g + 0_W) = (f, g). \end{equation*}

Thus \((0_V, 0_W)\) is an additive identity for \(V \times W\).

Additive inverse. Suppose \((f, g) \in V \times W\). Because \(V\) is a vector space, there exists \(f^{\prime} \in V\) with \(f + f^{\prime} = 0_V\); because \(W\) is a vector space, there exists \(g^{\prime} \in W\) with \(g + g^{\prime} = 0_W\). Then \((f^{\prime}, g^{\prime}) \in V \times W\) and

\begin{equation*} (f, g) + (f^{\prime}, g^{\prime}) = (f + f^{\prime},\ g + g^{\prime}) = (0_V, 0_W). \end{equation*}

Thus every element of \(V \times W\) has an additive inverse in \(V \times W\).

Multiplicative identity. For \((f, g) \in V \times W\),

\begin{equation*} 1(f, g) = (1f,\ 1g) = (f, g), \end{equation*}

using the multiplicative identity property in \(V\) and in \(W\).

Distributive properties. For \(\alpha \in \mathbf{F}\) and \((f_1, g_1), (f_2, g_2) \in V \times W\),

\begin{equation*} \begin{aligned} \alpha \bigl( (f_1, g_1) + (f_2, g_2) \bigr) &= \alpha (f_1 + f_2,\ g_1 + g_2) \\ &= \bigl( \alpha(f_1 + f_2),\ \alpha(g_1 + g_2) \bigr) \\ &= (\alpha f_1 + \alpha f_2,\ \alpha g_1 + \alpha g_2) \\ &= (\alpha f_1,\ \alpha g_1) + (\alpha f_2,\ \alpha g_2) \\ &= \alpha(f_1, g_1) + \alpha(f_2, g_2), \end{aligned} \end{equation*}

where the third equality uses the first distributive property in \(V\) and in \(W\). Also, for \(\alpha, \beta \in \mathbf{F}\) and \((f, g) \in V \times W\),

\begin{equation*} \begin{aligned} (\alpha + \beta)(f, g) &= \bigl( (\alpha + \beta) f,\ (\alpha + \beta) g \bigr) \\ &= (\alpha f + \beta f,\ \alpha g + \beta g) \\ &= (\alpha f,\ \alpha g) + (\beta f,\ \beta g) \\ &= \alpha(f, g) + \beta(f, g), \end{aligned} \end{equation*}

where the second equality uses the second distributive property in \(V\) and in \(W\).

All the properties required by 6.27 hold, so \(V \times W\), with the addition and scalar multiplication defined above, is a vector space over \(\mathbf{F}\).

Let \((V, \|\cdot\|)\) be a normed vector space and let \(N \colon V \to \mathbf{F}\) be defined by \(N(f) = \|f\|\). Recall from 6.36 that the metric on \(V\) is \(d(f,g) = \|f - g\|\), and the metric on \(\mathbf{F}\) is \(d(s,t) = |s - t|\).

First we prove the reverse triangle inequality: if \(f, g \in V\), then

\begin{equation*} \bigl| \|f\| - \|g\| \bigr| \le \|f - g\|. \end{equation*}

To see this, note that the triangle inequality (see 6.33) gives

\begin{equation*} \|f\| = \|(f - g) + g\| \le \|f - g\| + \|g\|, \end{equation*}

and hence \(\|f\| - \|g\| \le \|f - g\|\). Interchanging the roles of \(f\) and \(g\) gives \(\|g\| - \|f\| \le \|g - f\|\). By homogeneity (see 6.33) with \(\alpha = -1\) we have

\begin{equation*} \|g - f\| = \|(-1)(f - g)\| = |-1| \, \|f - g\| = \|f - g\|. \end{equation*}

Thus \(\|g\| - \|f\| \le \|f - g\|\) as well. Because \(\bigl| \|f\| - \|g\| \bigr|\) equals either \(\|f\| - \|g\|\) or \(\|g\| - \|f\|\), the displayed inequality holds.

Now we verify continuity at an arbitrary \(f \in V\), using the definition of continuity for metric spaces (see 6.10). Suppose \(\varepsilon > 0\). Take \(\delta = \varepsilon\). If \(g \in V\) satisfies \(\|g - f\| < \delta\), then the reverse triangle inequality gives

\begin{equation*} \bigl| N(g) - N(f) \bigr| = \bigl| \|g\| - \|f\| \bigr| \le \|g - f\| < \delta = \varepsilon. \end{equation*}

Thus \(N\) is continuous at \(f\). Because \(f \in V\) was arbitrary, \(N\) is continuous on \(V\).

In fact the estimate above shows more: \(N\) is Lipschitz with Lipschitz constant \(1\), and hence uniformly continuous on \(V\).

Note that \(N\) is not a linear map (indeed \(N\) takes only nonnegative real values, so it is not linear unless \(V = \{0\}\)), so 6.48 does not apply here; the direct argument above is needed.

Here \(B(f,r) = \{g \in V : \|f - g\| < r\}\) is the open ball and \(\overline{B}(f,r) = \{g \in V : \|f - g\| \le r\}\) is the closed ball (see 6.3, interpreted via the metric \(d(g,h) = \|g - h\|\) of 6.36), while \(\overline{B(f,r)}\) denotes the closure of the open ball (see 6.7). We prove the two inclusions separately, using the characterization of closure by limits of sequences given in 6.9(a).

First suppose \(g \in \overline{B(f,r)}\). By 6.9(a) there is a sequence \(g_1, g_2, \ldots\) in \(B(f,r)\) with \(\lim_{k \to \infty} g_k = g\), which in the normed vector space context means \(\lim_{k \to \infty} \|g_k - g\| = 0\). For each \(k \in \mathbf{Z}^+\) the triangle inequality gives

\begin{equation*} \|f - g\| \le \|f - g_k\| + \|g_k - g\| < r + \|g_k - g\|. \end{equation*}

Letting \(k \to \infty\) in the inequality \(\|f - g\| < r + \|g_k - g\|\) shows that \(\|f - g\| \le r\). Thus \(g \in \overline{B}(f,r)\), proving that \(\overline{B(f,r)} \subseteq \overline{B}(f,r)\).

Now suppose \(g \in \overline{B}(f,r)\), so \(\|f - g\| \le r\). For \(k \in \mathbf{Z}^+\) define

\begin{equation*} g_k = f + \Bigl(1 - \tfrac{1}{k}\Bigr)(g - f). \end{equation*}

Because \(1 - \frac{1}{k} \in [0,1)\), homogeneity of the norm gives

\begin{equation*} \|g_k - f\| = \Bigl(1 - \tfrac{1}{k}\Bigr) \|g - f\| \le \Bigl(1 - \tfrac{1}{k}\Bigr) r < r, \end{equation*}

so \(g_k \in B(f,r)\) for each \(k \in \mathbf{Z}^+\). Furthermore

\begin{equation*} \|g_k - g\| = \Bigl\| \Bigl(1 - \tfrac{1}{k}\Bigr)(g - f) - (g - f) \Bigr\| = \tfrac{1}{k} \|g - f\| \le \tfrac{r}{k}, \end{equation*}

which tends to \(0\) as \(k \to \infty\). Hence \(\lim_{k \to \infty} g_k = g\) with all \(g_k \in B(f,r)\), and so \(g \in \overline{B(f,r)}\) by 6.9(a). This proves that \(\overline{B}(f,r) \subseteq \overline{B(f,r)}\).

The two inclusions together give \(\overline{B(f,r)} = \overline{B}(f,r)\).

Note that the vector space structure is essential here; the corresponding statement is false in a general metric space. For example, if \(V\) is a set with at least two elements equipped with the discrete metric (\(d(g,h) = 1\) when \(g \ne h\)), then \(B(f,1) = \{f\}\), which is closed, so \(\overline{B(f,1)} = \{f\}\), whereas \(\overline{B}(f,1) = V\). The second half of the proof above used scalar multiplication to slide points of the closed ball inward, which is exactly what the discrete metric space lacks.

Recall the two functions in question. Suppose \(n \in \mathbf{Z}^+\). The second-to-last bullet point of 6.35 defines \(\|\cdot\|\) on \(\mathbf{F}^n\) by

\begin{equation*} \|(a_1, \ldots, a_n)\| = |a_1|^{1/2} + \cdots + |a_n|^{1/2}, \end{equation*}

and the last bullet point defines \(\|\cdot\|_{1/2}\) on \(\mathbf{F}^n\) by

\begin{equation*} \|(a_1, \ldots, a_n)\|_{1/2} = \bigl( |a_1|^{1/2} + \cdots + |a_n|^{1/2} \bigr)^2 . \end{equation*}

By 6.33, to show that a function is not a norm it suffices to exhibit a failure of any one of the three defining conditions.

First consider \(\|\cdot\|\) as defined in the second-to-last bullet point. We show that homogeneity fails. Let \(f = (1, 0, \ldots, 0) \in \mathbf{F}^n\) and let \(\alpha = 4\). Then

\begin{equation*} \|f\| = |1|^{1/2} = 1 \qquad \text{and} \qquad \|\alpha f\| = \|(4,0,\ldots,0)\| = |4|^{1/2} = 2 . \end{equation*}

Thus \(\|\alpha f\| = 2\), while \(|\alpha| \, \|f\| = 4 \cdot 1 = 4\). Hence \(\|\alpha f\| \ne |\alpha| \, \|f\|\), so the homogeneity condition in 6.33 fails and \(\|\cdot\|\) is not a norm on \(\mathbf{F}^n\).

More generally, for every \(\alpha \in \mathbf{F}\) and every \((a_1, \ldots, a_n) \in \mathbf{F}^n\) we have

\begin{equation*} \|\alpha(a_1, \ldots, a_n)\| = \sum_{k=1}^{n} |\alpha a_k|^{1/2} = |\alpha|^{1/2} \sum_{k=1}^{n} |a_k|^{1/2} = |\alpha|^{1/2} \, \|(a_1, \ldots, a_n)\| , \end{equation*}

so this function scales like \(|\alpha|^{1/2}\) rather than like \(|\alpha|\); the two agree only when \(|\alpha| \in \{0, 1\}\) or when the vector is \(0\).

Now consider \(\|\cdot\|_{1/2}\) as defined in the last bullet point, and suppose \(n > 1\). (This hypothesis is necessary: when \(n = 1\) we have \(\|(a_1)\|_{1/2} = \bigl(|a_1|^{1/2}\bigr)^2 = |a_1|\), which is the usual absolute value norm on \(\mathbf{F}\).) Note first that \(\|\cdot\|_{1/2}\) does satisfy homogeneity, since

\begin{equation*} \|\alpha(a_1, \ldots, a_n)\|_{1/2} = \Bigl( |\alpha|^{1/2} \sum_{k=1}^{n} |a_k|^{1/2} \Bigr)^{2} = |\alpha| \, \|(a_1, \ldots, a_n)\|_{1/2}, \end{equation*}

and it is positive definite, because the sum \(\sum_{k=1}^n |a_k|^{1/2}\) is \(0\) if and only if every \(a_k\) is \(0\). What fails is the triangle inequality. Let

\begin{equation*} f = (1, 0, 0, \ldots, 0) \qquad \text{and} \qquad g = (0, 1, 0, \ldots, 0), \end{equation*}

both of which lie in \(\mathbf{F}^n\) because \(n \ge 2\). Then

\begin{equation*} \|f\|_{1/2} = \bigl(|1|^{1/2}\bigr)^2 = 1, \qquad \|g\|_{1/2} = 1, \end{equation*}

while \(f + g = (1, 1, 0, \ldots, 0)\) and so

\begin{equation*} \|f + g\|_{1/2} = \bigl( |1|^{1/2} + |1|^{1/2} \bigr)^2 = 2^2 = 4 . \end{equation*}

Thus \(\|f + g\|_{1/2} = 4 > 2 = \|f\|_{1/2} + \|g\|_{1/2}\), so the triangle inequality in 6.33 fails. Hence \(\|\cdot\|_{1/2}\) is not a norm on \(\mathbf{F}^n\) when \(n > 1\).

Suppose \(V\) is a normed vector space and \(f_1, f_2, \ldots\) is a Cauchy sequence in \(V\). As recorded in Section 6C, in the context of a normed vector space the Cauchy condition (6.12) states that for every \(\varepsilon > 0\) there exists \(n \in \mathbf{Z}^+\) such that \(\|f_j - f_k\| < \varepsilon\) for all integers \(j \ge n\) and \(k \ge n\).

Apply this with \(\varepsilon = 1\): there exists \(n \in \mathbf{Z}^+\) such that

\begin{equation*} \|f_j - f_k\| < 1 \quad \text{for all integers } j \ge n \text{ and } k \ge n . \end{equation*}

In particular, if \(k \ge n\) then \(\|f_k - f_n\| < 1\), and hence the triangle inequality gives

\begin{equation*} \|f_k\| = \|(f_k - f_n) + f_n\| \le \|f_k - f_n\| + \|f_n\| < 1 + \|f_n\| . \end{equation*}

Thus all terms of the sequence from the \(n^{\text{th}}\) onwards have norm less than \(1 + \|f_n\|\).

Only finitely many terms remain, namely \(f_1, \ldots, f_{n-1}\) (there are none of these if \(n = 1\)). Define

\begin{equation*} M = \max \bigl\{ \|f_1\|, \ldots, \|f_{n-1}\|, \, 1 + \|f_n\| \bigr\}, \end{equation*}

which is a well-defined real number because it is the maximum of a finite nonempty set of real numbers.

We claim that the real number \(M + 1\) is greater than the norm of every element of the sequence. Indeed, if \(k < n\) then \(\|f_k\| \le M < M + 1\), and if \(k \ge n\) then \(\|f_k\| < 1 + \|f_n\| \le M < M + 1\). Hence

\begin{equation*} \|f_k\| < M + 1 \quad \text{for every } k \in \mathbf{Z}^+ , \end{equation*}

which is exactly the assertion that the Cauchy sequence \(f_1, f_2, \ldots\) is bounded.

This result is used in the proof of 6.47, where the boundedness of a Cauchy sequence \(T_1, T_2, \ldots\) in \(\mathcal{B}(V,W)\) guarantees that \(\sup\{\|T_k\| : k \in \mathbf{Z}^+\} < \infty\).

The two norms in the first bullet point of 6.34 are defined on \(\mathbf{F}^n\) by

\begin{equation*} \|(a_1, \ldots, a_n)\|_1 = |a_1| + \cdots + |a_n| \qquad \text{and} \qquad \|(a_1, \ldots, a_n)\|_\infty = \max\{|a_1|, \ldots, |a_n|\} . \end{equation*}

Step 1: both are norms. Positive definiteness is clear in each case, since each of \(|a_1| + \cdots + |a_n|\) and \(\max\{|a_1|, \ldots, |a_n|\}\) equals \(0\) exactly when \(a_1 = \cdots = a_n = 0\). Homogeneity follows from \(|\alpha a_m| = |\alpha| |a_m|\), which gives \(\sum_m |\alpha a_m| = |\alpha| \sum_m |a_m|\) and \(\max_m |\alpha a_m| = |\alpha| \max_m |a_m|\). For the triangle inequality, if \(a = (a_1, \ldots, a_n)\) and \(b = (b_1, \ldots, b_n)\), then \(|a_m + b_m| \le |a_m| + |b_m|\) for each \(m\); summing over \(m\) gives \(\|a + b\|_1 \le \|a\|_1 + \|b\|_1\), and taking the maximum over \(m\) gives

\begin{equation*} \|a + b\|_\infty \le \max_m \bigl( |a_m| + |b_m| \bigr) \le \max_m |a_m| + \max_m |b_m| = \|a\|_\infty + \|b\|_\infty . \end{equation*}

Step 2: the two norms are comparable. For every \(a = (a_1, \ldots, a_n) \in \mathbf{F}^n\) and every \(m \in \{1, \ldots, n\}\),

\begin{equation*} |a_m| \le \|a\|_\infty \le \|a\|_1 \le n \|a\|_\infty , \end{equation*}

where the middle inequality holds because the maximum of the nonnegative numbers \(|a_1|, \ldots, |a_n|\) is one of the terms in the sum \(|a_1| + \cdots + |a_n|\), and the last inequality holds because each of the \(n\) terms of that sum is at most \(\|a\|_\infty\).

Step 3: completeness. Let \(\|\cdot\|\) denote either \(\|\cdot\|_1\) or \(\|\cdot\|_\infty\), and suppose \(f_1, f_2, \ldots\) is a Cauchy sequence in \(\mathbf{F}^n\) with respect to \(\|\cdot\|\). Write

\begin{equation*} f_k = (a_{k,1}, \ldots, a_{k,n}) \quad \text{for each } k \in \mathbf{Z}^+ . \end{equation*}

Fix \(m \in \{1, \ldots, n\}\). By Step 2, for all \(j, k \in \mathbf{Z}^+\) we have

\begin{equation*} |a_{j,m} - a_{k,m}| \le \|f_j - f_k\|_\infty \le \|f_j - f_k\| , \end{equation*}

the last inequality holding trivially if \(\|\cdot\| = \|\cdot\|_\infty\) and by Step 2 if \(\|\cdot\| = \|\cdot\|_1\). Because \(f_1, f_2, \ldots\) is Cauchy with respect to \(\|\cdot\|\), this shows that \(a_{1,m}, a_{2,m}, \ldots\) is a Cauchy sequence in \(\mathbf{F}\). Since \(\mathbf{F}\) (which is \(\mathbf{R}\) or \(\mathbf{C}\)) with the absolute value is a complete metric space, there exists \(b_m \in \mathbf{F}\) with

\begin{equation*} \lim_{k \to \infty} a_{k,m} = b_m . \end{equation*}

Let \(b = (b_1, \ldots, b_n) \in \mathbf{F}^n\). We claim that \(\lim_{k \to \infty} f_k = b\) with respect to \(\|\cdot\|_1\) (and hence, by Step 2, also with respect to \(\|\cdot\|_\infty\)). Suppose \(\varepsilon > 0\). For each \(m \in \{1, \ldots, n\}\) there exists \(N_m \in \mathbf{Z}^+\) such that

\begin{equation*} |a_{k,m} - b_m| < \frac{\varepsilon}{n} \quad \text{for all } k \ge N_m . \end{equation*}

Let \(N = \max\{N_1, \ldots, N_n\}\), which is a positive integer because it is the maximum of finitely many positive integers. If \(k \ge N\), then

\begin{equation*} \|f_k - b\|_1 = \sum_{m=1}^{n} |a_{k,m} - b_m| < n \cdot \frac{\varepsilon}{n} = \varepsilon , \end{equation*}

and also \(\|f_k - b\|_\infty \le \|f_k - b\|_1 < \varepsilon\) by Step 2. Hence \(\lim_{k \to \infty} \|f_k - b\| = 0\) for either choice of \(\|\cdot\|\), which means \(\lim_{k \to \infty} f_k = b\) in \((\mathbf{F}^n, \|\cdot\|)\).

Thus every Cauchy sequence in \((\mathbf{F}^n, \|\cdot\|)\) converges to an element of \(\mathbf{F}^n\), so \((\mathbf{F}^n, \|\cdot\|)\) is complete. Being a complete normed vector space, it is a Banach space (see 6.37), for both \(\|\cdot\| = \|\cdot\|_1\) and \(\|\cdot\| = \|\cdot\|_\infty\).

Here \(b(X)\) is the subspace of \(\mathbf{F}^X\) consisting of the bounded functions from \(X\) to \(\mathbf{F}\), and for \(f \in b(X)\) we set \(\|f\| = \sup\{|f(x)| : x \in X\}\), which is a finite nonnegative real number because \(f\) is bounded and \(X\) is nonempty (see the third bullet point of 6.34).

Step 1: \(\|\cdot\|\) is a norm on \(b(X)\). If \(\|f\| = 0\), then \(|f(x)| \le 0\) for every \(x \in X\), so \(f(x) = 0\) for every \(x \in X\), that is, \(f = 0\); conversely \(\|0\| = 0\). If \(\alpha \in \mathbf{F}\) and \(f \in b(X)\), then

\begin{equation*} \|\alpha f\| = \sup_{x \in X} |\alpha| \, |f(x)| = |\alpha| \sup_{x \in X} |f(x)| = |\alpha| \, \|f\| . \end{equation*}

If \(f, g \in b(X)\), then for every \(x \in X\) we have

\begin{equation*} |(f + g)(x)| \le |f(x)| + |g(x)| \le \|f\| + \|g\| , \end{equation*}

and taking the supremum over \(x \in X\) gives \(\|f + g\| \le \|f\| + \|g\|\). Thus all three conditions of 6.33 hold.

Step 2: construction of a candidate limit. Suppose \(f_1, f_2, \ldots\) is a Cauchy sequence in \(b(X)\). Fix \(x \in X\). For all \(j, k \in \mathbf{Z}^+\),

\begin{equation*} |f_j(x) - f_k(x)| \le \sup_{y \in X} |f_j(y) - f_k(y)| = \|f_j - f_k\| , \end{equation*}

so \(f_1(x), f_2(x), \ldots\) is a Cauchy sequence of elements of \(\mathbf{F}\). Because \(\mathbf{F}\) is complete, this sequence converges; define

\begin{equation*} f(x) = \lim_{k \to \infty} f_k(x) . \end{equation*}

This defines a function \(f \colon X \to \mathbf{F}\).

Step 3: \(f\) is bounded. By Exercise 4 in this section, the Cauchy sequence \(f_1, f_2, \ldots\) is bounded: there exists \(M \in \mathbf{R}\) with \(\|f_k\| < M\) for every \(k \in \mathbf{Z}^+\). Hence \(|f_k(x)| < M\) for all \(x \in X\) and all \(k \in \mathbf{Z}^+\). Because the absolute value function on \(\mathbf{F}\) is continuous, for each \(x \in X\) we have

\begin{equation*} |f(x)| = \lim_{k \to \infty} |f_k(x)| \le M . \end{equation*}

Thus \(f\) is bounded, so \(f \in b(X)\) and \(\|f\| \le M\).

Step 4: \(f_k \to f\) in the norm of \(b(X)\). Suppose \(\varepsilon > 0\). Because \(f_1, f_2, \ldots\) is Cauchy, there exists \(n \in \mathbf{Z}^+\) such that

\begin{equation*} \|f_j - f_k\| < \frac{\varepsilon}{2} \quad \text{for all integers } j \ge n \text{ and } k \ge n . \end{equation*}

Now fix \(k \ge n\) and \(x \in X\). For every \(j \ge n\) we have \(|f_j(x) - f_k(x)| \le \|f_j - f_k\| < \frac{\varepsilon}{2}\). Letting \(j \to \infty\) and using \(\lim_{j \to \infty} f_j(x) = f(x)\) together with the continuity of the absolute value, we get

\begin{equation*} |f(x) - f_k(x)| = \lim_{j \to \infty} |f_j(x) - f_k(x)| \le \frac{\varepsilon}{2} . \end{equation*}

This holds for every \(x \in X\), so taking the supremum over \(x \in X\) gives

\begin{equation*} \|f - f_k\| \le \frac{\varepsilon}{2} < \varepsilon \quad \text{for every } k \ge n . \end{equation*}

(The function \(f - f_k\) lies in \(b(X)\) because \(f\) and \(f_k\) both do.) Hence \(\lim_{k \to \infty} \|f_k - f\| = 0\), which means \(\lim_{k \to \infty} f_k = f\) in \(b(X)\).

Therefore every Cauchy sequence in \(b(X)\) converges to an element of \(b(X)\), so \(b(X)\) is complete. Being a complete normed vector space, \(b(X)\) is a Banach space (see 6.37).

Note that the norm \(\|\cdot\|\) on \(b(X)\) is the norm of uniform convergence: Step 4 shows precisely that a Cauchy sequence in \(b(X)\) converges pointwise to a bounded function and that the convergence is in fact uniform on \(X\).

Recall that \(\ell^1\) is the vector space of sequences \((a_1, a_2, \ldots)\) of elements of \(\mathbf{F}\) such that \(\sum_{k=1}^{\infty} |a_k| < \infty\) (see the last bullet point of 6.32). For a sequence \(h = (c_1, c_2, \ldots)\) we write \(h(m) = c_m\) for its \(m^{\text{th}}\) coordinate. What is to be proved is the assertion made in the second bullet point of Example 6.39.

First note that \(\|\cdot\|_\infty\) really is a norm on \(\ell^1\). It is well defined and finite there, because if \((a_1, a_2, \ldots) \in \ell^1\) then \(|a_m| \le \sum_{k=1}^{\infty} |a_k| < \infty\) for each \(m\), so the sequence is bounded. Positive definiteness holds because \(\sup_k |a_k| = 0\) forces every \(a_k\) to be \(0\); homogeneity holds because \(\sup_k |\alpha a_k| = |\alpha| \sup_k |a_k|\); and the triangle inequality holds because \(|a_k + b_k| \le |a_k| + |b_k| \le \|(a_1,a_2,\ldots)\|_\infty + \|(b_1,b_2,\ldots)\|_\infty\) for each \(k\), after which we take the supremum over \(k\). Thus \((\ell^1, \|\cdot\|_\infty)\) is a normed vector space, and to show it is not a Banach space we must exhibit a Cauchy sequence in \(\ell^1\) that does not converge (with respect to \(\|\cdot\|_\infty\)) to an element of \(\ell^1\).

For \(k \in \mathbf{Z}^+\) define

\begin{equation*} f_k = \Bigl( 1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots, \tfrac{1}{k}, 0, 0, \ldots \Bigr), \end{equation*}

so the \(m^{\text{th}}\) coordinate of \(f_k\) is \(\frac{1}{m}\) if \(m \le k\) and is \(0\) if \(m > k\). Each \(f_k\) has only finitely many nonzero coordinates, so \(\sum_{m=1}^{\infty} |f_k(m)| = \sum_{m=1}^{k} \frac{1}{m} < \infty\) and hence \(f_k \in \ell^1\).

The sequence \(f_1, f_2, \ldots\) is Cauchy with respect to \(\|\cdot\|_\infty\). Indeed, if \(j > k \ge 1\), then the coordinates of \(f_j - f_k\) are \(0\) except in positions \(k+1, \ldots, j\), where they equal \(\frac{1}{k+1}, \ldots, \frac{1}{j}\); the largest of these in absolute value is \(\frac{1}{k+1}\), so

\begin{equation*} \|f_j - f_k\|_\infty = \frac{1}{k+1} . \end{equation*}

Hence for all \(j, k \ge n\) we have \(\|f_j - f_k\|_\infty \le \frac{1}{n+1}\) (this also covers \(j = k\), where the norm is \(0\), and \(j < k\) by symmetry). Given \(\varepsilon > 0\), choose \(n \in \mathbf{Z}^+\) with \(n > \frac{1}{\varepsilon}\); then \(\|f_j - f_k\|_\infty \le \frac{1}{n+1} < \varepsilon\) for all \(j \ge n\) and \(k \ge n\). So \(f_1, f_2, \ldots\) is a Cauchy sequence in \((\ell^1, \|\cdot\|_\infty)\).

Now suppose, seeking a contradiction, that this sequence converges in \((\ell^1, \|\cdot\|_\infty)\), say

\begin{equation*} \lim_{k \to \infty} \|f_k - g\|_\infty = 0 \quad \text{for some } g = (b_1, b_2, \ldots) \in \ell^1 . \end{equation*}

Fix \(m \in \mathbf{Z}^+\). For every \(k \ge m\) the \(m^{\text{th}}\) coordinate of \(f_k\) equals \(\frac{1}{m}\), so

\begin{equation*} \Bigl| \tfrac{1}{m} - b_m \Bigr| = |f_k(m) - b_m| \le \|f_k - g\|_\infty . \end{equation*}

Letting \(k \to \infty\), the right side tends to \(0\) while the left side does not depend on \(k\); hence \(\bigl| \frac{1}{m} - b_m \bigr| = 0\), that is, \(b_m = \frac{1}{m}\). Because \(m \in \mathbf{Z}^+\) was arbitrary,

\begin{equation*} g = \Bigl( 1, \tfrac{1}{2}, \tfrac{1}{3}, \ldots \Bigr) . \end{equation*}

But then

\begin{equation*} \sum_{m=1}^{\infty} |b_m| = \sum_{m=1}^{\infty} \frac{1}{m} = \infty \end{equation*}

because the harmonic series diverges, so \(g \notin \ell^1\), contradicting the assumption that \(g \in \ell^1\).

Therefore the Cauchy sequence \(f_1, f_2, \ldots\) has no limit in \(\ell^1\) with respect to \(\|\cdot\|_\infty\). Hence \((\ell^1, \|\cdot\|_\infty)\) is not complete, and so it is not a Banach space.

The failure is caused by using the wrong norm rather than by any defect of \(\ell^1\) itself: with its own norm \(\|(a_1, a_2, \ldots)\|_1 = \sum_{k=1}^{\infty} |a_k|\), the space \(\ell^1\) is a Banach space (see the second bullet point of 6.38 and Exercise 8 in this section). Indeed the sequence \(f_1, f_2, \ldots\) constructed above is not Cauchy with respect to \(\|\cdot\|_1\): if \(j > k\) then \(\|f_j - f_k\|_1 = \sum_{m=k+1}^{j} \frac{1}{m}\), and because the harmonic series diverges, for each \(k \in \mathbf{Z}^+\) there exists \(j > k\) with \(\sum_{m=k+1}^{j} \frac{1}{m} > 1\). Hence the Cauchy condition fails for \(\varepsilon = 1\).

Recall that \(\ell^1\) is the vector space of sequences \((a_1, a_2, \ldots)\) of elements of \(\mathbf{F}\) such that \(\sum_{k=1}^{\infty} |a_k| < \infty\), with coordinatewise addition and scalar multiplication.

Step 1: \(\|\cdot\|_1\) is a norm. If \(\|(a_1,a_2,\ldots)\|_1 = 0\), then \(\sum_{k=1}^\infty |a_k| = 0\); since every term is nonnegative, \(|a_k| = 0\) for each \(k\), so \((a_1,a_2,\ldots) = 0\). Conversely \(\|0\|_1 = 0\). For \(\alpha \in \mathbf{F}\) we have \(\sum_{k=1}^\infty |\alpha a_k| = |\alpha| \sum_{k=1}^\infty |a_k|\), giving homogeneity. For the triangle inequality, note that for each \(n \in \mathbf{Z}^+\),

\begin{equation*} \sum_{k=1}^{n} |a_k + b_k| \le \sum_{k=1}^{n} |a_k| + \sum_{k=1}^{n} |b_k| \le \|(a_1,a_2,\ldots)\|_1 + \|(b_1,b_2,\ldots)\|_1 ; \end{equation*}

letting \(n \to \infty\) gives \(\|(a_1,a_2,\ldots) + (b_1,b_2,\ldots)\|_1 \le \|(a_1,a_2,\ldots)\|_1 + \|(b_1,b_2,\ldots)\|_1\). Thus \(\|\cdot\|_1\) is a norm on \(\ell^1\) (and, incidentally, this shows \(\ell^1\) is closed under addition).

Step 2: completeness. Suppose \(f^1, f^2, \ldots\) is a Cauchy sequence in \(\ell^1\), where

\begin{equation*} f^n = (a_1^n, a_2^n, \ldots). \end{equation*}

Fix \(k \in \mathbf{Z}^+\). For all \(m, n \in \mathbf{Z}^+\) we have

\begin{equation*} |a_k^n - a_k^m| \le \sum_{j=1}^{\infty} |a_j^n - a_j^m| = \|f^n - f^m\|_1 . \end{equation*}

Hence \(a_k^1, a_k^2, \ldots\) is a Cauchy sequence in \(\mathbf{F}\). Because \(\mathbf{F}\) (either \(\mathbb{R}\) or \(\mathbb{C}\)) is complete, this sequence converges; call its limit \(a_k\). Let \(f = (a_1, a_2, \ldots)\).

Now let \(\varepsilon > 0\). Because \(f^1, f^2, \ldots\) is Cauchy, there exists \(N \in \mathbf{Z}^+\) such that

\begin{equation*} \|f^n - f^m\|_1 < \varepsilon \quad \text{for all } m, n \ge N . \end{equation*}

Fix \(n \ge N\) and fix \(M \in \mathbf{Z}^+\). For every \(m \ge N\),

\begin{equation*} \sum_{k=1}^{M} |a_k^n - a_k^m| \le \|f^n - f^m\|_1 < \varepsilon . \end{equation*}

The left side is a finite sum, so we may let \(m \to \infty\) term by term, obtaining

\begin{equation*} \sum_{k=1}^{M} |a_k^n - a_k| \le \varepsilon . \end{equation*}

This bound holds for every \(M \in \mathbf{Z}^+\), and the partial sums are increasing in \(M\), so

\begin{equation*} \sum_{k=1}^{\infty} |a_k^n - a_k| \le \varepsilon \quad \text{for all } n \ge N . \end{equation*}

In particular \(f^N - f \in \ell^1\). Because \(\ell^1\) is a vector space and \(f^N \in \ell^1\), we conclude that

\begin{equation*} f = f^N - (f^N - f) \in \ell^1 . \end{equation*}

Moreover the displayed inequality says exactly that \(\|f^n - f\|_1 \le \varepsilon\) for all \(n \ge N\). Since \(\varepsilon > 0\) was arbitrary, \(\lim_{n \to \infty} \|f^n - f\|_1 = 0\), which means \(f^n \to f\) in \(\ell^1\).

Thus every Cauchy sequence in \(\ell^1\) converges to an element of \(\ell^1\), so \(\ell^1\) with \(\|\cdot\|_1\) is a Banach space (6.37).