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%  Lecture 4
%  Source: handwritten notes (Mathpix-converted) + Kashlak STAT 571
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\section[Lecture 4 -- Lebesgue Measure; Non-Measurable Sets; Product Measure and Independence]{Lecture 4 \textemdash{} Lebesgue Measure; Non-Measurable Sets; Product Measure \& Independence (briefly)}
\label{sec:lec04}

Carath\'eodory extension and the Dynkin \(\pi\)-\(\lambda\) machinery from
Lecture~3 finally pay off here: we use them to build \emph{Lebesgue
measure} on the unit interval (and on \(\R^p\)), exhibit a subset that no
translation-invariant measure can size (the Vitali set), and then sketch
how the same toolkit produces \emph{product measures} and the
\emph{independence} of \(\sigma\)-fields.

\subsection{Lebesgue measure on \texorpdfstring{$(0,1]$}{(0,1]} and \texorpdfstring{$\R$}{R}}

Take \(\Omega=(0,1]\) (or \(\Omega=\R\)). Let
\(\Acal=\{(a,b]:0\le a<b\le 1\}\cup\{\emptyset\}\) and, more generally,
the collection of finite disjoint unions of half-open intervals. We want
a measure satisfying
\[
\lambda\bigl((a,b]\bigr) \;=\; b-a.
\]

\begin{proposition}{$\Acal$ is a $\pi$-system and a ring}{halfopen-ring}
The collection \(\Acal\) of finite disjoint unions of half-open intervals
\((a,b]\subseteq(0,1]\) (together with \(\emptyset\)) is closed under
finite intersections (a \(\pi\)-system) and under set difference and
finite unions (a ring).
\end{proposition}

\begin{proposition}{$\lambda$ is a pre-measure on $\Acal$}{lambda-premeasure}
The set function
\(\lambda(A)=\sum_{j=1}^{n}\lambda(I_j)\) for
\(A=\bigsqcup_{j=1}^{n}I_j\in\Acal\), where \(\lambda((a,b])=b-a\), is a
pre-measure on the ring \(\Acal\).
\end{proposition}

Combining the previous two propositions with Lecture~3 yields the main
construction.

\begin{theorem}{Lebesgue Measure}{lebesgue}
There exists a unique measure \(\lambda\) on the Borel \(\sigma\)-field
\(\Bcal=\sigma(\Acal)\) of \((0,1]\) (resp.\ of \(\R\)) such that
\(\lambda((a,b])=b-a\). It is obtained by Carath\'eodory extension from
\(\Acal\); uniqueness on \(\sigma(\Acal)\) follows from the
\(\pi\)-\(\lambda\) uniqueness theorem since \(\Acal\) is a
\(\pi\)-system. Writing \(\mathcal{M}_\lambda\) for the \(\sigma\)-field of
\emph{Lebesgue measurable} sets,
\[
\Acal \;\subset\; \Bcal \;=\; \sigma(\Acal)
       \;\subset\; \mathcal{M}_\lambda \;\subset\; \Pcal((0,1]),
\]
and \(\mathcal{M}_\lambda\) is the completion of \(\Bcal\) with respect to the
\(\lambda\)-null sets \(\Ncal_\lambda\). The inclusions
\(\Bcal\subsetneq\mathcal{M}_\lambda\subsetneq\Pcal((0,1])\) are strict.
\end{theorem}

\begin{remark}
Lebesgue measure is the \emph{only} translation-invariant measure on
\(\R\) (up to a multiplicative constant) assigning length \(b-a\) to
\((a,b]\). The same uniqueness statement holds on \(\R^n\).
\end{remark}

\subsection{Non-measurable sets: the Vitali construction}

We now show that the inclusion \(\mathcal{M}_\lambda\subsetneq\Pcal((0,1])\) is
strict by exhibiting a set \(H\subseteq(0,1]\) that lies outside
\(\mathcal{M}_\lambda\). The construction uses the Axiom of Choice and
addition modulo \(1\).

\begin{definition}{Addition modulo $1$}{add-mod-one}
For \(x,y\in(0,1]\) define
\[
x+y \;=\;
\begin{cases}
x+y     & \text{if } x+y\le 1,\\
x+y-1   & \text{if } x+y>1,
\end{cases}
\]
which can be visualised as wrapping \((0,1]\) into a circle. For
\(A\subseteq(0,1]\) and \(x\in(0,1]\), set
\(A+x=\{y\in(0,1]: y-x\in A\}\) (with subtraction also taken mod~\(1\)).
\end{definition}

\begin{lemma}{Translation invariance on Borel sets}{translation-invariance}
Let \(\Lcal=\{A\in\mathcal{M}_\lambda: \lambda(A+x)=\lambda(A)\text{ for all }
x\in(0,1]\}\). Then \(\Lcal\) is a \(\lambda\)-system, and since
\(\Acal\subset\Lcal\), Dynkin's \(\pi\)-\(\lambda\) theorem gives
\(\Bcal=\sigma(\Acal)\subseteq\Lcal\). In particular, \(\lambda\) is
translation invariant on every Borel set.
\end{lemma}

\begin{example}[Vitali set]
Define an equivalence relation on \((0,1]\) by \(x\sim y\iff
x-y\in\Q\). By the Axiom of Choice pick a set \(H\subset(0,1]\)
containing exactly one representative from each equivalence class. Then
\((H+r_1)\cap(H+r_2)=\emptyset\) for distinct \(r_1,r_2\in\Q\cap(0,1]\),
and
\[
(0,1] \;=\; \bigsqcup_{r\in\Q\cap(0,1]}(H+r).
\]
If \(H\) were Lebesgue measurable, countable additivity together with
translation invariance would give
\[
1 \;=\; \lambda((0,1]) \;=\; \sum_{r\in\Q\cap(0,1]}\lambda(H+r)
       \;=\; \sum_{r\in\Q\cap(0,1]}\lambda(H),
\]
which is \(0\) if \(\lambda(H)=0\) and \(\infty\) if
\(\lambda(H)>0\)\,---\,either way a contradiction. Hence
\(H\notin\mathcal{M}_\lambda\).
\end{example}

\begin{remark}
The construction shows \(\mathcal{M}_\lambda\subsetneq\Pcal((0,1])\). It also
illustrates a ``fun fact'': there is no analogue of Lebesgue measure in
infinite dimensions, since the only locally finite, translation-invariant
Borel measure on an infinite-dimensional separable Banach space is the
trivial measure.
\end{remark}

\subsection{Product measure, briefly}

Having constructed \(\lambda\) on \(\R\) using half-open intervals, the
same recipe works on \(\R^p\) using half-open rectangles.

\begin{definition}{Lebesgue measure on $\R^p$}{lebesgue-rp}
On \(\R^p\), define
\[
\lambda^{(p)}\bigl((a_1,b_1]\times\cdots\times(a_p,b_p]\bigr)
\;=\; \prod_{i=1}^{p}\lambda\bigl((a_i,b_i]\bigr)
\;=\; \prod_{i=1}^{p}(b_i-a_i).
\]
The collection of such half-open rectangles is a \(\pi\)-system, and the
extension theorem gives a unique measure on the Borel
\(\sigma\)-field~\(\Bcal(\R^p)\). It is the only translation-invariant
measure (up to scale) with the prescribed value on rectangles.
\end{definition}

\begin{definition}{Product measure}{product-measure}
Given \(\sigma\)-finite measure spaces \((\mathbb{X},\mathcal{X},\mu)\)
and \((\mathbb{Y},\mathcal{Y},\nu)\), the \emph{product measure space} is
\((\mathbb{X}\times\mathbb{Y},\,\mathcal{X}\times\mathcal{Y},\,\pi)\),
where the product \(\sigma\)-field
\(\mathcal{X}\times\mathcal{Y}=\sigma\bigl(\{A\times B:
A\in\mathcal{X},\,B\in\mathcal{Y}\}\bigr)\) and \(\pi\) is uniquely
defined by
\[
\pi(A\times B) \;=\; \mu(A)\,\nu(B), \qquad A\in\mathcal{X},\ B\in\mathcal{Y}.
\]
\end{definition}

\begin{remark}
The product of two Borel \(\sigma\)-fields satisfies
\(\Bcal(\mathbb{X})\times\Bcal(\mathbb{Y})\subseteq
\Bcal(\mathbb{X}\times\mathbb{Y})\); equality holds in ``nice'' (e.g.\
second-countable) settings, including
\(\mathbb{X}=\mathbb{Y}=\R\).
\end{remark}

\subsection{Independence}

Let \((\Omega,\Fcal,\mu)\) be a probability space.

\begin{definition}{Independence for sets}{indep-sets}
A countable collection \(\{A_i\}_{i\in I}\subseteq\Fcal\) is
\emph{independent} if for every finite \(J\subseteq I\),
\[
\mu\!\left(\bigcap_{j\in J}A_j\right) \;=\; \prod_{j\in J}\mu(A_j).
\]
\end{definition}

\begin{example}[A standard 52-card deck]
Draw one card uniformly. Let
\(A_1=\{\text{red}\}\), \(A_2=\{\text{heart or club}\}\),
\(A_3=\{\text{queen}\}\). Then
\[
\mu(A_1)=\tfrac{1}{2},\ \mu(A_2)=\tfrac{1}{2},\ \mu(A_3)=\tfrac{1}{13},
\]
and one checks
\(\mu(A_1\cap A_2)=\tfrac{1}{4}\),
\(\mu(A_1\cap A_3)=\tfrac{1}{26}\),
\(\mu(A_2\cap A_3)=\tfrac{1}{26}\),
\(\mu(A_1\cap A_2\cap A_3)=\tfrac{1}{52}\), so the three events are
independent.
\end{example}

\begin{definition}{Independence for $\sigma$-fields}{indep-sigma-fields}
A countable collection \(\{\Fcal_i\}_{i\in I}\) of sub-\(\sigma\)-fields
of \(\Fcal\) is \emph{independent} if every choice
\(\{A_i\in\Fcal_i:i\in I\}\) is an independent collection of sets in the
sense of \cref{def:indep-sets}.
\end{definition}

The next theorem is the main tool: independence on a generating
\(\pi\)-system already forces independence of the generated
\(\sigma\)-fields.

\begin{theorem}{Independence from $\pi$-systems}{indep-pi}
Let \(\Acal_1,\Acal_2\subseteq\Fcal\) be \(\pi\)-systems. If
\[
\mu(A_1\cap A_2) \;=\; \mu(A_1)\,\mu(A_2)
\quad\text{for all } A_1\in\Acal_1,\ A_2\in\Acal_2,
\]
then \(\sigma(\Acal_1)\) and \(\sigma(\Acal_2)\) are independent.
\end{theorem}