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\begin{document}
pecture 1:\\
Adam Koshiah\\
Measure: $\left(\Omega, \mathcal{F}^{\prime}\right) \leftarrow$ Measurkeble Spere\\
a preasure $\mu: \tau \rightarrow \mathbb{R}^{+}$

\begin{enumerate}
  \item $\mu(\phi)=0$
  \item $\mu *$ is countably aelelitive
\end{enumerate}

Leeture 1: What is a Measure?

How is $\mathbb{R}^{n}$ reloted to its norm?\\
09/03/26\\
$\Omega$ sample space.\\
$\stackrel{L}{R}$ on $\mathbb{R}^{2}$\\
$Q, Z, Q$

$$A \in \Omega_{A}:=\{x \in \Omega: x \in A\}$$

$A^{c} \subset \Omega:=\{x \in \Omega: x \notin A\}$\\
$\vec{x}^{c}=\Omega / A$\\
set sirptemme

$$x^{c}=\Omega \backslash A$$

union:\\
symmetric difference is the same thing as yor.\\
For a collection of sets:\\
\{ $\left.A_{i}\right\}, B_{i=1}^{\infty}$ they are pairuise disjoint ix $A_{1} \cap A_{j}=\phi, i \nexists_{j}$.

Note: Thene ove do signed recomes that refurn $\mathbb{R}$.\\
for a pairwise\\
disjoint collection\\
$\left\{A_{i}\right\}_{i=1}^{\infty}$ then\\
$\mu\left(\bigcup_{i=1}^{\infty} A_{i}\right)$\\
$=\sum_{i=1}^{\infty} \mu\left(A_{i}\right)$\\
Special Coses:\\
$(\Omega, \mathcal{F}, \mu)$\\
меобине sраме\\
(not be confred with mexomedue spree)

\begin{itemize}
  \item if $\mu(\Omega)=7$ then we say that $(\Omega, F, \mu)$ is a probability space onel $\mu$ is a probility measure
  \item if $\mu(\Omega)<\infty$\\
then $\mu$ is : finite measure
  \item if $\Omega=\bigcup_{i=1}^{\infty} A_{i}$ s.t. $\mu\left(A_{i}\right) \leqslant \infty \quad H_{i}$ then $\mu$ is a $\mathcal{J}$-finite measure\\
Example: $\Omega=\mathbb{R}, \mu([a, b])=b-a$\\
Then $\mathbb{R}=\bigcup_{i=1}^{\infty}[i-i, i] \cup[-i,-i+1]$\\
Notation: $P(\Omega)=$ Power set.\\
⇒ set all subects of $\Omega$\\
Note: $F \subset P(\Omega)$ and power set is offen too sig to be usefy.
\end{itemize}

Example: (conting Measure) $\Omega=\{1,2, \ldots n\}$ Then $\phi(\Omega)=2^{\Omega}$ is the set of possible subets of $\Omega$ (herve the wation $z^{\Omega}, \ldots$ from $z^{n}$ subets)

\begin{itemize}
  \item $\mu(\{1,3,7\})=3$, $\mu(\alpha \Omega)=1$
  \item define $\nu(A)=\frac{1}{n} \mu(A)$
\end{itemize}

In controst, consider the bitomigl $(n, p)$ distribtion, each poit jets a measure of $\binom{n}{i} p^{i}(1-p)^{n-i}$ for $p \in(0,1)$

\section*{s-field (s-algebra)}
→ what sets (subsets of $n$ ) am I allowed to measure.\\
Definition 1: For some set $\Omega$ contable derses a $\sigma$-field $\mathcal{F}$ is a collection of sets $A \subseteq \Omega$ s.t.\\
(1) $\phi, \Omega \in \mathcal{F}$\\
sob-intervals\\
(2) if $A \in F$ then $A^{c} \in F$\\
(3) for a countable collection of sets $\left\{A_{i}\right\}_{i=1}^{\infty}$ s.t. $A_{i} \in \mathcal{F}$ for\\
\includegraphics[max width=\textwidth, alt={}, center]{5c85dc7d-3850-49bf-8fca-ec10384f670c-03_155_292_1792_1426}

Then $\bigcup_{i=1}^{n} A_{i} \in \mathcal{F}$\\
consequence: combing $(2)$ and (3) gives you countable intersections.\\
(3*) $\bigcap_{i=1}^{\infty} x_{i} \in$ F\\
why? $\left(\Lambda_{i=1}^{\infty} A_{i}\right)^{c}=\bigcup_{i=1}^{\infty} A_{i}{ }^{c}$

Lecture 2: Existence Coratheodory 10/03/26 Exteryion Theorem

Idea: if the measure of $(a, b]$ is $b$ - $a$ for any $b>a$ Then, whet else con wel measure?\\
Storf: The set of all sich, intervals $(a, b]$ is not a $\sigma$-fiebl.

Definition: (semi-ring)\\
(they ere a semi-ring)\\
A is a collection of rubets of $\Omega$, is called a semi-ring when\\
$-\phi \in \mathscr{f}$

\begin{itemize}
  \item for any $A, B \in f, A \cap B \in f$ and
\end{itemize}

$$B \backslash A=\bigcup_{i=1}^{3} c_{i} \text { for } c_{i}: \in \mathscr{\theta}$$

"the set difference might not be in A, int could be created with - Finite union ol sets from \&"

Example: A all intervals $(a, b]$ is a semi-ring ( $c_{i}$ )

Defer: (Ring) $A$, a collection of subsets of $\Omega$, is called a Ring when

$$\begin{aligned}
& -\phi \in A \\
& - \text { for any } A, B \in A \\
& -B \mid A \in A \\
& -A \cup B \in A \quad \text { (finite cuion) }
\end{aligned}$$

fample: all finite uniong countrible union of welt spen intervals 4 which novel tate :) $(a, b] \cup[c, d] \in A$ to sigms Feels!

\section*{Definition: (Fielol)}
$A$ is : field if it $x$ a fring and $\Omega \in f$

Note: Fief + contable uniors = of field. Defn: (Set function)\\
$\mu: A \rightarrow \mathbb{R}^{+} \quad$ (not rec. a measure)\\
for $A, B \in D$, we say that

\begin{itemize}
  \item $\mu$ is increasing it $A \subset B \Rightarrow \mu(A) \leq \mu(B)$
  \item $M$ is avelelitive it for $A, B$ disjoint then $\mu(A \cup B)=\mu(A)+\mu(* \beta)$
  \item $\mu$ is confably areditive it for $\left[A_{i}\right]_{i}^{\infty}=$ s.t. $A_{i} \cap A_{j}=\varnothing \quad \theta_{i} \neq j$ (i.e pircuise sligint)
\end{itemize}

$$\mu\left(\bigcup_{i=1}^{\infty} A_{i}\right)=\sum_{i=1}^{\infty} \mu\left(A_{i}\right)$$

$$\bigcup_{i=1}^{\infty} A_{i} \in \neq \phi$$

\begin{itemize}
  \item $\mu$ is counfobly sub-additive if $\left[A_{i}\right]_{i}^{\infty}: 1$ are in $A$ anel $U^{\infty} A_{i} \in A$\\
(not recesserily it primise dissoint)\\
Then $\mu\left(\bigcup_{i=1}^{\infty} A_{i}\right) \leq \sum_{i=1}^{\infty} \mu\left(A_{i}\right)$\\
Often, a set funfion on $M$ on a ring A S.t. $\mu(\phi)=0$ and $\mu$ counfably ablelitine, then we say that $\mu$ is a pres-messise.
\end{itemize}

For a pre-measure $\mu$ on ame Ring A we have an outer measure.

$$\mu^{*}(E)=\inf S_{i} \mu\left(A_{i}\right) \text {, for any } E \subseteq \Omega$$

where the inf is taken over all collections of $A_{i}$ s.t. $E \subset \underset{i=1}{\cup} A_{i}$ ↓\\
(finite or countable)

Question: Con we "measure" (outer measure) any $E \subset \Omega$ ?

\section*{Not necessorily.}
\begin{itemize}
  \item Denote (suript $m$ ) $m$ to be the colleulton of all $\mu^{*}$ measurable sets.\\
where we say that $B \subseteq \Omega$ is $\mu^{*}$ measurable when
\end{itemize}

$$\mu^{*}(E \cap B)+\mu^{*}\left(E \cap B^{C}\right)=\mu^{*}(E)$$

$$\text { for all } E \subseteq \Omega$$

Theorem: (Corathédalony Extergion Thooreny\\
\includegraphics[max width=\textwidth, alt={}, center]{5c85dc7d-3850-49bf-8fca-ec10384f670c-06_466_870_1485_1175}

Let A bo a ring on $B^{C}$\\
$\Omega$ and $\mu$ be a pre-measure, then $\mu$ extenals to a measure an $O(A)$.\\
Moke: $v(A)$ is the $v$-fierrel then comes from extereling $A$ by; ulveling countable unions and $\Omega$ itsulf.\\
◯ Note: The "correct" extersion is the outer\\
◯ measure.

Proof: Assume $B \subseteq \Omega$ and $\mu^{*}(B)<\infty$\\
(Step 4) Prove stuff alout $\mu^{*}$

\begin{itemize}
  \item $\mu^{*}(\phi)=0$ as $\mu$ is a pre-measure
  \item $\mu^{*}$ is von-negative $\forall B \subseteq \Omega$ as $\mu$ is non-reg
  \item $\mu^{H}$ is morotone (increasing)
\end{itemize}

Let $B_{1}, B_{2} A A_{1}$ ool $B_{1} \subset A_{2}$ then for any $\left.\sum A_{i}\right\}$ s.t. $B_{z} \subseteq \cup A_{i}$\\
Then $B_{1} \subseteq \cup_{i} A_{i}$\\
$\therefore \mu^{*}\left(B_{1}\right) \leqslant \mu^{*}\left(B_{2}\right)$

\begin{itemize}
  \item $\mu^{*}$ is countably sadelitive
\end{itemize}

For $\left[B_{i}\right]_{i=1}^{\infty}$ and a given $c>0$, let $B_{i} \leq U_{i} A_{i j}$ for $A_{i j} \in A$ s.t.

$$\sum_{j} \mu\left(A_{i j}\right) \leqslant \mu^{*}\left(B_{i}\right)+\varepsilon Z^{-i}$$

As $U_{i=1}^{\infty} B_{i} \leq U_{i, j} A_{i j}$ and as $\mu^{*}$ is monotone and $\mu$ is sob-odelditive, then

$$\begin{aligned}
\mu^{*}\left(\bigcup_{i=1}^{\infty} B_{i}\right) & \leqslant \mu\left(\bigcup_{i, j} A_{i j}\right) \\
& \leqslant \sum_{i j} \mu\left(A_{i j}\right) \\
& \leqslant \sum_{i} \mu^{*}\left(B_{i}\right)+\varepsilon
\end{aligned}$$

(1) Take $\Sigma \rightarrow 0$ to get that $\mu^{*}$ is countrally subadblitine\\
(Step 2) chech that $\mu$ and $\mu^{*}$ coincide for all $A \in A$

For any $A \in A$ we have $\mu^{*}(A) \leqslant \mu(A)$ Secarsel $A \subseteq A$

For the reverse, if $A \subset \cup A_{i}$, then by countralle sub-additivity and

$$\begin{gathered}
\mu(A) \leqslant \sum_{i} \mu\left(A \cap A_{i}\right) \leqslant \sum_{i} \mu\left(A_{i}\right) \\
\therefore \mu(A) \leqslant \mu^{*}(A) \\
\therefore \mu(A)=\mu^{*}(A) \quad \theta A \in A
\end{gathered}$$

(step 7) check that $A \subset M$\\
∴ i. for any $A \in A$ we want to show that $A$ is $\mu^{*}$-meosubable.

$$\text { ∴ e. } \mu^{*}(E)=\mu^{*}(E \cap A)+\mu^{*}\left(E \cap A^{c}\right) \forall E \leq \Omega$$

Note $\mu^{*}(E \cap A)+\mu^{*}\left(E \cap A^{c}\right) \geqslant \mu^{*}$ (E)\\
as $\mu^{*}$ is subselalitive\\
Next, for some $\varepsilon>0$, cloose $\left[A_{i}\right]$ s.t.\\
$E \leqslant A_{i}$ and $\sum_{i} \mu\left(A_{i}\right) \leqslant \mu^{*}(E)+\sum$\\
Firthernore, $E \cap A \subseteq \cup A_{i}\left(A \cap A_{i}\right)$

$$E \cap A^{C} \subseteq U_{i}\left(A^{C} \cap A_{i}\right)$$

Thus, $\mu^{*}(E \cap A)+\mu^{*}\left(E \cap A^{c}\right) \leqslant$

$$\begin{aligned}
& \leqslant \sum_{i} \mu\left(A \cap A_{i}\right)+\sum_{i} \mu\left(A \cap A_{i}\right) \\
& =\sum_{i} \mu\left(A_{i}\right) \leqslant \mu^{*}(E)+\varepsilon
\end{aligned}$$

(Step 4) Slow that $M$ is a v-field

\begin{itemize}
  \item $\phi \in M$ since $\phi \in \mathbb{A}$
  \item $\Omega \in M$ since for any. $E \subseteq \Omega$\\
(a sigma-field requires $\Omega$ thes in it).
\end{itemize}

$$\mu^{*}(E \cap \Omega)+\mu^{*}(E \cap \phi)=\mu^{*}(E)$$

Next since, $A \cap B=\left(A^{c} \cup B^{c}\right)^{c}$, we will show M is utosed under intersections\\
For $b_{1}, B_{2} \in M$ arel any $E \subseteq \Omega \mu^{*}(E)=\mu^{*}(B, \cap \epsilon)$

$$+\mu^{*}\left(B_{1}{ }^{c} \cap E\right)$$

$=\mu^{*}\left(B_{2} \cap B_{1} \cap E\right)+\mu^{*}\left(B_{2} \cap B_{1}^{c} \cap E\right)+\mu^{*}\left(B_{2}^{c} \cap B_{1} \cap E\right)$\\
(Win subed it it inty) $+\mu^{*}\left(x_{2}{ }^{C} \cap B_{1}{ }^{C} \cap \epsilon\right)$

$$\begin{aligned}
& \geqslant \mu^{*}\left(B_{2} \cap B_{1} \cap E\right)+\mu^{*}\left(\left[B_{2} \cap B_{1}^{c} \cap E\right] \cup\left\{B_{2}^{c} \cap B_{1} \cap E\right\}\right] \\
& \left.\cup\left\{B_{2}^{c} \cap B_{1}^{c} \cap E\right\}\right] \\
& \left.=\mu^{*}\left(L B_{2} \cap B_{1}\right] \cap E\right)+\mu^{*}\left(\left[B_{2} \cap B_{1}\right]^{c} \cap E\right) \\
& \geqslant \mu^{*}(E) \Rightarrow \mu^{*}(E) \\
& \therefore B_{1} \cap B_{2} \in M
\end{aligned}$$

Lestly, rote that $B \backslash A=B \cap A^{c}$\\
∴ wer reed to show that $M$ is used wheler complementation because $\mu^{*}\left(E \cap B^{c}\right)+\mu^{*}\left(E \cap\left(B^{c}\right)^{c}\right)$\\
$\therefore M$ is a field.

$$=\mu^{*}(E)$$

To get to a $r$-fidel, let $\left\langle B_{i}\right\rangle$ in $M$ be countable and pairuise disjoint and $U_{i} B_{i} \in M$. Let $B=U_{i=1}^{\infty} B_{i}$\\
Then, $\mu^{*}(E)=\mu^{*}\left(E \cap B_{1}\right)+\mu^{*}\left(E \cap B_{1}^{C}\right)$ \(\square\)

$$\begin{aligned}
& =\mu^{*}\left(E \cap B_{1}\right)+\mu^{*}\left(E \cap B_{2}\right) \\
& +\mu^{*}\left(E \cap B_{1}^{c} \cap B_{2}\right) \\
& \mu^{*}(E)=\sum_{i=1}^{n} \mu^{*}\left(E \cap B_{i}\right)+\mu^{*}\left(E \cap\left[\hat{n} B_{i}^{c}\right]\right) \\
& \text { Then, by monotonicity, staddivity and } n \rightarrow \infty \\
& \mu^{*}(E) \geqslant \sum_{i=1}^{n} \mu^{*}\left(E \cap B_{i}\right)+\mu^{*}\left(E \cap B^{c}\right) \\
& \geqslant \mu^{*}(E \cap B)+\mu^{*}\left(E \cap B^{c}\right) \\
& { }^{1} \geqslant \mu^{*}(E)
\end{aligned}$$

$\therefore M$ is closed uneler counforles unions close $E=S$ and

$$\therefore \mu^{*}(E)=\sum_{i=1}^{\infty} \mu^{*}\left(E \cap B_{i}\right)$$

$\therefore \mu^{*}$ countasly additive.\\
Conelusion: $\mu^{*}$ is a set funfion $P(\Omega) \rightarrow \mathbb{R}^{+}$, but it's also a measure on $M$ and since $A \subset M$ then $\sigma(A) \subseteq m$\\
Lostl, $\mu^{*}$ is a measure on $M$ it is also a measure on $J(A)$\\
Note: $J(A) \subset M$ (sometines)

Lecture 3: Uniqueress and Dykin's $P_{i}$-ambods Question: $\mu_{1} / \mu_{2}$ on $\sigma(A)$ system.

$$\begin{aligned}
& \text { if } \mu_{1}(A)=\mu_{2}(A) \forall A \in A \\
& \text { Dors } \mu_{1}(B)=\mu_{2}(B) \forall B \in \sigma(A)
\end{aligned}$$

Def ( $\pi$-system)\\
a collection of sets $A$ is a $\pi$-system if for any $A, B \in A$ then $A \cap B \in A$

Det (7-system)\\
a collection of set $\mathcal{L}$ is a $\lambda$-system

\begin{itemize}
  \item $\Omega \in \mathcal{L}$
  \item $A, B \in \mathcal{L}$ s.t. $A C B$ then $B \mid A \in \mathcal{L}$
  \item $\left\{A_{i}\right\}_{i=1}^{\infty}$ pairwise misjoint, $A_{i} \in \mathcal{L}$\\
(1) then $\bigcup_{i=1}^{\infty} A_{i} \in \mathcal{L}$\\
similar\\
to $\sigma$-fiefd.
\end{itemize}

Note: a field is a $\pi$-system\\
Theorem (uniqueress of extension)\\
Let $\mu_{1}, \mu_{2}$ be $v$-finite measures on $v(A)$ where $A$ is a $\pi$-system. Then, if $\mu_{1}(A)=\mu_{2}(A) \forall A \in A$ then $\mu_{1}$ and $\mu_{2}$ agree on $\sigma(A)$

Theorem (Dywhin $\pi-\lambda$ theorem)\\
Let $A$ be a $\pi$ system, $\mathcal{L}$ be a lambala system and $A \subset \mathcal{L}$. Then $g(A) \subset \mathcal{L}$\\
Proof: Let $\mathcal{L}_{0}$ be the snallest 7 -system such thet $A \subset \mathcal{L}_{0}$, then $\mathcal{L}_{0} \subseteq \mathcal{L}$

Croal is to show that $\mathcal{L}_{0}$ is also a $\pi$-system and a collection of sets thet is both a $\pi$-system and a $\lambda$-system is a of-field. Then we recessarily were that $v(A) \leqslant L_{0} \leqslant L$\\
) Show that $\mathcal{L}_{0}$ is closed unaler intersections

→ Let $\mathcal{L}^{\prime}=\left[B \in \mathcal{L}_{0}: B \cap A \in \mathcal{L}_{0}, \forall A \in A\right]$\\
Then $\subset \mathcal{L}^{\prime}$ as $A$ is a $\pi$-system\\
Let's show that $\mathcal{L}^{\prime}$ is also a $\lambda$-system.

\begin{itemize}
  \item $\Omega \in L^{\prime}$ because $\notin L_{0}$
  \item it $B_{1}, B_{2} \in \mathcal{L}$ s.t. $B_{1} \subset B_{2}$ then for any $A \in A$ we have that $B_{1} \cap A, B_{2} \cap A \in \mathcal{L}_{0}$\\
Thus $\left(B_{2} \cap A\right) \backslash\left(B_{1} \cap A\right)=\left(B_{2} \mid B_{1}\right) \cap A \in \mathscr{L}_{0}$
\end{itemize}

$$\therefore B_{2} \mid B_{1} \in \mathcal{L}^{\prime}$$

\begin{itemize}
  \item if $\left[\beta_{i}\right]_{i=1}^{\infty} \in Z^{\prime}$ are pairute disjoint\\
then $A_{-}$any $A \in A, A \cap B_{i} \in \mathcal{L}_{0}$
\end{itemize}

$$\therefore \bigcup_{i=1}^{\infty}\left(A \cap B_{i}\right)=A \cap\left(\bigcup_{i=1}^{\infty} \sigma_{i}\right) \in \mathscr{L}_{0}$$

Hence, $\bigcup_{i=1}^{\infty} B_{i} \in \mathcal{L}^{\prime}$

By allinition $\mathcal{L}^{\prime}, \mathcal{L}^{\prime} \subseteq \mathcal{L}_{\sigma}$, but $\mathcal{L}_{0}$ is minimal\\
$\therefore Z_{0}=Z^{\prime} \therefore Z_{0}$ contains all intersections with elements of $A$\\
Lastly, let $\mathcal{L}^{\prime \prime}=\left\{B \in \mathcal{L}_{0}: B \cap C \in \mathcal{L}_{0} \forall H_{C} \in \mathcal{L}_{0}\right\}$ since $\mathcal{L}_{0}=\mathcal{L}^{\prime}, A \subset \mathcal{L}^{\prime \prime}$\\
Then do the same thing we did to $\mathcal{L}^{\prime}$ to $\mathcal{L}^{\prime \prime}$ to show that $\mathcal{L}^{\prime \prime}$ is a $\lambda$-system and thus $\mathcal{L}^{\prime \prime}=\mathcal{L}_{0}$\\
$\therefore \mathcal{L}_{0}$ is closed wheler intergections

$$\begin{aligned}
& \therefore \mathcal{L}_{0} \text { is - } \sigma \text {-field } \\
& \therefore v(A) \subseteq \mathcal{L}_{0} \subseteq \mathcal{L}
\end{aligned}$$

from of uniqueress. (finite measures)

\begin{itemize}
  \item $\mu_{1}(\Omega)=\mu_{2}(\Omega)<\infty$
  \item Let $\mathcal{L}=\left\{B \subset \Omega: \mu_{1}(B)=\mu_{2}(B)\right\}$\\
if $\mathcal{L}$ is a $\lambda$-system, we're clone!
\end{itemize}

$$A \subset \mathcal{L} \therefore \sigma(A) \subset \mathcal{L}$$

→ show that $\mathcal{L}$ is a $\mathcal{Z}$-system

\begin{itemize}
  \item $\Omega=\mathcal{Z}$ (by assumption)
  \item Next, if $A, B \in \mathcal{L}$ with $A \subset B$ then only valied $\mu_{1}(B \backslash A)+\mu_{1}(A)=\mu_{1}(B)$ subtretting, becarte.
\end{itemize}

$$=\mu_{2}(B) \quad \text { to finite measure. }$$

Hence $B \mid A \in \mathcal{L}$

\begin{itemize}
  \item $\left\{A_{i}\right\}_{i=1}^{\infty}$ are painvise dijjoint $A_{i} \in \mathcal{L}$
\end{itemize}

$$\begin{array}{ll}
\mu_{1}\left(\bigcup_{i=1}^{\infty} A_{i}\right)=\sum_{i=1}^{\infty} \mu_{1}\left(A_{i}\right) & =\sum_{i=1}^{\infty} \mu_{2}\left(A_{i}\right) \\
\therefore \bigcup_{i=1}^{\infty} A_{i} \in \mathcal{L} & =\mu_{2}\left(\hat{U}_{i=1} A_{i}\right) \cos
\end{array}$$

$\Rightarrow \mathcal{L}$ is a $\lambda$-system! and a $A \subset \mathcal{L}$\\
∴ contains $\sigma(A) \subset \mathcal{L}$\\
$\therefore \mu_{1}=\mu_{2}$ for all sets in $\sigma(A)$.\\
Proof ox uniqueness (o-finite measure)

\begin{itemize}
  \item For any $A \in A$ s.t. $\mu_{1}(A)=\mu_{2}(A)<\infty$ we abefine $\mathcal{L}_{A}$ to be all $B \subseteq \sqrt{2}$ s.t. $\mu_{1}(A \cap B)=\mu_{2}(A \cap B)$
\end{itemize}

Proceeling as in the proof for finite measures we can show that $\mathcal{L}_{A}$ is a 7 -system and $\therefore v(A) \subset \mathcal{L}_{A}$ (by Dynkin $\pi-\lambda$ )

\begin{itemize}
  \item By 0 -finiteress we decompose $\Omega=U_{i=1}^{\infty}, A_{i} \in A$ gnal $\mu_{1}\left(A_{*}\right)=\mu_{2}(N)<\infty$\\
for any $B \in O(A)$ and any $n \in \mathbb{N}$,
\end{itemize}

$$\begin{aligned}
& \mu_{1}\left(\sum_{i=1}^{\hat{U}}\left(B \cap A_{i}\right)\right)=\sum_{i=1}^{n} \mu_{1}\left(B \cap A_{i}\right)-\sum_{i, j} \mu_{1}\left(B \cap A_{i} \cap A_{j}\right)+ \\
& \text { also worles for } \mu_{2} \quad \text { oo (by inclusion- } \\
& \text { exclusion firmita }
\end{aligned}$$

since $A$ is a $\pi$-system, $A_{i} \cap A_{j} \in A$ as well as further, intersections

$$\therefore \mu_{1}\left(\hat{U}_{i=1}^{\hat{U}}\left(B \cap A_{i}\right)\right)=\mu_{2}\left(\hat{U}_{i=1}^{\hat{U}}\left(B \cap A_{i}\right)\right)$$

any $n \in \mathbb{N}$ - finite\\
Let $n \rightarrow \infty$ and arelvale

\section*{II \\
 (unite foster.)}
<Remerh>9 $\pi$-system is very naturel in probability theory or $n \equiv$ iand'\\
without\\
<Rement> ${ }_{1} \theta$-finitemess, uniqueness may fail!\\
$\Omega=(0,1 ; A$ be all finite unions of welt open intervels ( $a, b$ ]

\begin{itemize}
  \item $\mu$ is a set function that assigns 0 to $\varnothing$ and $\infty$ to any non-empty set\\
$\therefore \mu^{*}$ assigns $\infty$ to any subset of "require" I that is not $\phi$\\
-Counting measure: "Counts the \# of elements" also essign 0 to $\phi$ and $(a, b]$\\
Hewever $\left\{\frac{1}{4}, \frac{1}{2}, \frac{3}{4}\right\} \rightarrow 3$\\
Doer not coincide with $\mu^{*}$\\
◯ Conyleteness: we want to "complete" a reasure.\\
◯ If $E$ and $A$ only differ on a oet of measie o then we want both $E$ and $A$ to be measuralle as' was the sare measure.
\end{itemize}

Def: Symmetric Difference\\
Sor sets $A, B, A \Delta B=(A \mid B) \cup(B \mid A)$\\
\includegraphics[max width=\textwidth, alt={}, center]{5c85dc7d-3850-49bf-8fca-ec10384f670c-16_264_556_404_1334}

For a measure space $(x, \mathcal{F}, \mu)$\\
spere signs measure\\
(otter measure) fielor\\
$\mu^{*}(B)=\inf \{\mu(A): B \subset A, A \in \mathcal{F}\}$\\
Then, the $\mu$-null sets are $\mathcal{F}_{\mu}<P(x)$

$$\text { s.t. } \mu^{*}(N)=0 \quad \forall \quad N \in N_{\mu}$$

we say that a measure space is complete\\
if $N_{\mu} \subset \mathcal{F}$\\
Exercise.\\
Slow that $\mathcal{N}_{\mu}$ is a ring)

If $(x, \tilde{F}, \mu)$ is not complete, we can conplete\\
it by replacing F with

$$\mathcal{F} v N_{\mu}=\left\{A \cup N: A \in \mathcal{F}, N \in N_{\mu}(3\}\right.$$

Prop 3.3 .2 in Dudley\\
This completion is equal to $\mathcal{E S S} x$ : FAEF s.t.

$$\left.A \Delta S \in N_{\mu}\right\}$$

and this is the smulkest or-field that contains both $\mathcal{F}$ and $N_{\mu}$\\
$\left(x, \mathcal{F} \vee N_{\mu}, \bar{\mu}\right)$ where $\bar{\mu}(A O N)=\mu(A)$.

\section*{Leeture 4: Lebesgue measure $12 / 03 / 28$}
$\Omega=$ Exi) $(0,17$ or $\Omega=R$\\
we want $\lambda((a, b])=6-a$\\
Ar now, $\lambda$ is a pre-measure on $A$ where $A$ is any finite union of halt-pen intervals.

$$(a, b] \cup(c, d), a<b<c<d$$

Claim: $A$ is a $\pi$-yystem\\
$\Rightarrow[a, b] \cap c c, d]=\sum \varnothing$ if $6<c$ aléa $[c, s]$ if $a \lll b<d$\\
"pi-system boopenly reaver you con ab intersuctiong "\\[0pt]
[Hhelf-apen, infervel, if you induseed, you get arother uffrogen infival

Claim: A is a ring\\
Recoll, a ring Las $\phi$ nol is closed under set satraction and finite unions.\\
"there ove differend ways to come ye with Lebesgne nerve"

\begin{itemize}
  \item $e_{1}=b \quad(a, a]=\phi$
  \item $(a, b] \backslash(c, d]$
  \item by definition. (finite uniors condition)
\end{itemize}

$$\begin{array}{lll}
( & 7 & 7 \\
a & b & d
\end{array}$$

$\beta=$ borel $\sigma$-fiend

$$B(\mathbb{R})=v \text { (oppen sets) }$$

we can generete $\beta$ for $A$. Thet is, $\beta=\sigma(A)$ we also heve $M_{7}=$ set of all hebesgue peosible sets.\\
Question: $\beta=M_{\lambda}$ ? No! BCM in fact, $M_{2}$ is the compation "strickly subset" of $\beta$ with $N_{7} \leftarrow$ Lebergue Null sets.\\
Poith of Interest:

$$I\left((s, b \cdot 7)=b-a^{c} \text { clanbols is the only }\right. \text { measure that eloce this. }$$

Step I: 7 is a pre-mesure on a ring $A$\\
Step 2: Corathésolong says that $\lambda$ is a measure of $\sigma(A)=\beta$\\
steg 3: A is a $\pi$-system\\
∴ for any reasure $\mu((a, b])=b-a$\\
$M$ coincides with $A$ on $D$\\
the there any $A \subset P_{p}(0,17)$ s.t

\begin{displayquote}
A is not M-measirabla\\
\includegraphics[max width=\textwidth, alt={}, center]{5c85dc7d-3850-49bf-8fca-ec10384f670c-18_363_513_2359_1598}
\end{displayquote}

Example: Vitali sef

$$\Omega=(0,1] \text {, for } x, y \in(0,1]$$

define actelition mod 1

$$x+y=\left\{\begin{array}{lll}
x+y & \text { if } & x+y \leq 1 \\
x+y-1 & \text { if } & x+y>1
\end{array}\right.$$

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{5c85dc7d-3850-49bf-8fca-ec10384f670c-19_541_406_940_1508}
\end{center}

Defite $\mathcal{L}$ to contain all 7 -measurable\\
sets $A \leq[0,1]$ s.t. $\lambda(A)=\lambda(A+x)$\\
For any $x \in(0,1]$\\
where $A+x=\{y \in[0,1]: y-x \in A\}$\\
$A+x$ means shift $A$ by $x$\\
crim $\mathcal{L}$ is a 7-system\\
since $A$ from robre is s.t. $A \subset \mathcal{L}$

$$\begin{aligned}
& \lambda(a, b])=b-a \\
& \lambda((a, b]+x)=\lambda((a+x, b+x])=b-a \\
& \therefore \theta(A)=\beta \subset \mathcal{L}\binom{\text { Dynk, b } \pi+7}{\text { from becture } 3} \\
& \text { we say that } x \sim y \text { if } x-y \in \mathbb{D} \text { ever }
\end{aligned}$$

Next, we say that $x \sim y$ if $x-y \in Q$ "every borel ost

$$\begin{aligned}
& \text { e.g } \frac{1}{\sqrt{2}} \text { and } \frac{1}{\sqrt{2}}+\frac{1}{100} \\
& \text { same! } \\
& \text { checompose (0, 17 into }
\end{aligned}$$

∴ we can clecompose (0) (7) into disjoint equinterne. is stick imporiant with respect to sebeyle reasure"

Decive $H \subset C 0,17$ s.t. H confails over elenect from each equinalence class.\\
lue can do this if we assume the axiom of choice)\\
No two points in $H$ are equivalent\\
i.e. $r_{1}, r_{2}, \&$ then $H+r_{1} \neq H+r_{2}$ untess $r_{1}=r_{2}$ actually, $\left(1++r_{1}\right) \cap\left(H+r_{2}\right) \doteq \phi$ for $r_{1} \neq r_{2}$\\
$\therefore(0,1]=U_{\vec{k} r \in \mathbb{Q}}(H+r)$\\
∴ by countable addivity $I=\exists((0,1])=\sum_{c \in \mathbb{Q}} \lambda(H+r)$\\
at, becose $\lambda$-measure is trensintion invariant $\exists\left(H+C_{1}\right)=\lambda\left(H+C_{2}\right)$

$$\begin{aligned}
& I=\{\lambda(H) \\
& \text { if } \lambda(H)=0 \text { then } 1=0 \text { contraliction! } \\
& \text { if } \lambda(H)=\infty \text { then } 1=\infty
\end{aligned}$$

Conclusion: $M_{A} \subset P((0,1])$\\
Fin Fut: Lebergue Measure on $\mathbb{R}$ is the only travilation inveriant measure.

\begin{itemize}
  \item same for $\mathbb{R}^{n}$.
  \item there is no $D$-elinersional Lebesgue neasere. → i.e. Nes tranglation Invariant measure likelihoseds don't extst.
\end{itemize}

Product Measures.\\
We construct $\lambda$ on $\mathbb{R}$ by considering the intervels $(a, b]$\\
For $\mathbb{R}^{p}$ we can do the same thing with $p$-dimensional rectangles

$$\left.\begin{array}{rl}
\lambda^{(2)}((a, b] \times(c, d]) & =(b-a) \times(d-c) \\
& =\lambda(b a, b) \times \cdots \\
& \lambda((a, b]) \lambda((c, d]))
\end{array}\right\}$$

Note: the set of such rectangle in $\mathbb{R}^{p}$ form a $\pi$-system\\
Generally: For two measure spaces $\left(\mathbb{X}, X_{T}^{*}, \mu\right)^{K}$

$$\sigma \text {-fical on }$$

$(Y, Y, U)$ we can dutine\\
$(\mathbb{A} \times \boldsymbol{Y}, x \times y, \sigma)$ where $\pi(A \times B)=\mu(A) v(B)$\\
for $A \in X$ and $B \in Y$\\
Question: How are these related?

$$\begin{aligned}
& \rightarrow B(x)+B(\mathbb{y}) \\
& \rightarrow B(\mathbb{x}+\mathbb{y})
\end{aligned}$$

Ifrom addley $(4.1 .7), B(X) \times B(Y) \subset B(X+Y)$

\begin{itemize}
  \item but these are "ugually" equal to each other? (Linore precisely, equality occurs when the measurable spacel are "second count-ble") $(x, x)$, speen are
\end{itemize}

Trecependence artully probability Let $(\Omega, \mathcal{F}, \mu)$ be a neasure space bef: (Inalypardences for sets)\\
Fol a countable collection of sefs $A_{i}, i \in I$, we say that this lollection is independent if for all finite JCI, we have

Eample:

$$\left\langle\mu\left(\bigcap_{j \in J} A_{j}\right)=\prod_{j \in J} \mu\left(A_{j}\right)\right\rangle$$

$A_{1}=$ [olsow s. real cand]

$$\begin{aligned}
& A_{2}=\{\text { sorno a leart or club }\} \\
& A_{3}=[\text { draw a queen }] \\
& \mu\left(A_{1}\right)=\frac{1}{2} \quad \mu\left(A_{1} \cap A_{2}\right)=\frac{1}{4}\left(=\frac{1}{2} \times \frac{1}{2}\right) \\
& \mu\left(A_{2}\right)=\frac{1}{2} \quad \mu\left(A_{1} \cap A_{3}\right)=\frac{1}{26}\left(=\frac{1}{2} \times \frac{1}{13}\right) \\
& \mu\left(A_{3}\right)=\frac{1}{13} \quad \mu\left(A_{2} \cap A_{3}\right)=\frac{1}{26} \\
& \quad \mu\left(A_{1} \cap A_{2} \cap A_{3}\right)=\frac{1}{\sqrt{2}}
\end{aligned}$$

Definition (independert signs fielels)\\
For a countable collection of or-fieble $F_{i} \subset \mathcal{F}, i \in I$, we say the collection $I$ irelependent if any set of sets $\left\{A_{i} \in \mathcal{F}_{i}: i \in I\right\}$ is indepentent\\
Theorem: Let $k_{1}, A_{2} \subset \mathcal{F}$ be T-systems.\\
If $\mu\left(A_{1} \cap A_{2}\right)=\mu\left(A_{1}\right) \mu\left(A_{2}\right)$\\
For any $A_{1} \in A_{1}$ and $A_{2} \in A_{2}$, then $\sigma\left(A_{1}\right)$ onel $v\left(A_{2}\right)$ are independent

Prot: Fix on $A_{1} \in A_{1}$ and define two meagures for $B \in \mathcal{F}$

$$\begin{aligned}
& \nu_{1}(B)=\mu\left(A_{1} \cap B\right) \\
& \nu_{2}(B)=\mu\left(A_{1}\right) \mu(B)
\end{aligned}$$

By assumption, $V_{1}\left(A_{2}\right)=V_{2}\left(A_{2}\right)$

\section*{Eor any $A_{2} \in A_{2}$}
Then by iniqueness of extension $\nu_{1}$ and $\nu_{2}$ Lure + coincidar on $\sigma\left(A_{2}\right)$.

$$\therefore \mu\left(A_{1} \cap B_{2}\right)=\mu\left(A_{1}\right) \mu\left(B_{2}\right)$$

for any $B_{2} \in \sigma(A / 2)$\\
Now es the same angment again, bit by fixing sone $B_{2} \in V\left(A_{2}\right)$ to get tha $\mu\left(B_{1} \cap B_{2}\right)=\mu\left(B_{1}\right) \mu\left(B_{2}\right)$ for any $B_{i} \in \sigma\left(A_{i}\right), i=1,2$.


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