pecture 1: Adam Koshiah Measure: $\left(\Omega, \mathcal{F}^{\prime}\right) \leftarrow$ Measurkeble Spere a preasure $\mu: \tau \rightarrow \mathbb{R}^{+}$ 1. $\mu(\phi)=0$ 2. $\mu *$ is countably aelelitive 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) - if $\mu(\Omega)=7$ then we say that $(\Omega, F, \mu)$ is a probability space onel $\mu$ is a probility measure - if $\mu(\Omega)<\infty$ then $\mu$ is : finite measure - 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. 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) - $\mu(\{1,3,7\})=3$, $\mu(\alpha \Omega)=1$ - define $\nu(A)=\frac{1}{n} \mu(A)$ 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 ![](https://cdn.mathpix.com/cropped/5c85dc7d-3850-49bf-8fca-ec10384f670c-03.jpg?height=155&width=292&top_left_y=1792&top_left_x=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 $-\phi \in \mathscr{f}$ - for any $A, B \in f, A \cap B \in f$ and $$B \backslash A=\bigcup_{i=1}^{3} c_{i} \text { for } c_{i}: \in \mathscr{\theta}$$ "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 - $\mu$ is increasing it $A \subset B \Rightarrow \mu(A) \leq \mu(B)$ - $M$ is avelelitive it for $A, B$ disjoint then $\mu(A \cup B)=\mu(A)+\mu(* \beta)$ - $\mu$ 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) $$\mu\left(\bigcup_{i=1}^{\infty} A_{i}\right)=\sum_{i=1}^{\infty} \mu\left(A_{i}\right)$$ $$\bigcup_{i=1}^{\infty} A_{i} \in \neq \phi$$ - $\mu$ 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. 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.} - Denote (suript $m$ ) $m$ to be the colleulton of all $\mu^{*}$ measurable sets. where we say that $B \subseteq \Omega$ is $\mu^{*}$ measurable when $$\mu^{*}(E \cap B)+\mu^{*}\left(E \cap B^{C}\right)=\mu^{*}(E)$$ $$\text { for all } E \subseteq \Omega$$ Theorem: (Corathédalony Extergion Thooreny ![](https://cdn.mathpix.com/cropped/5c85dc7d-3850-49bf-8fca-ec10384f670c-06.jpg?height=466&width=870&top_left_y=1485&top_left_x=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 $\mu^{*}$ - $\mu^{*}(\phi)=0$ as $\mu$ is a pre-measure - $\mu^{*}$ is von-negative $\forall B \subseteq \Omega$ as $\mu$ is non-reg - $\mu^{H}$ is morotone (increasing) 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 $B_{1} \subseteq \cup_{i} A_{i}$ $\therefore \mu^{*}\left(B_{1}\right) \leqslant \mu^{*}\left(B_{2}\right)$ - $\mu^{*}$ is countably sadelitive 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 - $\phi \in M$ since $\phi \in \mathbb{A}$ - $\Omega \in M$ since for any. $E \subseteq \Omega$ (a sigma-field requires $\Omega$ thes in it). $$\mu^{*}(E \cap \Omega)+\mu^{*}(E \cap \phi)=\mu^{*}(E)$$ 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 - $\Omega \in \mathcal{L}$ - $A, B \in \mathcal{L}$ s.t. $A C B$ then $B \mid A \in \mathcal{L}$ - $\left\{A_{i}\right\}_{i=1}^{\infty}$ pairwise misjoint, $A_{i} \in \mathcal{L}$ (1) then $\bigcup_{i=1}^{\infty} A_{i} \in \mathcal{L}$ similar to $\sigma$-fiefd. 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. - $\Omega \in L^{\prime}$ because $\notin L_{0}$ - 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 $\left(B_{2} \cap A\right) \backslash\left(B_{1} \cap A\right)=\left(B_{2} \mid B_{1}\right) \cap A \in \mathscr{L}_{0}$ $$\therefore B_{2} \mid B_{1} \in \mathcal{L}^{\prime}$$ - if $\left[\beta_{i}\right]_{i=1}^{\infty} \in Z^{\prime}$ are pairute disjoint then $A_{-}$any $A \in A, A \cap B_{i} \in \mathcal{L}_{0}$ $$\therefore \bigcup_{i=1}^{\infty}\left(A \cap B_{i}\right)=A \cap\left(\bigcup_{i=1}^{\infty} \sigma_{i}\right) \in \mathscr{L}_{0}$$ 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) - $\mu_{1}(\Omega)=\mu_{2}(\Omega)<\infty$ - Let $\mathcal{L}=\left\{B \subset \Omega: \mu_{1}(B)=\mu_{2}(B)\right\}$ if $\mathcal{L}$ is a $\lambda$-system, we're clone! $$A \subset \mathcal{L} \therefore \sigma(A) \subset \mathcal{L}$$ → show that $\mathcal{L}$ is a $\mathcal{Z}$-system - $\Omega=\mathcal{Z}$ (by assumption) - 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. $$=\mu_{2}(B) \quad \text { to finite measure. }$$ Hence $B \mid A \in \mathcal{L}$ - $\left\{A_{i}\right\}_{i=1}^{\infty}$ are painvise dijjoint $A_{i} \in \mathcal{L}$ $$\begin{array}{ll} \mu_{1}\left(\bigcup_{i=1}^{\infty} A_{i}\right)=\sum_{i=1}^{\infty} \mu_{1}\left(A_{i}\right) & =\sum_{i=1}^{\infty} \mu_{2}\left(A_{i}\right) \\ \therefore \bigcup_{i=1}^{\infty} A_{i} \in \mathcal{L} & =\mu_{2}\left(\hat{U}_{i=1} A_{i}\right) \cos \end{array}$$ $\Rightarrow \mathcal{L}$ 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) - 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. $\mu_{1}(A \cap B)=\mu_{2}(A \cap B)$ 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 $\pi-\lambda$ ) - 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 $n \in \mathbb{N}$, $$\begin{aligned} & \mu_{1}\left(\sum_{i=1}^{\hat{U}}\left(B \cap A_{i}\right)\right)=\sum_{i=1}^{n} \mu_{1}\left(B \cap A_{i}\right)-\sum_{i, j} \mu_{1}\left(B \cap A_{i} \cap A_{j}\right)+ \\ & \text { also worles for } \mu_{2} \quad \text { oo (by inclusion- } \\ & \text { exclusion firmita } \end{aligned}$$ 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.)} 9 $\pi$-system is very naturel in probability theory or $n \equiv$ iand' without ${ }_{1} \theta$-finitemess, uniqueness may fail! $\Omega=(0,1 ; A$ be all finite unions of welt open intervels ( $a, b$ ] - $\mu$ 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. Def: Symmetric Difference Sor sets $A, B, A \Delta B=(A \mid B) \cup(B \mid A)$ ![](https://cdn.mathpix.com/cropped/5c85dc7d-3850-49bf-8fca-ec10384f670c-16.jpg?height=264&width=556&top_left_y=404&top_left_x=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} A is not M-measirabla ![](https://cdn.mathpix.com/cropped/5c85dc7d-3850-49bf-8fca-ec10384f670c-18.jpg?height=363&width=513&top_left_y=2359&top_left_x=1598) Example: Vitali sef $$\Omega=(0,1] \text {, for } x, y \in(0,1]$$ define actelition mod 1 $$x+y=\left\{\begin{array}{lll} x+y & \text { if } & x+y \leq 1 \\ x+y-1 & \text { if } & x+y>1 \end{array}\right.$$ ![](https://cdn.mathpix.com/cropped/5c85dc7d-3850-49bf-8fca-ec10384f670c-19.jpg?height=541&width=406&top_left_y=940&top_left_x=1508) 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. - same for $\mathbb{R}^{n}$. - there is no $D$-elinersional Lebesgue neasere. → i.e. Nes tranglation Invariant measure likelihoseds don't extst. 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 $(4.1 .7), B(X) \times B(Y) \subset B(X+Y)$ - but these are "ugually" equal to each other? (Linore precisely, equality occurs when the measurable spacel are "second count-ble") $(x, x)$, speen are 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 $B_{i} \in \sigma\left(A_{i}\right), i=1,2$.