$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ > [!theorem] Caratheodory Extension Theorem > Let $\mu_0$ be a [[pre-measure]] over the [[measurable space#^algebra|algebra]] $\AAA_0$. Then there exists a unique [[measure space|measure]] $\mu: \AAA\to [0,\infty)$ on the generated $\sigma$-[[measurable space|algebra]] $\AAA$ such that $\mu_0(A) = \mu(A)$ for all $A\in \AAA_0$. This theorem is completely legendary; its what makes measure theory tick. In topology, if I hand you $\RR^2$ and ask you to draw an open set, you draw some connected blob and _have confidence that your intuition is completely general_. In measure theory, this lets you do the same thing. # Uniqueness The uniqueness portion of the proof uses [[pi lambda system|Dynkin's Theorem]]. Suppose there were two distinct extensions $\mu_1$ and $\mu_2$ of $\mu_0$. They all agree on $A\in \AAA_0$, so we can let $ \LL = \{B \in \AAA : \mu_1(B) = \mu_2(B)\} $ Observe that $\AAA\subset \LL$, and we know $\AAA$ is a $\pi$-system because it is an algebra. > [!claim] > > $\LL$ is a $\lambda$-system. By Dynkin's theorem, we conclude that $\AAA\subset \LL \subset \AAA$, thus $\mu_1 = \mu_2$ on all of $\AAA$. > [!proof]- Direct verification of claim > $\Omega \in \LL$ because $\Omega\in \mathcal{A}$ and $\mu_1(\Omega) = \mu_2(\Omega)$. If $A\subset B$ are in $\LL$, then $\mu_1(B\setminus A) = \mu_1(B) - \mu_1(A) = \mu_2(B) - \mu_2(A) = \mu_2(B\setminus A)$, so $B\setminus A$ is in $\LL$. Finally, if $A_1, A_2, \dots$ are in $\LL$ and $A_i \uparrow A$, then by countable additivity, > > $ > \mu_1(A) = \lim \mu_1(A_i) = \lim \mu_2(A_i) = \mu_2(A) > $ > > so $A$ is in $\LL$. # Existence The construction of the Caratheodory extension requires lots of machinery. First, we'll upgrade $\mu_0$ to _all_ subsets of $\Omega$ (this _overshoots_ what we're trying to do!) before restricting it to our desired $\sigma$-algebra. ## Outer Measure > [!definition] Outer Measure > The ==**outer measure**== $\mu^*$ over $\Omega$ is a function $\mu^*: 2^\Omega \to [0,\infty]$ such that: > > - $\mu^*(\emptyset) = 0$. > - ==**Monotonicity**==: if $A \subseteq B$, then $\mu^*(A) \leq \mu^*(B)$. > - ==**Countable Subadditivity**==: if $A_1, A_2, \dots \subseteq \Omega$, then > $ > \mu^*\left(\bigcup A_i\right) \leq \sum \mu^*(A_i). > $ > [!example] Outer Measures are easy to create > Here's a slightly stupid way to construct an outer measure. Let $\rho: \mathcal{E} \to [0,\infty]$ be any function, where $\mathcal{E}\subset 2^\Omega$. Then, > > $ > \mu^*(A) = \inf\left(\sum_{i=1}^\infty \rho(K_i) : K_i\in \mathcal{E}, A \subseteq \bigcup_{i = 1}^\infty K_i\right) > $ > > is an outer measure. This is called the ==**outer measure generated by $\rho$**==. > [!proof]- > $\mu^*(\emptyset) = 0$ is clear. Monotonicity is clear; if $A\subseteq B$, any collection $\{K_i\}$ that works for $B$ works for $A$. > Countable subadditivity is a good exercise in $\NN$-strength logic. First off, if any of the $\mu^*(A_i)$ are $\infty$, then the inequality is trivial. On the other hand, countable absolutely convergent sums commute with infimums, so > > $ > \sum \mu^* (A_i) = \inf\left(\sum_{i=1}^\infty \sum_{j = 1}^\infty \rho(K_{i,j}) : A_i \subseteq \bigcup_{j = 1}^\infty K_{i,j}\right). > $ > > Alright fine, but clearly $K_{i,j}$ taken altogether is a countable collection of sets that covers $\bigcup A_i$. So we're done. For general $\rho$, this $\mu^*$ is probably meaningless; in particular, it probably isn't compatible with $\rho$ at all. But for $\rho = \mu_0$, $\mu^*$ is the correct upgrade. > [!theorem] Outer Measures generated by Pre-Measures > The outer measure $\mu^*$ generated by a pre-measure $\mu_0$ satisfies $\mu^*(A) = \mu_0(A)$ for all $A \in \AAA_0$. > [!proof]- > Obviously $\mu_0(A)\geq \mu^*(A)$; let $K_0 = A$. To show $\mu_0(A)\leq \mu^*(A)$, lets fix any collection of sets $A_1,\cdots$ which cover $A$. We want to show > > $ > \mu_0(A) \leq \sum \mu_0(A_i). > $ > > This looks almost exactly like the definition of $\mu_0$; we just need to get rid of the overlaps. Fine; let > > $ > B_i = A\cap\left(A_1\cup \cdots A_i\right),\qquad C_i = B_i \setminus B_{i-1}. > $ > > Then, its clear that each $C_i\in \AAA_0$ and form a disjoint partition of $A$. For all finite $n$, > > $ > \sum_{i = 1}^n \mu_0(A_i) \geq \sum_{i = 1}^n \mu_0(C_i) > $ > > This inequality holds in the limit, where the RHS is $\mu_0(A)$ by countable additivity. In the case of the [[Lebesgue Measure]], at this step we let $\lambda^*$ be the outer measure generated by $\lambda_0$. Thus, we are now guaranteed that intervals have the correct length. Here's another observation: > [!example] Outer Measures are Infimums of Open Sets > For every subset $A\subset \RR$ and $\eps > 0$, there exists an open set $U$ such that $A\subset U$ and $\lambda^*(U)\leq \lambda^*(A)\leq \lambda^*(U) + \eps$. Indeed, if $\lambda^*(A) = \infty$, $U = \RR$ works. Otherwise, let $\{I_i\}$ be a countable collection of open intervals that cover $A$ with length at most $\lambda^*(A) + \eps$. Then, $U = \bigcup I_i$ works by countable subadditivity. ## Caratheodory Measureability > [!definition] Caratheodory Measurable > A set $A$ is ==**Caratheodory measurable**== with respect to outer measure $\mu^*$ if for all $E \subseteq \Omega$, > > $ > \mu^*(E) = \mu^*(E\cap A) + \mu^*(E\setminus A). > $ > > The $\sigma$-algebra of Caratheodory measurable sets is denoted $\AAA^{\text{cm}}$. Of course, implicit in this definition is a bunch of detail-checking to show that $\AAA^{\text{cm}}$ is a $\sigma$-algebra. > [!idea] Tips on verification > > - By subadditivity, we already know $\mu^*(E) \leq \mu^*(E\cap A) + \mu^*(E\setminus A)$. $A$ is measurable iff $\mu^*(E) \geq \mu^*(E\cap A) + \mu^*(E\cap A^c)$. > - If $\mu^*(E) = \infty$, then the inequality is trivial. > - We can actually inductively iterate this. Let $E\subset \Omega$, and let $A_1,\cdots A_n$ be disjoint measurable sets. Then, > $ > \mu^*(E\cap (A_1\cup \cdots \cup A_n)) = \sum_{i = 1}^n \mu^*(E\cap A_i). > $ > - In particular, $\mu^*$ is obviously finitely additive on $\AAA^{\text{cm}}$. > [!proof]- Caratheodory is an Algebra > $\AAA^{\text{cm}}$ is clearly closed under complements, and contains $\emptyset$. Suppose $A$ and $B$ are Caratheodory measurable. Then, for any $E$, by subadditivity > > $ > \mu^*(E\cap (A\cup B)) \leq \mu^*(E\cap A) + \mu^*(E\cap B\setminus A) > $ > > Now, $A$ is measurable, so $\mu^*(E\cap A) = \mu^*(E) - \mu^*(E\setminus A)$. $B$ is measurable, so $m^*(E\setminus A) = m^*(E\setminus (A\cup \setminus B)) + m^*((E\setminus A)\cap B)$. Combining, > > $ > \mu^*(E\cap (A\cup B)) \leq \mu^*(E) - \mu^*(E\setminus (A\cup \setminus B)) > $ > > Thus $A\cup B$ is measurable; thus finite unions are measurable. > [!proof]- Caratheodory is a $\sigma$-Algebra > Alright, game time. Give me a bunch of random measurable sets $A_1,A_2,\cdots$; I want to show their union, $A = \bigcup A_i$, is measurable. First, I'm going to make them disjoint. Let > > $ > B_n = A_n \setminus \bigcup_{i = 1}^{n-1} A_i. > $ > > Clearly the $B_n$ are disjoint measurable sets, and $\bigcup A_i = \bigcup B_i$. Next, pick any stupid set $E$. In order to show the Caratheodory property, I'm going to truncate our sum at $N$; let > > $ > C_N = \bigcup_{i = 1}^N B_i > $ > > We know that $C_N$ is measurable, so > > $ > \mu^*(E) = \mu^*(E \cap C_N) + \mu^*(E\setminus C_N) > $ > > By countable additivity on $E \cap C_N$ and monotonicity on $E\setminus C_N$, we find > > $ > \mu^*(E) \geq \sum_{n\leq N} \mu^*(E\cap B_n) + \mu^*(E\setminus A) > $ > > Fine. This is true for all $N$ though, so we can take the limit: > > $ > \mu^*(E) \geq \sum_{n\geq 1} \mu^*(E\cap B_n) + \mu^*(E\setminus A)\geq \mu^*(E\cap A) + \mu^*(E\setminus A) > $ > > by countable subadditivity. Over $\RR$, these are the ==**Lebesgue measurable**== sets, $\LL(\RR)$. > [!theorem] Caratheodory compatible with Pre-Measure > We should make sure that if $\mu^*$ is generated by a pre-measure $\mu_0$, then all $A\in \AAA_0$ are Caratheodory measurable. > [!proof]- > For all $\eps > 0$, there exists a covering $E\subset \bigcup K_i$ using sets $K_i \in \AAA_0$ such that > > $ > \mu^*(E) + \eps \geq \sum \mu_0(K_i) = \sum \mu_0(K_i \cap A) + \sum \mu_0(K_i \setminus A)\geq \mu^*(E\cap A) + \mu^*(E\setminus A). > $ Thus, we find that the Borel $\sigma$-algebra $\BBB(\RR)$ is contained in $\LL(\RR)$. In any case, to create a measure space on $\Omega$, simply restrict _any_ outer measure $\mu^*$ to the Caratheodory measurable sets. > [!theorem] This is actually a Measure > The restriction of $\mu^*$ to $\AAA^{\text{cm}}$ is countably additive. > [!proof] > Suppose $A_1, A_2,\cdots$ are disjoint measurable sets. Then for all $n$, > > $ > \mu^*\left( \bigcup A_i \right) \geq \sum_{i = 1}^n \mu^*(A_i) > $ > > by finite additivity. Taking the limit, we conclude. ## Restriction Thus, we have constructed a measure $\mu^*$ on $\AAA^{\text{cm}}$; in particular, $\AAA_0 \subset \AAA^{\text{cm}}$, thus $\sigma(\AAA_0) = \AAA\subset \AAA^{\text{cm}}$. We thus can speak unambiguously of the measure $\mu: \AAA \to [0,\infty]$ obtained by restricting $\mu^*$ to $\AAA$. # Properties 1. All measure spaces constructed in this way are [[complete measure space#^caratheodory-complete|complete]]. > [!problem] Littlewood's First Principle > Suppose $E\subset \RR$ with $\lambda^*(E) < \infty$. Show that $E$ is measurable iff for all $\eps > 0$, there exists a _finite_ union of intervals $I_1, \cdots, I_N$ such that > > $ > \lambda^*\left( E\Delta \bigcup I_i \right) < \eps. > $ > > where $\Delta$ means symmetric difference. > [!solution]- > Suppose $E$ is measurable, and fix $\eps$. By definition of the outer measure, there exists a countable cover of intervals $I_i$ with measure arbitrarily close to $\mu^*(E)$: > > $ > \sum \mu(I_i) < \mu^*(E) + 0.001\eps. > $ > > By countable subadditivity and monotonicity, we also know > > $ > \mu^*(E) < \mu\left(\bigcup I_i\right) < \mu^*(E) + 0.001\eps > $ > > We know countable unions of open sets are generated by $\AAA_0$ and thus in $\AAA^{cm}$; > > $ > \mu^*\left( \bigcup I_i \setminus E\right) = \mu^*\left( \bigcup I_i \right) - \mu^*(E) < 0.001\eps. > $ > > Finally, the sum on the LHS is strictly increasing thus this inequality is true for any prefix $N$. Thus, we can take $N$ large enough such that > > $ > \mu^*\left( E \setminus \bigcup_{i=1}^N I_i \right) < \mu^*\left( \bigcup_{i\geq N} I_i\right) < 0.001\eps. > $ > > This concludes. > On the other hand, suppose you give me a $A\subset \RR$ and $\eps > 0$. I'll pick a finite union of intervals called $U\in \AAA_0$ such that $\mu^*(A\Delta U) < 0.001\eps$. Then for all $E$, > > $ > \mu^*(A\setminus E) \leq \mu^*(A\cap (E^c \cap U^c)) + \mu^*(A\cap (E^c \cap U)) \leq 0.001\eps + \mu^*(U\setminus E) > $ > > and > > $ > \mu^*(A\cap E) \leq \mu^*(A\cap (E\cap U^c)) + \mu^*(A\cap (E\cap U)) \leq 0.001\eps + \mu^*(U\cap E) > $ > > Adding it all up, > > $ > \mu^*(A\setminus E) + \mu^*(A\cap E) \leq 0.002\eps + \mu^*(U) \leq 0.003\eps + \mu^*(A) > $ > > by countable subadditivity.