Saturday, September 10, 2016

Carathéodory's Theorem

I like this theorem because it allows us to construct measures from outer measures—complete measures, in fact.

Claim (Carathéodory's Theorem): If $\mu^*$ is an outer measure on a nonempty set $X$, then the collection $\mathcal M$ of $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra, and $\mu^*|_{\mathcal M}$ is a complete measure.

Proof: idk how 2 explain it.. i just feel it in me, u know what im saying? $\blacksquare$

I'm just kidding. The way people usually prove things does not constitute a valid proof. lol

Proof: Observe that if $E\in\mathcal M$, then $E^c\in\mathcal M$ since $\mu^*$-measurability is symmetric in $E$ and $E^c$.

Let $A,B\in\mathcal M$ and let $E\subseteq X$. Then$$\begin{align}\mu^*\left(E\right)&=\mu^*\left(E\cap A\right)+\mu^*\left(E\cap A^c\right)\\&=\mu^*\left(E\cap A\cap B\right)+\mu^*\left(E\cap A\cap B^c\right)+\mu^*\left(E\cap A^c\cap B\right)+\mu^*\left(E\cap A^c\cap B^c\right).\end{align}$$Since $A\cup B=(A\cap B)\cup(A\cap B^c)\cup(B\cap A^c)$, it is the case that$$\mu^*\left(E\cap\left(A\cup B\right)\right)\leq\mu^*\left(E\cap A\cap B\right)+\mu^*\left(E\cap A\cap B^c\right)+\mu^*\left(E\cap A^c\cap B\right),$$which implies that $\mu^*(E)\geq\mu^*(E\cap(A\cup B))+\mu^*(E\cap(A\cup B)^c)$. Therefore, $\mathcal M$ is an algebra.

Observe that if $A,B\in\mathcal M$ and $A\cap B=\varnothing$, then$$\mu^*\left(A\cup B\right)=\mu^*\left(\left(A\cup B\right)\cap A\right)+\mu^*\left(\left(A\cup B\right)\cap A^c\right)=\mu^*\left(A\right)+\mu^*\left(B\right).$$Therefore, $\mu^*$ is finitely additive on $\mathcal M$.

Let $\{A_j\}_1^\infty$ be a disjoint collection of elements of $\mathcal M$, let $B_n=\bigcup_1^n A_j$, let $B=\bigcup_1^\infty A_j$, and let $E\subseteq X$. Then$$\mu^*\left(E\cap B_n\right)=\mu^*\left(E\cap B_n\cap A_n\right)+\mu^*\left(E\cap B_n\cap A_n^c\right)=\mu^*\left(E\cap A_n\right)+\mu^*\left(E\cap B_{n-1}\right).$$By induction, it is the case that $\mu^*(E\cap B_n)=\sum_1^n\mu^*(E\cap A_j)$, which implies that$$\mu^*\left(E\right)=\mu^*\left(E\cap B_n\right)+\mu^*\left(E\cap B_n^c\right)\geq\sum_1^n\mu^*\left(E\cap A_j\right)+\mu^*\left(E\cap B^c\right)\geq\mu^*\left(\bigcup_1^n E\cap A_j\right)+\mu^*\left(E\cap B^c\right).$$Letting $n$ tend to infinity yields that $\mu^*(E)\geq\mu^*(E\cap B)+\mu^*(E\cap B^c)$. Therefore, $\mathcal M$ is a $\sigma$-algebra.

Observe that if $E=B$, then $\mu^*(B)=\sum_1^\infty\mu^*(A_j)$. Therefore, $\mu^*$ is countably additive on $\mathcal M$, which implies that $\mu^*|_{\mathcal M}$ is a measure.

Let $A,E\subseteq X$ such that $\mu^*(A)=0$. Then$$\mu^*(E)\leq\mu^*(E\cap A)+\mu^*(E\cap A^c)=\mu^*(E\cap A^c),$$which implies that $A\in\mathcal M$. Therefore, $\mu^*|_{\mathcal M}$ is complete. $\blacksquare$