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. $\blacksquare$

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Post Prelims: New Beginning, Pt. 1

Here's fresh, new math—stuff I've never posted about. It's a bit of measure theory—an indispensable generalization of the notion of "length", "area", "volume", etc., which will later on allow us to extend the Riemann integral to the Lebesgue integral (and thus take our calculus to the "next level", if I may). Real-world applications of these concepts are plentiful in quantum mechanics. In fact, it feels like all the math I do now has applications only in similar fields at around that level, which makes it somewhat difficult to motivate. But someone has to do it, right?

Claim. Let $\left\{X_j\right\}_{j=1}^\infty$ be a collection of sets, let $\mathcal E_j\subseteq\mathcal P\left(X_j\right)$, let $\mathcal M_j$ be the $\sigma$-algebra on $X_j$ generated by $\mathcal E_j$, and let $X_j\in\mathcal E_j$. Then $\bigotimes_{j=1}^\infty\mathcal M_j$ is generated by$$\mathcal F=\left\{\prod_{j=1}^\infty E_j:E_j\in\mathcal E_j\right\}.$$Proof. Recall that $\bigotimes_{j=1}^\infty\mathcal M_j$ is generated by$$\mathcal G=\left\{\pi_j^{-1}\left(E_j\right):j\in J\text{ and }E_j\in\mathcal E_j\right\},$$where $\pi_k:\prod_{j=1}^\infty X_j\to X_k$ is defined canonically. An element of $\mathcal F$ has the form $\prod_{j=1}^\infty E_j=E_1\times E_2\times\cdots,$ where $E_j\in\mathcal E_j$. Since $\pi_k^{-1}\left(E_k\right)=X_1\times X_2\times\cdots\times E_k\times\cdots\in\mathcal G$, it is the case that$$\prod_{j=1}^\infty E_j=E_1\times E_2\times\cdots=\bigcap_{j=1}^\infty\pi_j^{-1}\left(E_j\right)\in\mathcal M\left(\mathcal G\right).$$Hence, $\mathcal M\left(\mathcal F\right)\subseteq\mathcal M\left(\mathcal G\right).$ Conversely, an element of $\mathcal G$ has the form $\pi_k^{-1}\left(E_k\right)=X_1\times X_2\times\cdots\times E_k\times\cdots$, where $E_k\in\mathcal E_k$. Since $X_j\in\mathcal E_j$, it is the case that $\pi_k^{-1}\left(E_k\right)\in\mathcal F$. Hence, $\mathcal M\left(\mathcal G\right)\subseteq\mathcal M\left(\mathcal F\right)$. $\blacksquare$

Note that it is crucial that $\left\{X_j\right\}_{j=1}^\infty$ is a countable collection and that $X_j\in\mathcal E_j$ for $j=1,2,\dots$.

Later on, I will introduce the notion of a measure, define Borel $\sigma$-algebras, and do some computations on $\mathbb R$.

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