Theorem 14.7.4 (Carathéodory, [Theorem 1.11, Fol99]). Let $X$ be a set, $\mu^{*}: 2^{X} \to [0, \infty]$ be an outer measure, and $\cm \subset 2^{X}$ be the collection of all $\mu^{*}$-measurable sets, then:
For any $F \subset X$ and $\seq{E_n}\subset \cm$ pairwise disjoint,
\[\mu^{*}\paren{F \cap \bigsqcup_{n \in \natp}E_n}= \sum_{n \in \natp}\mu^{*}(F \cap E_{n})\]$\cm$ is a $\sigma$-algebra.
$(X, \cm, \mu^{*}|_{\cm})$ is a complete measure space.
Proof. (1): Let $N \in \nat$, then
As this holds for all $N \in \nat$,
(2): Firstly, $\emptyset, X \in \cm$ since $\mu^{*}(\emptyset) = 0$. For any $A \in \cm$, $A^{c} \in \cm$ as well because the definition of $\mu^{*}$-measurability is symmetric.
Let $A, B \in \cm$ and $F \subset X$, then
so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by Lemma 13.1.5.
(3): By (1), $\mu|_{\cm}$ is a measure. Let $N \in \cm$ with $\mu^{*}(N) = 0$, then for any $E \subset N$ and $F \subset X$,