Proof of Caratheodory Extension Theorem

Throughout, $\mu^{*}$ is an outer measure on $X$, $\mathcal{M}$ is the family of Carathéodory measurable sets, and we use the splitting condition in the form: $E\in\mathcal{M}$ if and only if $\mu^{*}(A)\ge\mu^{*}(A\cap E)+\mu^{*}(A\setminus E)$ for every $A\subseteq X$, the reverse inequality being automatic from subadditivity (apply countable subadditivity to the sequence $A\cap E, A\setminus E,\varnothing,\varnothing,\dots$, using $\mu^{*}(\varnothing)=0$).

Step 1 ($\mathcal{M}$ contains $X$ and is closed under complements). $X\in\mathcal{M}$ since $A\cap X=A$ and $A\setminus X=\varnothing$. The defining condition is symmetric in $E$ and $X\setminus E$, because $A\cap(X\setminus E)=A\setminus E$ and $A\setminus(X\setminus E)=A\cap E$; hence $E\in\mathcal{M}$ implies $X\setminus E\in\mathcal{M}$ (complements).

Step 2 (closure under finite unions and finite additivity). Let $E,F\in\mathcal{M}$ and $A\subseteq X$. Splitting $A$ by $E$, then splitting both pieces by $F$:

$$\mu^{*}(A)=\mu^{*}(A\cap E\cap F)+\mu^{*}((A\cap E)\setminus F)+\mu^{*}((A\setminus E)\cap F)+\mu^{*}((A\setminus E)\setminus F).$$

The first three sets cover $A\cap(E\cup F)$ (indeed they partition it), so by subadditivity their outer measures sum to at least $\mu^{*}(A\cap(E\cup F))$, while the fourth set is $A\setminus(E\cup F)$. Hence $\mu^{*}(A)\ge\mu^{*}(A\cap(E\cup F))+\mu^{*}(A\setminus(E\cup F))$, so $E\cup F\in\mathcal{M}$. With Step 1, $\mathcal{M}$ is closed under finite unions, finite intersections, and differences. Moreover, if $E,F\in\mathcal{M}$ are disjoint, splitting $A\cap(E\cup F)$ by $E$ gives

$$\mu^{*}(A\cap(E\cup F))=\mu^{*}(A\cap E)+\mu^{*}(A\cap F)\qquad\text{for every }A\subseteq X,$$

and by induction the analogous identity holds for finitely many pairwise disjoint members of $\mathcal{M}$.

Step 3 (countable unions and countable additivity). Let $(E_m)_{m\in\mathbb{N}}$ be a sequence in $\mathcal{M}$ and $E=\bigcup_m E_m$. Replacing $E_m$ by $E_m\setminus\bigcup_{l<m}E_l$ — members of $\mathcal{M}$ by Step 2 with the same union — we may assume the $E_m$ are pairwise disjoint. Fix $A\subseteq X$ and $k\in\mathbb{N}$. Splitting $A$ by $F_k=\bigcup_{m\le k}E_m\in\mathcal{M}$ and using the finite additivity identity of Step 2 and monotonicity ($A\setminus F_k\supseteq A\setminus E$):

$$\mu^{*}(A)=\mu^{*}(A\cap F_k)+\mu^{*}(A\setminus F_k)\ \ge\ \sum_{m\le k}\mu^{*}(A\cap E_m)+\mu^{*}(A\setminus E).$$

Letting $k\to\infty$ (the partial sums are nondecreasing; sums in $[0,\infty]$ as in Measure, Measure Space, and Probability Measure) and then applying countable subadditivity to $A\cap E=\bigcup_m(A\cap E_m)$:

$$\mu^{*}(A)\ \ge\ \sum_{m}\mu^{*}(A\cap E_m)+\mu^{*}(A\setminus E)\ \ge\ \mu^{*}(A\cap E)+\mu^{*}(A\setminus E)\ \ge\ \mu^{*}(A),$$

the last inequality again by subadditivity. Hence all inequalities are equalities: $E\in\mathcal{M}$, and $\mathcal{M}$ satisfies the three properties of a $\sigma$-algebra, proving claim 1.

Taking $A=E$ in the displayed chain of equalities gives

$$\mu^{*}(E)=\sum_{m}\mu^{*}(E_m)$$

for every sequence of pairwise disjoint members of $\mathcal{M}$ with union $E$; together with $\mu^{*}(\varnothing)=0$, the restriction of $\mu^{*}$ to $\mathcal{M}$ is a measure, proving claim 2. $\blacksquare$