proof of Carathéodory’s lemma

A set $S\subseteq X$ is $\mu$-measurable if and only if

$$\mu (E)\ge \mu (E\cap S)+\mu (E\cap S^c)$$(1)

for every $E\subseteq X$.As this inequality is clearly satisfied if $S=\mathrm{\varnothing }$ and is unchanged when $S$ is replaced by $S^c$, then $𝒜$ contains the empty set and is closed under taking complements of sets.To show that $𝒜$ is a $\sigma$-algebra, it only remains to show that it is closed under taking countable unions of sets. Choose any sets $A,B\in 𝒜$ and $E\subseteq X$. Then,

The first two inequalities here follow from applying (1) with $A$ and then $B$ in place of $S$, and the third uses the subadditivity of $\mu$ together with $A\cup (A^c\cap B)=A\cup B$. So (1) is satisfied with $A\cup B$ in place of $S$, showing that $𝒜$ is closed under taking pairwise unions and is therefore an algebra of sets on $X$. If $A,B$ are disjoint sets in $𝒜$ then replacing $E$ by $E\cap (A\cup B)$ and $S$ by $A$ in (1) gives $\mu (E\cap (A\cup B))\ge \mu (E\cap A)+\mu (E\cap B)$. As the reverse inequality follows from subadditivity of $\mu$, this implies that

$$\mu (E\cap (A\cup B))=\mu (E\cap A)+\mu (E\cap B).$$

So, the map $A\mapsto \mu (E\cap A)$ is an additive set function on $𝒜$. In particular, taking $E=X$ shows that $\mu$ is additive on $𝒜$.

Now choose a sequence $A_i\in 𝒜$, and set $B_i\equiv {\bigcup }_{j=1}^i{A}_j$ which are in the algebra $𝒜$. To prove that $𝒜$ is a $\sigma$-algebra it needs to be shown that $A\equiv {\bigcup }_i{A}_i={\bigcup }_i{B}_i$ is itself in $𝒜$.First, as $B_i\in 𝒜$ and $A^c\subseteq B_i^c$,

$$\mu (E)\ge \mu (E\cap B_i)+\mu (E\cap B_i^c)\ge \mu (E\cap B_i)+\mu (E\cap A^c).$$

As $C_i\equiv B_i\setminus B_{i-1}$ are pairwise disjoint sets in $𝒜$ satisfying ${\bigcup }_{j=1}^i{C}_j=B_i$ the additivity of $C\mapsto \mu (E\cap C)$ on $𝒜$ gives

$$\mu (E)\ge \sum _{j=1}^i\mu (E\cap C_j)+\mu (E\cap A^c).$$

So, letting $i$ increase to infinity, the subadditivity of $\mu$ applied to ${\bigcup }_j(E\cap C_j)=E\cap A$ gives

$$\mu (E)\ge \sum _j\mu (E\cap C_j)+\mu (E\cap A^c)\ge \mu (E\cap A)+\mu (E\cap A^c).$$

This shows that $A$ is $\mu$-measurable and so $𝒜$ is a $\sigma$-algebra.

It only remains to show that the restriction of $\mu$ to $𝒜$ is a measure, for which it needs to be shown that $\mu$ is countably additive on $𝒜$. So, choose any pairwise disjoint sequence $A_i\in 𝒜$ and set $A={\bigcup }_i{A}_i$. The following inequality

$$\sum _{j=1}^i\mu (A_j)=\mu \left(\bigcup _{j=1}^i{A}_j\right)\le \mu (A)\le \sum _j\mu (A_j)$$

follows from the additivity of $\mu$ on $𝒜$, the requirement that $\mu$ is increasing and from the countable subadditivity of $\mu$. Letting $i$ increase to infinity gives $\mu (A)={\sum }_j\mu (A_j)$ and $\mu$ is indeed countably additive on $𝒜$.