proof of Carathéodory’s lemma

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

$\mu(E)\geq\mu(E\cap S)+\mu(E\cap S^{c})$

(1)

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

$\begin{split}\displaystyle\mu(E)&\displaystyle\geq\mu(E\cap A)+\mu(E\cap A^{c}% )\\ &\displaystyle\geq\mu(E\cap A)+\mu(E\cap A^{c}\cap B)+\mu(E\cap A^{c}\cap B^{c% })\\ &\displaystyle\geq\mu(E\cap(A\cup B))+\mu(E\cap A^{c}\cap B^{c})\end{split}$

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 $\mathcal{A}$ is closed under taking pairwise unions and is therefore an algebra of sets on $X$ . If $A, B$ are disjoint sets in $\mathcal{A}$ then replacing $E$ by $E\cap(A\cup B)$ and $S$ by $A$ in (1) gives $\mu(E\cap(A\cup B))\geq\mu(E\cap A)+\mu(E\cap B)$ . 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 $\mathcal{A}$ . In particular, taking $E=X$ shows that $\mu$ is additive on $\mathcal{A}$ .

Now choose a sequence $A_{i}\in\mathcal{A}$ , and set $B_{i}\equiv\bigcup_{j=1}^{i}A_{j}$ which are in the algebra $\mathcal{A}$ . To prove that $\mathcal{A}$ is a $\sigma$ -algebra it needs to be shown that $A\equiv\bigcup_{i}A_{i}=\bigcup_{i}B_{i}$ is itself in $\mathcal{A}$ . First, as $B_{i}\in\mathcal{A}$ and $A^{c}\subseteq B_{i}^{c}$ ,

$\mu(E)\geq\mu(E\cap B_{i})+\mu(E\cap B_{i}^{c})\geq\mu(E\cap B_{i})+\mu(E\cap A% ^{c}).$

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

$\mu(E)\geq\sum_{j=1}^{i}\mu(E\cap C_{j})+\mu(E\cap A^{c}).$

So, letting $i$ increase to infinity, the subadditivity of $\mu$ applied to $\bigcup_{j}(E\cap C_{j})=E\cap A$ gives

$\mu(E)\geq\sum_{j}\mu(E\cap C_{j})+\mu(E\cap A^{c})\geq\mu(E\cap A)+\mu(E\cap A% ^{c}).$

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

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

$\sum_{j=1}^{i}\mu(A_{j})=\mu\left(\bigcup_{j=1}^{i}A_{j}\right)\leq\mu(A)\leq% \sum_{j}\mu(A_{j})$

follows from the additivity of $\mu$ on $\mathcal{A}$ , 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 $\mathcal{A}$ .

Title	proof of Carathéodory’s lemma
Canonical name	ProofOfCaratheodorysLemma
Date of creation	2013-03-22 18:33:25
Last modified on	2013-03-22 18:33:25
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 28A12
Related topic	CaratheodorysLemma
Related topic	OuterMeasure