Caratheodory's Theorem

Statement

Consider an out measure $\mu$ on $X$. Define

${\mathcal {M}}=\{A\subseteq X:A\ {\text{is}}\ \mu -{\text{measurable}}\}$.

Then ${\mathcal {M}}$ is a $\sigma$-algebra and $\mu ^{*}$ is a measure on ${\mathcal {M}}$.

Proof

First, observe that ${\mathcal {M}}$ is closed under complements due to symmetry in the meaning of $\mu$-measurability. Now, we show if $A,B$ then $A\cup B\in {\mathcal {M}}$.

Suppose $E\subseteq X$. Then

$\mu ^{*}(E)=\mu ^{*}(E\cap A)=\mu ^{*}(E\cap A^{c})$

$=\mu ^{*}(E\cap A\cap B)+\mu ^{*}(E\cap A\cap B^{c})+\mu ^{*}(E\cap A^{c}\cap B)+\mu ^{*}(E\cap A^{c}\cap B^{c})$

and by subadditivity

$\geq \mu ^{*}(E\cap A^{c}\cap B^{c})+\mu ^{*}(E\cap (A\cup B))=\mu ^{*}(E\cap (A\cup B)^{c})+\mu ^{*}(E\cap (A\cup B))$

But certainly, since $E\subseteq (E\cap (A\cup B)^{c})\cup (E\cap (A\cup B)$ the inequality in the other direction also holds, and we conclude

$E=\mu ^{*}(E\cap (A\cup B)^{c})+\mu ^{*}(E\cap (A\cup B)$

hence $A\cup B\in {\mathcal {M}}$ and we have ${\mathcal {M}}$ is an algebra.