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.

Now, suppose $A,B\in {\mathcal {M}}$ and $A,B\in M$ are disjoint. Then

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

so $\mu ^{*}$ is finitely additive.

Next, we show ${\mathcal {M}}$ is closed under countable disjoint unions. Given a disjoint sequence of sets $\{B_{i}\}_{i=1}^{\infty }\subseteq {\mathcal {M}}$, for all $A\subseteq X$, by countable subadditivity,

$\mu ^{*}(A)\leq \sum _{i=1}^{\infty }\mu ^{*}(A\cup B_{i})+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c}).$

It remains to show the other inequality direction. Since ${\mathcal {M}}$ is closed under finite unions, $\cup _{i=1}^{\infty }B_{i}\in {\mathcal {M}}$. By using the definition of $\mu$-measurability,

$\mu ^{*}(A)=\mu ^{*}(A\cup (\cup _{i=1}^{\infty }B_{i}))+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c})$

and by monotonicity,

$=\sum _{i=1}^{n}\mu ^{*}(A\cap B_{i})+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c}).$

Since this holds for all $n$,

$\mu ^{*}(A)\geq \sum _{i=1}^{\infty }\mu ^{*}(A\cap B_{i})+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c})$

which proves

$\mu ^{*}(A)=\sum _{i=1}^{\infty }\mu ^{*}(A\cap B_{i})+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c})$

$\geq \mu ^{*}(\cup _{i=1}^{\infty }(A\cap B_{i}))+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c})$

$=\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i}))+\mu ^{*}(A\cap (\cup _{i=1}^{\infty }B_{i})^{c}).$

Similarly as before, the other inequality direction is proved by monotonicity, so we conclude equality and we have that $\cup _{i=1}^{\infty }B_{i}\in {\mathcal {M}}$.

The only thing that remains to be shown is closure under countable unions. Consider a sequence of sets $\{C_{i}\}_{i=1}^{\infty }$ not necessarily disjoint. Define

$D_{1}=C_{1},D_{2}=C_{2}\ C_{1},....,D_{n}=C_{n}\ (\cup _{i=1}^{n-1}C_{i}).$

Their unions are equal and $\cup _{i=1}^{infty}D_{i}$ is disjoint, hence contained in ${\mathcal {M}}$.

This finishes all parts to the proof.

$\square$