# Caratheodory's Theorem

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Let ${\displaystyle X}$ be a set and ${\displaystyle \mu ^{*}:2^{X}\to [0,\infty ]}$ an outer measure on ${\displaystyle X}$, we call a subset ${\displaystyle A\subseteq X}$ to be ${\displaystyle \mu ^{*}}$-measurable if

${\displaystyle \mu ^{*}(E)=\mu ^{*}(E\cap A)+\mu ^{*}(E\cap A^{C})}$

for all ${\displaystyle E\in 2^{X}}$. This should remind the reader of a full measure since this is similar to disjoint additivity. Thus, the natural question arises of the significance of these ${\displaystyle \mu ^{*}-}$measurable sets. Caratheodory's Theorem provides and answer by proving that if ${\displaystyle {\mathcal {M}}}$ is the collection of ${\displaystyle \mu ^{*}-}$measurable sets, then ${\displaystyle (X,{\mathcal {M}},\mu ^{*})}$ is indeed a Measure Space.

## Statement

Consider an outer measure ${\displaystyle \mu ^{*}}$ on ${\displaystyle X}$.

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

Then ${\displaystyle {\mathcal {M}}}$ is a ${\displaystyle \sigma }$-algebra and ${\displaystyle \mu ^{*}}$ is a measure on ${\displaystyle {\mathcal {M}}}$[1].

## Proof

First, observe that ${\displaystyle {\mathcal {M}}}$ is closed under complements due to symmetry in the meaning of ${\displaystyle \mu ^{*}}$-measurability. Now, we show if ${\displaystyle A,B}$ then ${\displaystyle A\cup B\in {\mathcal {M}}}$[2] to conlcude that ${\displaystyle {\mathcal {M}}}$ is an algebra.

Suppose ${\displaystyle E\subseteq X}$. Then

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

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

and by subadditivity

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

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

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

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

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

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

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

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

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

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

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

and by monotonicity,

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

Since this holds for all ${\displaystyle n\in \mathbb {N} }$,

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

which proves

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

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

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

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

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

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

Their unions are equal and ${\displaystyle \cup _{i=1}^{\infty }D_{i}}$ is disjoint, hence contained in ${\displaystyle {\mathcal {M}}}$, proving it to be a ${\displaystyle \sigma -}$algebra.

This finishes all parts to the proof.

${\displaystyle \square }$

1. Craig, Katy. MATH 201A Lectures 6,7. UC Santa Barbara, Fall 2020.
2. Folland, Gerald B., Real Analysis: Modern Techniques and Their Applications, second edition, §1.4