Theorem 5.5.4 (Carathéodory Construction).label Let $X$ be a set, $\mu^{*}:2^{X}\to [0,\infty]$ be an outer measure, and $\Si\suf 2^{X}$ be the collection of all $\mu^{*}$-measurable sets, then:
- (1)
For any $F\suf X$ and $\curl{E_n}_{n\in\N^+}\suf\Si$ pairwise disjoint
\begin{align*}\mu^{*}\parens{F\cap \dcups{}{n\in\N^+}E_n}=\sums{}{n\in\N^+}\mu^{*}(F\cap E_{n})\end{align*} - (2)
$\Si$ is a $\s$-algebra.
- (3)
$(X,\Si,\mu^{*}|_{\Si})$ is a complete measure space.
Proof.
- (1)
Let $N\in\N^{+}$ then
\begin{align*}\mu^{*}\parens{F\cap \dcups{}{n\in\N^+}E_n}\geq \mu^{*}\parens{F \cap\dcups{}{n\leq N}E_n}=\sums{}{n\leq N}\mu^{*}(F\cap E_{n})\end{align*}Since this holds for all $N\in\N^{+}$,
\begin{align*}\mu^{*}\parens{F\cap \dcups{}{n\in\N^+}E_n}\geq\sums{}{n\in\N^+}\mu^{*}(F\cap E_{n})\end{align*}By (OM2) we are done.
- (2)
Firstly, $\emp,X\in\cali{M}$ since $\mu^{*}(\emp)=0$. $\cali{M}$ is closed under complements since the definition of $\mu*$-measurability is symmetric. Let $A,B\in\cali{M},F\suf X$ then
\begin{align*}\mu^{*}(F)&=\mu^{*}(F\cap A)+\mu^{*}(F\del A)\\&=\mu^{*}(F\cap A\cap B)+\mu^{*}(F\cap A\del B)+\mu^{*}(F\del A\cap B)+\mu^{*}(F\del A\del B)\\&=\mu^{*}(F\cap B)+\mu^{*}(F\cap A\del B)+\mu^{*}(F\del (A\cup B))\\&=\mu^{*}(F\cap (A\cup B))+\mu^{*}(F\del (A\cup B))\end{align*}[1] so $\cali{M}$ is an algebra. From (1) we have $\cali{M}$ is closed under countable disjoint unions so $\cali{M}$ is a $\s$-algebra by Lemma 4.1.4.
- (3)
By (1), $\mu^{*}|_{\cali{M}}$ is a measure. Let $N\suf X$ with $\mu^{*}(N)=0$ and $F\suf X$ then
\begin{align*}\mu^{*}(F)\leq \mu^{*}(F\cap N)+\mu^{*}(F\cap N^{c})\leq \mu^{*}(N)+\mu^{*}(F)=\mu^{*}(F)\end{align*}hence $N\in\cali{M}$. Therefore $\mu^{*}|_{\cali{M}}$ is a complete measure.
$\square$
- both $\leq$ direction of last equality is subadditivity on $A\cup B$keyboard_return
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