# proof of capacity generated by a measure

For a finite measure space $(X,\mathcal{F},\mu)$, define

 $\displaystyle\mu^{*}\colon\mathcal{P}(X)\to\mathbb{R}_{+},$ $\displaystyle\mu^{*}(S)=\inf\left\{\mu(A)\colon A\in\mathcal{F},\ A\supseteq S% \right\}.$

We show that $\mu^{*}$ is an $\mathcal{F}$-capacity and that a subset $S\subseteq X$ is $(\mathcal{F},\mu^{*})$-capacitable (http://planetmath.org/ChoquetCapacity) if and only if it is in the completion (http://planetmath.org/CompleteMeasure) of $\mathcal{F}$ with respect to $\mu$.

Note, first of all, that $\mu^{*}(S)=\mu(S)$ for any $S\in\mathcal{F}$. That $\mu^{*}$ is increasing follows directly from the definition. If $A_{n}\in\mathcal{F}$ is a decreasing sequence of sets then $A=\bigcap_{n}A_{n}$ is also in $\mathcal{F}$ and, by continuity from above (http://planetmath.org/PropertiesForMeasure) for measures,

 $\mu^{*}(A_{n})=\mu(A_{n})\rightarrow\mu(A)=\mu^{*}(A)$

as $n\rightarrow\infty$.

Now suppose that $S_{n}$ is an increasing sequence of subsets of $X$ and set $S=\bigcup_{n}S_{n}$. Then, $\mu^{*}(S_{n})\leq\mu^{*}(S)$ for each $n$ and, hence, $\lim_{n\rightarrow\infty}\mu^{*}(S_{n})\leq\mu^{*}(S)$.

To prove the reverse inequality, choose any $\epsilon>0$ and sequence $A_{n}\in\mathcal{F}$ with $S_{n}\subseteq A_{n}$ and $\mu(A_{n})\leq\mu^{*}(S_{n})+2^{-n}\epsilon$. Then, $A_{m}\cap A_{n}\supseteq S_{n}$ whenever $m\geq n$ and, therefore,

 $\mu(A_{n}\setminus A_{m})=\mu(A_{n})-\mu(A_{m}\cap A_{n})\leq\mu(A_{n})-\mu^{*% }(S_{n})\leq 2^{-n}\epsilon.$

Additivity of $\mu$ then gives

 $\mu\left(\bigcup_{m\leq n}A_{m}\right)\leq\mu(A_{n})+\sum_{m=1}^{n-1}\mu(A_{m}% \setminus A_{n})\leq\mu^{*}(S_{n})+\epsilon.$

So, by continuity from below for measures,

 $\mu^{*}(S)\leq\mu\left(\bigcup_{n}A_{n}\right)=\lim_{n\rightarrow\infty}\mu% \left(\bigcup_{m\leq n}A_{m}\right)\leq\lim_{n\rightarrow\infty}\mu^{*}(S_{n})% +\epsilon.$

Choosing $\epsilon$ arbitrarily small shows that $\mu^{*}(S_{n})\rightarrow\mu^{*}(S)$ and, therefore, $\mu^{*}$ is indeed an $\mathcal{F}$-capacity.

Now suppose that $S$ is in the completion of $\mathcal{F}$ with respect to $\mu$, so that there exists $A,B\in\mathcal{F}$ with $A\subseteq S\subseteq B$ and $\mu(B\setminus A)=0$. Then,

 $\mu^{*}(A)=\mu(A)=\mu(B)\geq\mu^{*}(S)$

and $S$ is indeed $(\mathcal{F},\mu^{*})$-capacitable. Conversely, let $S$ be $(\mathcal{F},\mu^{*})$-capacitable. Then, there exists $A_{n},B_{n}\in\mathcal{F}$ such that $A_{n}\subseteq S\subseteq B_{n}$ and

 $\mu(A_{n})\geq\mu^{*}(S)-1/n,\ \mu(B_{n})\leq\mu^{*}(S)+1/n.$

Setting $A=\bigcup_{n}A_{n}$ and $B=\bigcap_{n}B_{n}$ gives $A\subseteq S\subseteq B$ and

 $\mu(A)\geq\mu^{*}(S)\geq\mu(B).$

So $\mu(B\setminus A)=\mu(B)-\mu(A)=0$, as required.

Title proof of capacity generated by a measure ProofOfCapacityGeneratedByAMeasure 2013-03-22 18:47:55 2013-03-22 18:47:55 gel (22282) gel (22282) 5 gel (22282) Proof msc 28A12 msc 28A05