proof of capacity generated by a measure

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

${\mu }^{*}:𝒫(X)\to {ℝ}_{+},$

${\mu }^{*}(S)=inf\{\mu (A):A\in ℱ,A\supseteq S\}.$

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

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

$${\mu }^{*}(A_n)=\mu (A_n)\to \mu (A)={\mu }^{*}(A)$$

as $n\to \mathrm{\infty }$.

Now suppose that $S_n$ is an increasing sequence of subsets of $X$ and set $S={\bigcup }_n{S}_n$. Then, ${\mu }^{*}(S_n)\le {\mu }^{*}(S)$ for each $n$ and, hence, ${lim}_{n\to \mathrm{\infty }}{\mu }^{*}(S_n)\le {\mu }^{*}(S)$.

To prove the reverse inequality, choose any $ϵ>0$ and sequence $A_n\in ℱ$ with $S_n\subseteq A_n$ and $\mu (A_n)\le {\mu }^{*}(S_n)+2^{-n}ϵ$. Then, $A_m\cap A_n\supseteq S_n$ whenever $m\ge n$ and, therefore,

$$\mu (A_n\setminus A_m)=\mu (A_n)-\mu (A_m\cap A_n)\le \mu (A_n)-{\mu }^{*}(S_n)\le 2^{-n}ϵ.$$

Additivity of $\mu$ then gives

$$\mu \left(\bigcup _{m\le n}{A}_m\right)\le \mu (A_n)+\sum _{m=1}^{n-1}\mu (A_m\setminus A_n)\le {\mu }^{*}(S_n)+ϵ.$$

So, by continuity from below for measures,

$${\mu }^{*}(S)\le \mu \left(\bigcup _n{A}_n\right)=\underset{n\to \mathrm{\infty }}{lim}\mu \left(\bigcup _{m\le n}{A}_m\right)\le \underset{n\to \mathrm{\infty }}{lim}{\mu }^{*}(S_n)+ϵ.$$

Choosing $ϵ$ arbitrarily small shows that ${\mu }^{*}(S_n)\to {\mu }^{*}(S)$ and, therefore, ${\mu }^{*}$ is indeed an $ℱ$-capacity.

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

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

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

$$\mu (A_n)\ge {\mu }^{*}(S)-1/n,\mu (B_n)\le {\mu }^{*}(S)+1/n.$$

Setting $A={\bigcup }_n{A}_n$ and $B={\bigcap }_n{B}_n$ gives $A\subseteq S\subseteq B$ and

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

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

Title	proof of capacity generated by a measure
Canonical name	ProofOfCapacityGeneratedByAMeasure
Date of creation	2013-03-22 18:47:55
Last modified on	2013-03-22 18:47:55
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 28A12
Classification	msc 28A05