capacity generated by a measure

Any finite measure (http://planetmath.org/SigmaFinite) can be extended to a set function on the power set of the underlying space. As the following result states, this will be a Choquet capacity.

Theorem.

Let $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{,}\mu \mathrm{)}$ be a finite measure space. Then,

${\mu }^{*}:𝒫(X)\to {ℝ}_{+},$

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

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

Note that, as well as being a capacity, ${\mu }^{*}$ is also an outer measure (see here (http://planetmath.org/ConstructionOfOuterMeasures)), which does not require the finiteness of $\mu$.Clearly, ${\mu }^{*}(A)=\mu (A)$ for all $A\in ℱ$, so ${\mu }^{*}$ is an extension of $\mu$ to the power set of $X$, and is referred to as the outer measure generated by $\mu$.

Recall that a subset $S\subseteq X$ is in the completion of $ℱ$ with respect to $\mu$ if and only if there are sets $A,B\in ℱ$ with $A\subseteq S\subseteq B$ and $\mu (B\setminus A)=0$ which, by the above theorem, is equivalent to the capacitability of $S$.