# proof of Choquet’s capacitability theorem

Let $(X,\mathcal{F})$ be a paved space such that $\mathcal{F}$ is closed under finite unions and finite intersections, and let $I$ be an $\mathcal{F}$-capacity (http://planetmath.org/ChoquetCapacity). We prove the capacitability theorem, which states that all $\mathcal{F}$-analytic (http://planetmath.org/AnalyticSet2) sets are $(\mathcal{F},I)$-capacitable (http://planetmath.org/ChoquetCapacity). The idea is to deduce it from the following special case.

###### Lemma.

With the above notation, every set in $\mathcal{F}_{\sigma\delta}$ is $(\mathcal{F},I)$-capacitable.

Recall that $\mathcal{F}_{\sigma\delta}$ is the collection of countable intersections of countable unions in $\mathcal{F}$ and, since countable unions and intersections of analytic sets are analytic, all such sets are analytic. According to the capacitability theorem they should then be capacitable, and the lemma is indeed a special case.

Supposing that the lemma is true, then the capacitability theorem can be deduced as follows. For an $\mathcal{F}$-analytic set $A\subseteq X$, there is a compact paved space (http://planetmath.org/PavedSpace) $(K,\mathcal{K})$ and $S\in(\mathcal{F}\times\mathcal{K})_{\sigma\delta}$ such that $A=\pi_{X}(S)$, where $\pi_{X}$ is the projection map from $X\times K$ to $X$. Letting $\mathcal{G}$ be the closure under finite unions and finite intersections of the paving $\mathcal{F}\times\mathcal{K}$, then the composition $I\circ\pi_{X}$ is a $\mathcal{G}$-capacity, and the projection of any $(\mathcal{G},I\circ\pi_{X})$-capacitable set onto $X$ is itself $(\mathcal{F},I)$-capacitable (see extending a capacity to a Cartesian product (http://planetmath.org/ExtendingACapacityToACartesianProduct)). In particular, $S\in\mathcal{G}_{\sigma\delta}$ so, by the lemma, is $(\mathcal{G},I\circ\pi_{X})$-capacitable. Therefore, $A=\pi_{X}(S)$ is $(\mathcal{F},I)$-capacitable. It only remains to prove the lemma.

###### Proof of lemma.

If $S\in\mathcal{F}_{\sigma\delta}$ then there exists $S_{m,n}\in\mathcal{F}$ such that

 $S=\bigcap_{n}\bigcup_{m}S_{m,n}.$

For any positive integers $m_{1},m_{2},\ldots,m_{k}$ let us write

 $S(m_{1},m_{2},\ldots,m_{k})\equiv\left(\bigcap_{n\leq k}\bigcup_{m\leq m_{n}}S% _{m,n}\right)\bigcap\left(\bigcap_{n>k}\bigcup_{m}S_{m,n}\right).$

In particular, $S()=S$ and, $I(S())=I(S)$. For any $\epsilon>0$ and $k\in\mathbb{N}$ suppose that we have chosen positive integers $m_{1},\dots,m_{k-1}$ such that $I(S(m_{1},\ldots,m_{k-1}))>I(S)-\epsilon$. Since $I$ is a capacity and $S(m_{1},\ldots,m_{k})$ increases to $S(m_{1},\ldots,m_{k-1})$ as $m_{k}$ increases to infinity,

 $I(S(m_{1},\ldots,m_{k}))\rightarrow I(S(m_{1},\ldots,m_{k-1}))$

as $m_{k}$ tends to infinity. So, by choosing $m_{k}$ large enough, we have

 $I(S(m_{1},\ldots,m_{k}))>I(S)-\epsilon.$

Then, by induction, we can find an infinite sequence $m_{1},m_{2},\ldots$ such that this inequality holds for every $k$. Setting

 $\displaystyle A_{k}\equiv\bigcap_{n\leq k}\bigcup_{m\leq m_{n}}S_{m,n}\in% \mathcal{F},$ $\displaystyle A\equiv\bigcap_{n}\bigcup_{m\leq m_{n}}S_{m,n}=\bigcap_{k}A_{k}% \in\mathcal{F}_{\delta},$

then $A\subseteq S$. Furthermore, $A_{k}$ contains $S(m_{1},\ldots,m_{k})$ and decreases to $A$ as $k$ tends to infinity. As $I$ is an $\mathcal{F}$-capacity this gives

 $I(A)=\lim_{k\rightarrow\infty}I(A_{k})\geq\lim_{k\rightarrow\infty}I(S(m_{1},% \ldots,m_{k}))\geq I(S)-\epsilon.$

So $S$ is $(\mathcal{F},I)$-capacitable, as required. ∎

Title proof of Choquet’s capacitability theorem ProofOfChoquetsCapacitabilityTheorem 2013-03-22 18:47:51 2013-03-22 18:47:51 gel (22282) gel (22282) 4 gel (22282) Proof msc 28A05 msc 28A12