proof of Choquet’s capacitability theorem

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

Lemma.

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

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

Proof of lemma.

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

$$S=\bigcap _n\bigcup _m{S}_{m,n}.$$

For any positive integers $m_1,m_2,\mathrm{\dots },m_k$ let us write

$$S(m_1,m_2,\mathrm{\dots },m_k)\equiv \left(\bigcap _{n\le k}\bigcup _{m\le m_n}{S}_{m,n}\right)\bigcap \left(\bigcap _{n>k}\bigcup _m{S}_{m,n}\right).$$

In particular, $S()=S$ and, $I(S())=I(S)$. For any $ϵ>0$ and $k\in \mathbb{N}$ suppose that we have chosen positive integers $m_1,\mathrm{\dots },m_{k-1}$ such that $I(S(m_1,\mathrm{\dots },m_{k-1}))>I(S)-ϵ$. Since $I$ is a capacity and $S(m_1,\mathrm{\dots },m_k)$ increases to $S(m_1,\mathrm{\dots },m_{k-1})$ as $m_k$ increases to infinity,

$$I(S(m_1,\mathrm{\dots },m_k))\to I(S(m_1,\mathrm{\dots },m_{k-1}))$$

as $m_k$ tends to infinity. So, by choosing $m_k$ large enough, we have

$$I(S(m_1,\mathrm{\dots },m_k))>I(S)-ϵ.$$

Then, by induction, we can find an infinite sequence $m_1,m_2,\mathrm{\dots }$ such that this inequality holds for every $k$.Setting

$A_k\equiv {\displaystyle \bigcap _{n\le k}}{\displaystyle \bigcup _{m\le m_n}}{S}_{m,n}\in ℱ,$

$A\equiv {\displaystyle \bigcap _n}{\displaystyle \bigcup _{m\le m_n}}{S}_{m,n}={\displaystyle \bigcap _k}{A}_k\in {ℱ}_{\delta },$

then $A\subseteq S$. Furthermore, $A_k$ contains $S(m_1,\mathrm{\dots },m_k)$ and decreases to $A$ as $k$ tends to infinity. As $I$ is an $ℱ$-capacity this gives

$$I(A)=\underset{k\to \mathrm{\infty }}{lim}I(A_k)\ge \underset{k\to \mathrm{\infty }}{lim}I(S(m_1,\mathrm{\dots },m_k))\ge I(S)-ϵ.$$

So $S$ is $(ℱ,I)$-capacitable, as required.∎