Theorem 4.1.2362C826EB1B48

Proof. Capacitability of $f(\widetilde A)$. Replacing $K_{ik}$ by $K_{i1}\cup \cdots \cup K_{ik}$, we may assume that $K_{ik}$ increases with $k$. Put $A=f(\widetilde A)$ and fix $\varepsilon >0$. We choose integers $h_1,h_2,\ldots $ inductively. Put $\widetilde A_0=\widetilde A$, and suppose $\widetilde A_{p-1}$ has been defined. Since $\widetilde A_{p-1}\subset \bigcup _ kK_{pk}$, the sets $\widetilde A_{p-1}\cap K_{ph}$ increase with $h$ and have union $\widetilde A_{p-1}$, so their images under $f$ increase to $f(\widetilde A_{p-1})\subset L$. By part (3) of theorem 4.1.19, we can choose $h_ p$ such that $\widetilde A_ p=\widetilde A_{p-1}\cap K_{p,h_ p}$ satisfies

Hence $C_ B^*(f(\widetilde A_ p))>C_ B^*(A)-\varepsilon $ for every $p$.

The sets $L_ p=K_{1,h_1}\cap \cdots \cap K_{p,h_ p}$ are compact, because compact subsets of the Hausdorff space $X$ are closed; they decrease and contain $\widetilde A_ p$, so the compact sets $f(L_ p)\subset L$ decrease and, by part (1) of theorem 4.1.19 and the monotonicity of $C_ B^*$, satisfy $C_ B(f(L_ p))=C_ B^*(f(L_ p))\geq C_ B^*(f(\widetilde A_ p))>C_ B^*(A)-\varepsilon $. Moreover $\bigcap _ pf(L_ p)=f(\bigcap _ pL_ p)$: if $y\in f(L_ p)$ for every $p$, the sets $f^{-1}(y)\cap L_ p$ form a decreasing sequence of nonempty compact sets, since $f^{-1}(y)$ is closed, and hence have a common point. Since $\bigcap _ pL_ p=\bigcap _ iK_{i,h_ i}\subset \widetilde A$, the compact set $\bigcap _ pf(L_ p)$ is contained in $A$, and by proposition 4.1.18 its capacity is $\lim _ pC_ B(f(L_ p))\geq C_ B^*(A)-\varepsilon $. As $\varepsilon $ is arbitrary, $A$ is capacitable.

Borel sets. If $L=\varnothing $ there is nothing to prove, so let $L\neq \varnothing $. Call a subset of $L$ analytic if it has the form $f(\widetilde A)$ considered above; this is the notion of descriptive set theory and is unrelated to complex analytic sets. In such a representation we may assume that $X$ is nonempty, since the empty set is the image of the empty subset of a one-point space under a constant map. We may also assume that $X=Y\times L$ for a nonempty compact Hausdorff space $Y$ and that $f$ is the projection onto $L$. Indeed, put $Y=X$. The map $g\colon X\to Y\times L$, $g(x)=(x,f(x))$, is continuous and injective, so $g(\widetilde A)=\bigcap _ i\bigcup _ kg(K_{ik})$ with compact sets $g(K_{ik})$, and the projection of $g(\widetilde A)$ onto $L$ is $f(\widetilde A)$.

Analytic sets are closed under countable intersections and countable unions. To see this, let $A_ j=\pi (\widetilde A_ j)$, $j\geq 1$, be analytic, with $\widetilde A_ j\subset Y_ j\times L$ as above and $\pi $ denoting projections onto $L$. The space $X=(\prod _ jY_ j)\times L$ is compact by lemma 2.7.2 and Hausdorff, as a product of Hausdorff spaces. Let $\widehat A_ j\subset X$ be the inverse image of $\widetilde A_ j$ under the continuous map $\pi _ j\colon (y,z)\mapsto (y_ j,z)$. If $\widetilde A_ j=\bigcap _ i\bigcup _ kK^{(j)}_{ik}$ with compact sets $K^{(j)}_{ik}\subset Y_ j\times L$, then $\widehat A_ j=\bigcap _ i\bigcup _ k\pi _ j^{-1}(K^{(j)}_{ik})$. Each $K^{(j)}_{ik}$ is closed, being a compact subset of a Hausdorff space, so $\pi _ j^{-1}(K^{(j)}_{ik})$ is a closed and hence compact subset of $X$. Thus each $\widehat A_ j$, and therefore also $\bigcap _ j\widehat A_ j$, is a countable intersection of countable unions of compact subsets of $X$.

A point $z\in L$ lies in $\bigcap _ jA_ j$ if and only if for every $j$ there is $y_ j\in Y_ j$ with $(y_ j,z)\in \widetilde A_ j$, that is, if and only if $(y,z)\in \bigcap _ j\widehat A_ j$ for some $y=(y_ j)_ j\in \prod _ jY_ j$. Hence the projection of $\bigcap _ j\widehat A_ j$ onto $L$ is $\bigcap _ jA_ j$, and countable intersections of analytic sets are analytic. Since every $Y_ i$ is nonempty, the same reasoning shows that the projection of each $\widehat A_ j$ onto $L$ is $A_ j$.

For unions, write $\widehat A_ j=\bigcap _ n\bigcup _ kH_{jnk}$ with compact $H_{jnk}\subset X$, and consider, in the product of $X$ with the one-point compactification $\{ 1,2,\ldots \} \cup \{ \infty \} $ of the positive integers (a compact Hausdorff space, homeomorphic to $\{ 1/j:j\geq 1\} \cup \{ 0\} \subset \mathbb R$), the set

A point $(x,j)$ belongs to it if and only if $j\neq \infty $ and $x\in \widehat A_ j$, so its projection onto $L$ is $\bigcup _ jA_ j$. Each $H_{jnk}\times \{ j\} $ is compact, so countable unions of analytic sets are analytic.

Closed subsets of the compact metric space $L$ are compact, hence analytic, and an open set $U\subset L$ is the union of the compact sets $\{ z\in L:\operatorname {dist}(z,L\setminus U)\geq 1/n\} $, hence analytic. The family of sets $E\subset L$ such that both $E$ and $L\setminus E$ are analytic is closed under complements, and under countable unions, because the complement of a countable union is the countable intersection of the complements. It is therefore a $\sigma $-algebra containing all open subsets of $L$, so every Borel subset of $L$ is analytic and hence capacitable.

Increasing compact sets. Let $A\subset L$ be Borel. Since $A$ is capacitable, there are compact sets $K_ j\subset A$ with $C_ B(K_ j)\to C_ B^*(A)$. The compact sets $K_1\cup \cdots \cup K_ j\subset A$ increase, and by monotonicity their capacities increase to $C_ B^*(A)$. $\square$