Proposition 14

Let K be a compact space and let f be a capacity on K. If A is the intersection of a decreasing sequence $(A_n)$ of subsets of K, each of which is a countable union of closed sets, then A is capacitable.

It is enough to show that, for each $a < f(A)$, there is a closed set $C \subset A$ such that $f(C) \geq a$. Let us first show that there exists a sequence $(B_n)_{n \geq 1}$ of closed sets such that $B_n \subset A_n$ and such that, if we define a sequence $(C_n)$ inductively by the conditions $C_0 = A, C_n = C_{n-1} \cap B_n$ for $n \geq 1$, then $f(C_n) > a$ for each $n \geq 0$. Suppose the $B_i$ have been defined for $i < n$; by hypothesis we have $C_{n-1} \subset A \subset A_n$ and $f(C_{n-1}) > a$; since $A_n$ is the union of an increasing sequence of closed sets $D_j$, it follows from (CAII) that

Hence there is an index $j$ such that $f(C_{n-1} \cap D_j) > a$, and we may take $B_n = D_j$.

Now let $C = \bigcap_n C_n$. Since $A = \bigcap_n A_n$ and $B_n \subset A_n$, we have

the set C is therefore compact and contained in A.

Let $B'_n = B_1 \cap B_2 \cap \cdots \cap B_n$; $(B'_n)$ is a decreasing sequence of compact subsets of K; as $C_n \subset C_i \subset B_i$ for $i < n$ we also have $C_n \subset B'_n$. By (CAIII), $f(C) = \inf_n f(B'_n)$, and since $C \subset C_n \subset B'_n$ we also have $f(C) = \inf_n f(C_n) \geq a$. This completes the proof.