Proposition 15

Let $\varphi$ be a continuous mapping of a Hausdorff space K into a Hausdorff space X, and let f be a capacity on X. If for each subset A of K we put $g(A) = f(\varphi(A))$, then g is a capacity on K; moreover, if A is capacitable with respect to g, then $\varphi(A)$ is capacitable with respect to f.

It is clear that $g$ satisfies axioms $(\mathrm{CA}_I)$ and $(\mathrm{CA}_{II})$. On the other hand, let $(C_n)$ be a decreasing sequence of compact subsets of $K$, and let $C = \bigcap_n C_n$; the sets $\varphi(C)$ are compact, and their intersection is $\varphi(C)$; for this intersection certainly contains $\varphi(C)$, and if $x \in \varphi(C_n)$ for all $n$, then the sets $\overline{\varphi^1}(x) \cap C_n$ form a decreasing sequence of non-empty compact subsets of $K$, and therefore their intersection is not empty. We have therefore $f(\varphi(C)) = \inf_n f(\varphi(C_n))$, that is $g(C) = \inf_n g(C_n)$; thus $g$ satisfies axiom $(\mathrm{CA}_{III})$ and is therefore a capacity on $K$.

Now let $A$ be a subset of $K$ which is capacitable with respect to $g$; if $a < f(\varphi(A)) = g(A)$, then there is a compact set $C \subset A$ such that $g(C) \geq a$; thus $\varphi(C)$ is a compact set contained in $\varphi(A)$, and $f(\varphi(C)) \geq a$. This shows that $\varphi(A)$ is capacitable with respect to $f$, and completes the proof of Proposition 15 and hence of Theorem 5.