Proposition 3

— Every Souslin space (in particular, every Polish or Lusin space) is strongly Radon.

Let $T$ be a Souslin space; since $T$ is a Lindelöf space (TG, IX, Appendix I, Cor. of Prop. 1),(2) it suffices to show that $T$ is Radon (Prop. 1). Let $I$ be a function defined on $\mathscr{B}(T)$, positive, countably additive and bounded. We extend $I$ to $\mathscr{P}(T)$ by setting, for every subset $A$ of $T$,

Let us show that this extension is a capacity on $T$ (GT, IX, §6, No. 9). It is clear that the relation $A \subset A'$ implies $I(A) \leq I(A')$. Let $(A_n)$ be an increasing sequence of subsets of $T$, and let $A = \bigcup_n A_n$. The set of Borel sets that contain $A_n$ being stable for countable intersections, there exists for each $n$ a Borel set $B_n$ such that $A_n \subset B_n$ and $I(A_n) = I(B_n)$ (cf. the proof of Prop. 2). Set $C_n = \bigcap_{p \geq n} B_p$; $C_n$ is Borel, and $A_n \subset C_n \subset B_n$, therefore $I(A_n) = I(C_n)$. On the other hand, the sequence $(C_n)$ is increasing. Let $C = \bigcup_n C_n$: the relation $A \subset C$ implies that

(2) Every Souslin space has a countable base for open sets (GT, IX, §6, No. 2, Prop. 4), hence is Lindelöf (GT, I, §9, Exer. 14).

whence the equality $I(A) = \lim_n I(A_n)$ is immediate. Consequently, I is a capacity.

If $(H_n)$ is a decreasing sequence of closed sets in T, obviously $I\left( \bigcap_n H_n \right) = \inf_n I(H_n)$. It follows that every Souslin subset F of T is capacitable for I (TG, IX, §6, No. 10, Prop. 15). In particular, every Borel set A of T is capacitable (loc. cit., §6, No. 3, Prop. 10).(3) In other words,

where K runs over the set of compact sets contained in A; we have proved that I is inner regular.