Corollary 2

— If X is metrizable, then every Souslin subset A of X (GT, IX, §6, No. 2) is μ-measurable for every measure μ on X.

By virtue of Prop. 3, it suffices to verify that every relatively compact Souslin set A is μ-integrable. Now, such a set A is capacitable for $|\mu|^*$ (GT, IX, §6, No. 9, Th. 5). Therefore, for every $\varepsilon > 0$ there exists a compact subset K of A such that $|\mu|^*(A) \leq |\mu|^*(K) + \varepsilon = |\mu|(K) + \varepsilon$. Let U be a relatively compact open set in X containing A such that

Then $|\mu|^*(U - K) = |\mu|(U) - |\mu|(K) \leq 2\varepsilon$, therefore $|\mu|^*(A - K) \leq 2\varepsilon$, which proves that A is μ-integrable (§4, No. 6, Cor. 1 of Th. 4).