Proposition 9

— Let T be a Souslin space (GT, IX, §6, No. 2, Def. 2), X a Hausdorff space, $\pi$ a continuous mapping of T onto X, and $\nu$ a bounded measure on X. Then there exists a bounded measure $\mu$ on T such that $\pi(\mu) = \nu$.

The hypotheses obviously imply that X is a Souslin space.

Let us consider the set function $c : A \mapsto \nu^\bullet(\pi(A))$ on $\mathfrak{P}(T)$. The relation $A \subset B$ implies $c(A) \leq c(B)$; if $(A_n)$ is an increasing sequence of subsets of T, and if $A = \bigcup_{n \in \mathbf{N}} A_n$, then $c(A) = \sup_n c(A_n)$ from the fact that $\nu^\bullet$ is an encumbrance. Finally, let $A \subset T$ and let $\varepsilon$ be a number $> 0$; choose an open subset G of X containing $\pi(A)$, such that $\nu^\bullet(G) \leq \nu^\bullet(\pi(A)) + \varepsilon$ (§ 1, No. 9, Prop. 13); the open subset $H = \overline{\pi^{-1}(G)}$ of T contains A, and $c(H) \leq c(A) + \varepsilon$. The function c is therefore a right-continuous capacity on T (TG, IX, §6, No. 10, Def. 9)(1) and the theorem on capacitability (loc. cit., Th. 6) implies the equality $c(T) = \sup_K c(K)$, where K runs over the set of compact subsets of T. Prop. 8 then implies the existence of the desired measure $\mu$.