Corollary 2

If T is a completely regular Souslin (resp. Lusin) space, then the space $\mathscr{M}_+^b(T)$ is Souslin (resp. Lusin).

For, consider a Polish space P and a continuous mapping $f$ of P onto T (GT, IX, §6, No. 2, Def. 2). Let $\tilde{f}$ be the continuous mapping $\mu \mapsto f(\mu)$ of $\mathscr{M}_+^b(P)$ into $\mathscr{M}_+^b(T)$; the space $\mathscr{M}_+^b(P)$ is Polish by Prop. 10, and $\tilde{f}$ is surjective (§2, No. 4, Prop. 9); the space $\mathscr{M}_+^b(T)$ is therefore Souslin. Similarly, if T is Lusin, then $f$ may be assumed to be injective (GT, loc. cit., No. 4, Prop. 12); then $\tilde{f}$ is injective (§2, No. 4, Prop. 8), and so $\mathscr{M}_+^b(T)$ is Lusin (GT, loc. cit., No. 4, Prop. 12).

Let T be a completely regular Souslin space (recall that for this, it suffices that T be Souslin and regular (TG, App. 1, Cor. of Prop. 2)), and let H be a compact subset of $\mathscr{M}_+^b(T)$; then H is compact and Souslin, hence metrizable, for the tight topology (loc. cit., App. 1, Cor. 2 of Prop. 3).