Theorem 1 (Prokhorov)

— Let T be a completely regular space, and let H be a subset of $\mathscr{M}^b(T)$ that satisfies Prokhorov’s condition; then H is relatively compact in $\mathscr{M}^b(T)$ for the tight topology.

We can suppose that T is a subspace of a compact space X; let i be the canonical injection of T into X. We can on the other hand suppose that H is closed in $\mathscr{M}^b(T)$, by Prop. 11. It will then suffice to show that every ultrafilter $\mathfrak{U}$ on H converges in $\mathscr{M}^b(T)$.

We shall begin with the case that $H \subset \mathscr{M}_+^b(T)$. The total masses of the measures $\mu \in H$ being bounded by hypothesis, $i(\mu)$ converges vaguely with respect to $\mathfrak{U}$, in $\mathscr{M}_+(X)$, to a measure $\nu \in \mathscr{M}_+(X)$ (Ch. III, §1, No. 9, Cor. 2 of Prop. 15); by Prop. 8 of No. 3, everything comes down to proving that $\nu$ is concentrated on T. Now, let $\varepsilon$ be a number $> 0$, and let $K_\varepsilon$ be a compact subset of T satisfying the formula (6). Since $X - K_\varepsilon$ is open in X, we have, by Prop. 6 of No. 3 applied in X, the inequalities

since $\varepsilon > 0$ is arbitrary, the theorem is established in this special case.

Let us pass to the general case; for every measure $\mu$ on T, set

since $\mu = a_1(\mu) - a_2(\mu) + ia_3(\mu) - ia_4(\mu)$, it will suffice to show that the mappings $a_j$ ($j = 1, 2, 3, 4$) converge tightly with respect to $\mathfrak{U}$. But the set $H_j$ of measures $a_j(\mu)$, where $\mu$ runs over H, satisfies Prokhorov’s condition by virtue of the relation $|a_j(\mu)| \leq |\mu|$, and is contained in $\mathscr{M}_+^b(T)$; it is therefore relatively compact in $\mathscr{M}_+^b(T)$ by the special case, and the theorem then follows at once.