# Prokhorov's Theorem, tightness implies convergence of a subsequence

I'm trying to apply Prokhorov's Theorem on the measure space $$M$$, defined as the space of non-negative, finite measures on $$[0,T] \times \mathbb R^d$$ with the weak topology in the following sense:

Given a tight sequence of probability measures $$P_n$$ on $$M$$, not elements of $$M$$ but rather probability measures over $$M$$, can I conclude that there is a weakly convergent subsequence? Do I need $$M$$ to be complete and separable?

Yes, any tight sequence $$(P_n)_{n\in\mathbb N}$$ of Radon probabilities on $$M$$ has a subsequence that converges weakly to Radon probability and there is no need for any additional assumption on $$M$$.
It can be shown that the space of nonnegative finite Radon measures on a Polish space is always a Polish space under the weak topology (cf. for example Prokhorov's original paper, thm. 1.11, or Bourbaki's Integration, chap. IX, §5, prop. 10). So under the weak topologies, your space $$M$$ is Polish and the space $$N$$ of nonnegative finite Radon measures on $$M$$ is also Polish and in particular is metrizable.
The metrizability of $$N$$ tells you that a tight sequence of elements of $$N$$ has a subsequence that converges weakly in $$N$$ if and only if it has an adherent point in $$N$$ for the weak topology, the existence of which is given by the relative compactness property given by Prokhorov's theorem.
• @FlorianEnte A bounded complex measure on the Borel $\sigma$-algebra of a Polish space is always Radon. Nov 7, 2023 at 11:08