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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?

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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.

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  • $\begingroup$ Throughout your answer you speak of Radon measures. What allows us to consider only Radon measures? $\endgroup$ Nov 7, 2023 at 10:40
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    $\begingroup$ @FlorianEnte A bounded complex measure on the Borel $\sigma$-algebra of a Polish space is always Radon. $\endgroup$ Nov 7, 2023 at 11:08

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