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Let $(E,\mathcal{X})$ be a topological space and denote by $\mathcal{F}$ its collection of Borel subsets referred to $\mathcal{X}$. Now let $\mathcal{P}$ be the set of all probabilities on $(E,\mathcal{F})$ and suppose the subset $\mathcal{M}\subset\mathcal{P}$ is tight.

My question is whether $\mathcal{M}$ is relatively compact with respect to weak-${\ast}$ convergence, that is to say, for any sequence $\{P_n\}\subset\mathcal{M}$, we can find a convergent subsequence $\{P_{n_k}\}$, i.e. there exists some $P\in\mathcal{P}$ such that for every $\mathcal{X}$-continuous and bounded function $f$ one has

$$\lim_{k\to\infty}\int_{E} fdP_{n_k}=\int_{E} fdP$$

Thx for the reply!

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  • $\begingroup$ In very general and reasonable assumptions regarding the topology (Polish space is probably more than enough) this is the contents of Prokhorov's theorem. $\endgroup$
    – Asaf
    Sep 21, 2014 at 19:29
  • $\begingroup$ "Polish" means separable and complete, but not necessarily a metric space I suppose $\endgroup$
    – CodeGolf
    Sep 21, 2014 at 19:38
  • $\begingroup$ In fact according to Chapter I of Billingsley <<Convergence of probability>>, $E$ is alway assumed to be a metric space. $\endgroup$
    – CodeGolf
    Sep 21, 2014 at 19:45
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    $\begingroup$ You should carefully distinguish (relative) compactness and sequential compactness. $\endgroup$
    – Jochen Wengenroth
    Sep 22, 2014 at 6:54
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    $\begingroup$ Bogachev's Measure Theory volume 2 is a good source for results like this in their most general form, and Theorem 8.6.7 addresses this question for completely regular spaces. $\endgroup$
    – Dan
    Sep 23, 2014 at 12:53

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