Equip the space of Borel probability measures on a fixed closed subset X of the s-dimensional Euclidean space with the topology induced by weak convergence of probability measures. In this setting, does compact equal bounded and closed (with respect to the Prokhorov metric)? If not, is there a simple counterexample?
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$\begingroup$ If I'm using the right definition, the Kantorovich (aka transportation or Wasserstein) metric doesn't even induce the weak topology: convergence in the metric is equivalent to weak convergence plus convergence of first moments. So any sequence of measures that converges weakly but has unbounded first moments should give a counterexample. $\endgroup$– Nate EldredgeFeb 1, 2016 at 22:59
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$\begingroup$ Thank you for the hint, it was supposed to be the Prokhorov metric. $\endgroup$– Tobias LaslopFeb 1, 2016 at 23:06
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$\begingroup$ You mean this Prokhorov metric? It's a bounded metric, so the "bounded" part is trivial, and certainly not every weakly closed set of probability measures is compact. Or have I got the wrong definition again? $\endgroup$– Nate EldredgeFeb 1, 2016 at 23:08
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$\begingroup$ If $X$ is compact then the set of all Borel probability measures is weakly compact (and also weakly sequentially compact). This is a simple version of Prokhorov's theorem (or, in functional analytic terms, Alaoglu's theorem, functional analysts call the weak topology weak* topology). Of course, closed subsets of compact sets are again compact. $\endgroup$– Jochen WengenrothFeb 2, 2016 at 7:21
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