Given a compact metrizable space $X$ endowed with its Borel sigma algebra, call $P$ the set of probability measures on the resulting measurable space. Endow $P$ with the topology of strong convergence i.e. a net $(m_{\alpha})$ converges to a point $m$ iff $m_{\alpha}(B)$ converges to $m(B)$ for any measurable set B. I would like to know whether P is compact and/or Polish. Of course we know that this is the case with weak convergence.
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No. Let $X=[0,1]$, $\mu_n=\delta_{1/n}$. Then $(\mu_n)$ has no strongly convergent subsequence.
(If $\mu_{n_j}\to\mu$ strongly then $\mu((0,1])=1$ but $\mu((1/n,1])=0$ for all $n$.)
answered Oct 2, 2019 at 12:08
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$\begingroup$ Thank you very much for the counterexample, how about separability and metrizability? $\endgroup$– ensueCommented Oct 2, 2019 at 12:30
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1$\begingroup$ A similar argument should show that $\{\delta_x : x \in X\}$ is an uncountable discrete set and therefore $P$ is not separable. $\endgroup$ Commented Oct 2, 2019 at 12:32
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1$\begingroup$ More specifically, for each $x \in X$ let $U_x = \{\mu \in P : \mu(\{x\}) > 1/2\}$. Then the $U_x$ are uncountably many disjoint open sets. $\endgroup$ Commented Oct 2, 2019 at 12:47
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$\begingroup$ Ok the argument for non separability is clear, thank you very much. $\endgroup$– ensueCommented Oct 2, 2019 at 13:42