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Let $X$ be a metric space, $\mathcal{P}(X)$ be the the set of all Borel probability measures on $X$. I endow $\mathcal{P}(X)$ with the weak* topology.

Question 1: Do I need to impose some further requirement on $X$ for $\mathcal{P}(X)$ to be metrizable?

Let us say that $\mathcal{P}(X)$ (with the weak* topology) is metrizable, and so consider $\mathcal{P}(\mathcal{P}(X))$. Let $\mu \in \mathcal{P}(\mathcal{P}(X))$ and define, for every $B$ borel subset of $X$, $$ p_\mu(B)=\int_{\mathcal{P}(X)}\sigma(B)\mu(d\sigma). $$

This is null on the empty set, non-negative and countably additive, so a probability measure on $X$.

Question 2: Am I right or am I missing some nuances?

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  • $\begingroup$ If $X$ is separable and metric then $\mathcal{P}(X)$ is also separable and metric Wiki. $\endgroup$ – Florian Sep 23 '14 at 14:09
  • $\begingroup$ @Florian My question 1, put more explicitly, was whether metrizability of $X$ (with no other requirement) is sufficient for metrizability of $\mathcal{P}(X)$. $\endgroup$ – fadeaway Sep 23 '14 at 16:46
  • $\begingroup$ I don't know for sure, but I doubt that metrizability of $X$ is sufficient. $\endgroup$ – Florian Sep 23 '14 at 17:31
  • $\begingroup$ For the definition of $p_{\mu}(B)$ I think you need that the evaluation map $f_B\colon \mathcal{P}(X)\to \mathbb{R}$ defined by $f_B(\sigma)=\sigma(B)$ is $\mu$-measurable -- is that obvious? $\endgroup$ – Florian Sep 23 '14 at 17:50

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