$P(X)$ is locally compact if $X$ is? Let us assume, as in here Measurable structure on the space of probability measures that $X$ is a locally compact Polish space.  Then can the same thing be said of $P(X)$, its probability measures with weak * convergence?  I am only missing local compactness and completeness relative to one metrization now, as I realized that appeal to the 1 point compactification of X helps. (This process gives a bigger space that is actually a metric space.)
 A: That $P(X)$ is Polish is true and can be found in many books; see e.g. Chapter 17 of Kechris' Classical descriptive set theory. One way to prove it is to observe that, denoting by $\widehat X=X\cup\{\infty\}$ the one-point compactification of $X$, the space $P(X)$ can be identified with $G=\{ \mu\in P(\widehat X);\; \mu(\{\infty\})=0\}$, which is $G_\delta$ in the compact metrizable space $P(\widehat X)$, and hence Polish.
On the other hand, $P(X)$ is not locally compact unless $X$ is compact.
For simplicity, consider $X=[0,\infty)$. Take any neighbourhood $\mathcal U$ of $\delta_0$ in $P(X)$. Then you will have no problem to show that one can find $\varepsilon >0$ such that $(1-\varepsilon)\delta_0+ \varepsilon \delta_a\in\mathcal U$ for every $a\in\mathbb R$.
Now take a sequence $(a_n)$ tending to $\infty$ and put $\mu_n:=(1-\varepsilon)\delta_0+\varepsilon\delta_{a_n}$. Then $(\mu_n)$ is a sequence in $\mathcal U$ which has no subsequence converging in $P(X)$. (The sequence $(\mu_n)$ is of course convergent in $P([0,\infty])$, but the limit $(1-\varepsilon)\delta_0+\varepsilon\delta_\infty$ is not in $P(X)$). So $\mathcal U$ is certainly not compact.
Exactly the same argument gives the result for a general locally compact and non-compact $X$.
