Metrization of the weak*-topology on a set of probability measures Let $X$ denote a metric space. One can assume that $X$ is Polish if that helps, but I was trying to avoid to assume that $X$ is compact. Let $P(X)$ denote the set of Borel probability measures on $X$. The weak*-topology on $P(X)$ is defined as usual: a net $(p_{\alpha})$ in $P(X)$ converges to $p \in P(X)$ iff $|\int fdp_{\alpha}-\int fdp|$ converges to $0$. 
Question: Is there any metric that metrizes the weak*-topology on $P(X)$ and has convex open balls?  
 A: On a separable metric space $(X,d)$, weak convergence of probability measures is equivalent to convergence with respect to the Lévy–Prokhorov metric defined by 
$$ \beta(\mu,\nu) = \sup \left( \int_X f\ d(\mu-\nu): \|f\|_{BL}\leq 1\right),$$
where $$\|f||_{BL}=\sup_x |f(x)|+\sup_{x\neq y}|f(x)-f(y)|/d(x,y).$$  
See Theorem 11.3.3 of R. M. Dudley's Real Analysis and Probability. 

Let's show that $B=\{\nu: \beta(\mu,\nu)\leq\varepsilon      \}$ is a convex set. Let $\nu_1,\nu_2\in B$, $\alpha\in(0,1)$, and $f$ with $\|f\|_{BL}\leq 1$. Then 
\begin{eqnarray*}
\int f d(\mu-[\alpha\nu_1+(1-\alpha)\nu_2])&=&\alpha\int f d(\mu-\nu_1)+(1-\alpha)\int f d(\mu-\nu_2)\\[5pt]
&\leq&\alpha\beta(\mu,\nu_1)+(1-\alpha)\beta(\mu,\nu_2)\\[5pt]
&\leq &\alpha \varepsilon +(1-\alpha)\varepsilon\\[5pt]
&=&\varepsilon.  
\end{eqnarray*}
Taking the supremum over such $f$ gives $\beta(\mu,\alpha\nu_1+(1-\alpha)\nu_2)\leq \varepsilon$, so the closed ball is convex.
The open ball $\{\nu: \beta(\mu,\nu)<\varepsilon\}$ is the increasing union 
of the convex sets $\{\nu: \beta(\mu,\nu)\leq\varepsilon-1/n\}$, for $n>1/\varepsilon$, and so is itself convex. 
