TITLE: Metrizability of the space of probability measures endowed with the topology of setwise convergence
QUESTION [5 upvotes]: Let $X$ be a separable completely metrizable space, let $\mathscr{B}(X)$ denote the Borel $\sigma$-algebra on $X$, and let $\mathscr{P}(X)$ denote the space of all probability measures on $(X, \mathscr{B}(X))$.
Let $\tau$ denote the topology of setwise convergence on $\mathscr{P}(X)$, i.e. the smallest topology on $\mathscr{P}(X)$ such that for every $B \in \mathscr{B}(X)$ the mapping $\nu \mapsto \nu(B)$ on $\mathscr{P}(X)$ is continuous.
Question: Is $\tau$ metrizable?

REPLY [5 votes]: It's not metrizable.
The simple argument given in this answer shows that the finitely supported measures are dense in $\tau$.  
However, assuming $X$ is uncountable, they are not sequentially dense.  Let $\mu_n$ be any sequence of finitely supported probability measures, and let $\mu$ be any atomless Borel probability measure.  (Such a $\mu$ exists for any uncountable $X$; for instance, $X$ contains a copy of the Cantor set, which we can equip with Cantor measure.  Or for an explicit example, just consider $X = [0,1]$ and $\mu$ Lebesgue measure.)  Let $A_n$ be the finite set which is the support of $\mu_n$ and let $A = \bigcup_n A_n$, which is a countable set.  We have $\mu_n(A) = 1$ for every $n$, but $\mu(A) = 0$ since $\mu$ is atomless.  So $\mu_n$ does not converge setwise to $\mu$.
Thus $\mu$ is in the closure of the finitely supported measures, but isn't the limit of any sequence of them.  This shows $\tau$ is not first countable.