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?



Metrizability of the space of probability measures endowed with the topology of setwise convergence
It's not metrizable.


The simple argument given in [this answer](https://mathoverflow.net/questions/207103/topology-of-setwise-convergence-of-measures) 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.


