# Metrizability of the space of probability measures endowed with the topology of setwise convergence

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?

• I don't think the topology you describe in the body is what's usually called the topology of "setwise convergence"; the latter would be the smallest topology that makes the mappings $\nu \mapsto \nu(A)$ continuous. There is a lot of information about these topologies in Section 4.7 of Bogachev's Measure Theory. Commented Jun 2, 2016 at 16:32
• For the topology of setwise convergence as usually defined, no, it's not metrizable. This answer shows that the finitely supported measures are dense in this topology, but they're certainly not sequentially dense. Commented Jun 2, 2016 at 17:50
• Thanks @NateEldredge, I think the definition you give is the one I'm interested in. I've updated the question. Would you mind posting an answer? Commented Jun 2, 2016 at 22:28

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.