# Is the topology of weak+Hausdorff convergence Polish?

Let $$X$$ be a compact metric space, $$P_X$$ the set of Borel probability measures on $$X$$, and $$K_X$$ the set of non-empty closed subsets of $$X$$. I will define the "topology of weak+Hausdorff convergence" to be the metrisable topology on $$P_X$$ whose convergence is given by \begin{align*} & \mu_n \overset{\text{w+H}}\to \mu \\ &\ \ \ \, \Longleftrightarrow \\ (\mu_n \to \mu \text{ weakly, and } \, \mathrm{supp}&\,\mu_n \to \mathrm{supp}\,\mu \text{ in Hausdorff distance})\text{.} \end{align*} This convergence is indeed metrisable, with a metric being given by $$d_{\text{w+H}}(\mu,\nu) = d_\text{w}(\mu,\nu) + d_\text{H}(\mathrm{supp}\,\mu,\mathrm{supp}\,\nu)$$ where $$d_\text{w}$$ is any metrisation of the topology of weak convergence, and $$d_\text{H}$$ is Hausdorff distance.

[This is a fairly natural topology to consider when one is interested in probability measures simultaneously at the "probabilistic" level of describing the distribution of likelihoods and at the "deterministic" level of prescribing what is and is not possible.]

Although $$d_\text{w}$$ and $$d_\text{H}$$ each generate compact topologies, it is easy to see that $$d_{\text{w+H}}$$ is not even complete: just take $$\mu_n$$ to be a sequence of fully supported probability measures converging weakly to a Dirac mass; then $$\mu_n$$ is $$d_{\text{w+H}}$$-Cauchy but not (weak+Hausdorff)-convergent as $$n \to \infty$$. Still, it would be nice at least to be able to say that the topology of weak+Hausdorff is Polish, meaning that it is both separable and admits a complete metrisation. (So, for example, even though the Euclidean metric on $$\mathbb{R} \setminus \mathbb{Q}$$ is not complete, it is known that the topology on $$\mathbb{R} \setminus \mathbb{Q}$$ generated by the Euclidean metric is Polish.)

Is the topology of weak+Hausdorff convergence Polish?

As in the answer to Comparison of several topologies for probability measures, separability should be fairly easy: e.g. let $$\mathcal{K}$$ be a Hausdorff-dense countable subset of $$K_X$$, and for each $$C \in \mathcal{K}$$ let $$\mathcal{P}_C$$ be a weakly dense countable subset of the set of probability measures supported on $$C$$; then $$\bigcup_{C \in \mathcal{K}} \mathcal{P}_C$$ should be (weak+Hausdorff)-dense.

So the remaining question is that of complete metrisability: Is there a complete metric on $$P_X$$ topologically equivalent to $$d_{\text{w+H}}$$?

My first guess: The $$\infty$$-Wasserstein metric $$W_\infty(\mu,\nu) = \inf\{\mathbb{P}\text{-}\mathrm{ess}\sup d(\boldsymbol{\cdot},\boldsymbol{\cdot}) \, : \, \mathbb{P} \text{ is a coupling of } \mu \text{ and } \nu \}$$ generates the topology of weak-Hausdorff convergence (perhaps generalising the answer to The infinity Wasserstein distance $W_\infty$ and the weak topology) and is complete?

• This is an idea, not a complete answer: The space $P_X$ with your topology is homeomorphic to a subspace of $P_X \times K_X$ equipped with the product of the weak/Hausdorff topologies. The homeomorphism is given by $I(\mu)=(\mu,\mathrm{supp}(\mu))$. The Polish subspaces of a Polish space are precisely the $G_\delta$ sets. Hence, to prove your space is Polish, it suffices to show that $I(P_X)$ is a $G_\delta$ set in $P_X \times K_X$. But I am not sure if it is, or how to show it.
– Dan
Jun 24, 2022 at 15:45
• @Dan Nice idea! I’ll think about whether I can prove that the graph of $\mathrm{supp}$ is $G_\delta$. Jun 24, 2022 at 18:31
• This works: write $I(P_X)=\cap_n A_n$ with $A_n=\{(\mu,S): \int_X d(x,S) \mu(dx) < 1/n \}$. These are open sets.
– Dan
Jun 25, 2022 at 2:19
• @Dan I think this (at least, in its current form) doesn't work: $\bigcap_n A_n$ contains $I(P_X)$ but also includes all points of the form $(\mu,S)$ with $S$ being any closed set containing $\mathrm{supp}\,\mu$. Jun 25, 2022 at 23:17
• Good point, my mistake!
– Dan
Jun 26, 2022 at 2:52

The set $$MS=\{(\mu,K)\in P_X\times K_X:\mathrm{supp}(\mu)=K\}$$ is of type $$G_\delta$$ in $$P_X\times K_X$$ and hence the weak+Hausdorff topology on $$P_X$$ is Polish.
Indeed, fix any countable base $$\{U_n\}_{n\in\mathbb N}$$ of the topology of $$X$$ and observe that $$MS=\bigcap_{n\in\mathbb N}\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}.$$ So, it remains to show that for every $$n\in\mathbb N$$ the set $$MS_n:=\{(\mu, K)\in P_X\times K_X:\mu(U_n)>0\;\Leftrightarrow\;U_n\cap K\ne\emptyset\}$$is of type $$G_\delta$$ in $$P_X\times K_X$$.
Observe that $$(P_X\times K_X)\setminus MS_n=MS_n'\cup MS_n''$$ where $$MS_n':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)>0\;\wedge\;U_n\cap K=\emptyset\}$$ and $$MS_n'':=\{(\mu,K)\in P_X\times K_X: \mu(U_n)=0\;\wedge\;U_n\cap K\ne\emptyset\}.$$ The known properties of the topologies on the spaces $$P_X$$ and $$K_X$$ ensure that the set $$\{\mu\in P_X:\mu(U_n)>0\}$$ is open in $$P_X$$ and the set $$\{K\in K_X:K\cap U_n\ne \emptyset\}$$ is open in $$K_X$$. Then $$MS_n'$$ and $$M_s''$$ are intersections of open and closed sets, so are of type $$F_\sigma$$ in $$P_X\times K_X$$. Then $$MS_n$$ is of type $$G_\delta$$ in $$P_X\times K_X$$ and so is the set $$MS=\bigcap_{n\in\mathbb N}MS_n$$.
• Nice! I guess you don't actually need to take the complement of $\mathit{MS}_n$, as you can express $\mathit{MS}_n$ itself as the union of the open set $\{\mu(U_n)>0 \ \wedge \ U_n \cap K \neq \emptyset\}$ and the closed set $\{\mu(U_n)=0 \ \wedge \ U_n \cap K = \emptyset\}$. Jul 11, 2022 at 9:53