Let $X$ be a compact metric space and denote $\mathcal M(X)$ the set of probability measures on $X$. For $\mu\in\mathcal M(X)$ we write $\operatorname{supp} \mu$ for the *support of $\mu$*. As is well known, if a sequence of measures $(\mu\_n)\_{n=1}^\infty$ weak$^\*$ converges to some $\mu\in\mathcal M(X)$, then
$$
\operatorname{supp}\mu\subset \lim\_{n\to\infty}(\operatorname{supp}\mu\_n).
$$
and the inverse is not necessarily true.


We can define the following metric on $\mathcal{M}(X)$:


$$ d^{PH}(\mu,\mu\_n)=\text{max}\{d^P(\mu,\mu\_n),d^H(\mu,\mu\_n)\}$$
where $d^P(\mu,\mu\_n)$ is the Prokhorov metric on $\mathcal{M}(X)$ and $d^H(\mu,\mu\_n)$ is a metric between $\mu$ and $\mu\_n$ defined by the Hausfdorff distance between $\operatorname{supp}\mu$ and $\operatorname{supp}\mu\_n$.


Finally denote as $d^{TV}(\mu, \mu\_n)$ the total variation metric on $\mathcal{M}(X)$, given by 
$$d^{TV}(\mu, \mu\_n)= \sup\_{A \in \mathcal F}|\mu\_n (A) - \mu(A)| $$ where $\mathcal F$ is the underlying $\sigma$-algebra on $X$.


I have the following question: Is it true that the topology on $\mathcal{M}(X)$ generated by $d^{TV}(\cdot)$ is strictly finer than the topology generated by $d^{PH}(\cdot)$ and how could one prove this? Is the topology generated by $d^{PH}(\cdot)$ separable?



Comparison of several topologies for probability measures
The topology generated by $d^{PH}$ is, in general, neither coarser nor finer than $d^{TV}$. The Hausdorff distance between the supports of two measures, even when combined with the Prokhorov distance, has very little to do with the total variation distance between them.


To see one direction, let $X$ have at least two points $x,y$ and let $\mu\_n = \frac{n-1}{n} \delta\_x + \frac{1}{n} \delta\_y$, $\mu = \delta\_x$. Then the support of $\mu\_n$ is $\{x,y\}$ while the support of $\mu$ is $\{x\}$, so the Hausdorff distance between their supports is $d(x,y)$ for every $n$. Hence $\mu\_n$ does not converge to $\mu$ in $d^{PH}$. On the other hand, $d^{TV}(\mu\_n, \mu) = 1/n$ so $\mu\_n$ does converge to $\mu$ in $d^{TV}$.


For the other direction, take $X = [0,1]$ and let $\mu\_n = \frac{1}{n} \sum\_{k=1}^n \delta\_{k/n}$, with $\mu$ being Lebesgue measure. Then $\mu\_n \to \mu$ weakly, and the Hausdorff distance between $\{1/n, 2/n, \dots, 1\}$ and $[0,1]$ is $1/n$. So $\mu\_n \to \mu$ in $d^{PH}$. But $\mu\_n$ is singular to $\mu$ for every $n$, so $d^{TV}(\mu\_n, \mu) = 1$, and $\mu\_n$ does not converge to $\mu$ in $d^{TV}$.


I think that $d^{PH}$ is separable. Fix a countable dense set $\{x\_1, x\_2, \dots\}$ in $X$ and consider the set $M\_0$ of all probability measures $\nu$ of the form $\nu = \sum\_{i=1}^m q\_i \delta\_{x\_i}$ where the values $q\_i$ are nonnegative rationals that sum to $1$. This is dense in the weak topology, and given a sequence $\nu\_n$ in $M\_0$ converging weakly to some $\mu$, you ought to be able to modify the measures $\nu\_n$ so as to ensure that all their supports (which are finite sets) are Hausdorff close to that of $\mu$. I leave the details to you, if you wish to work it out.


