Let $X$ be a compact metric space (without isolated points). The $\infty$-Wasserstein distance $W\_\infty$ on the space of Borel probability measures on $X$ can be described as $$W\_\infty(\mu,\nu) = \inf\{r>0 \mid \mu(U)\le\nu(U\_r)\,\forall \text{ open } U\subseteq X\},$$ where $U\_r=\{x\in X \mid d(x,U)\le r\}$. The topology induced by this distance is in general finer than the weak (or vague) topology. For example, if $X=[0,1]$ and $\delta\_x$ denotes the point mass at $x$, the measures $\mu\_n=\frac{n-1}{n}\delta\_0+\frac{1}{n}\delta\_1$ converge weakly to $\delta\_0$, but $W\_\infty(\mu\_n,\delta\_0)=1$ for every $n$. On the other hand, $W\_\infty$ dominates the Levy-Prokhorov distance, so $W\_\infty$-convergence does imply weak convergence.


**Question**: is $W\_\infty$-convergence equivalent to weak convergence if one restricts to *fully supported* measures?


(By *fully supported* I mean those measures $\mu$ such that $\mu(U)>0$ for every nonempty open set $U\subseteq X$.)


I believe this is the case at least for $X=[0,1]$, but I just want to make sure it is not obviously ridiculous!


This is a duplicate of a [question I asked](https://math.stackexchange.com/questions/4257317/the-infinity-wasserstein-distance-w-infty-and-the-weak-topology) on stackexchange, but perhaps it is too specialised for that site.



The infinity Wasserstein distance $W_\infty$ and the weak topology
First, I interpret your condition that "$X$ has no isolated points" in the following ways: First, every ball $B(x, \varepsilon)$ has non-empty interior. This means, in particular, that we can find arbitrarily fine partitions of $X$ with sets that each have non-empty interior. And second, if we have $X = \Omega\_1 \stackrel{.}{\cup} \Omega\_2$ for two non-empty sets $\Omega\_1, \Omega\_2$, then $\inf\_{x\_1 \in \Omega\_1, x\_2 \in \Omega\_2} d(x\_1, x\_2) = 0$. I am not entirely sure if these conditions can be stated in simpler terms, but they don't seem unreasonable to me.


If this is what you had in mind (it's certainly satisfied for $[0, 1]$ or convex and compact subsets of $\mathbb{R}^d$), then I believe the two notions of convergence are indeed equivalent, see the proposed proof below. It's rather lengthy, so I hope I didn't miss anything. Please let me know if the proof seems incomplete.


Let $\mu \in \mathcal{P}(X)$ be fully supported and $(\mu\_k)\_{k \in \mathbb{N}}$ be a sequence in $\mathcal{P}(X)$ such that $\mu\_k$ converges weakly to $\mu$. In particular, $d\_P(\mu\_k, \mu) \rightarrow 0$ for the Prokhorov metric $d\_P$.


Let $\mathcal{F}\_1, \mathcal{F}\_2, ...$ be a sequence of refinements of partitions of $X$, i.e., $\mathcal{F}\_N = \{\Omega\_{N, 1}, \dots, \Omega\_{N, N}\}$, $X = \stackrel{.}{\cup}\_{i=1}^N \Omega\_{N, i}$, such that $\delta(N) := \max\_{i=1, \dots, N} \max\_{x, y \in \Omega\_{N, i}}d(x, y) \rightarrow 0$ for $N \rightarrow 0$, each $\Omega\_{N, i}$ has non-empty interior and choose some fixed $x\_{N, i} \in \Omega\_{N, i}$. Note that by the "connectedness condition" on X, for any $J \subsetneq \{1, \dots, N\}$, it holds $\min\_{j \in J} \min\_{i \not\in J} d(x\_i, x\_j) \leq 2 \delta(N)$.


Define $\bar\mu^N := \sum\_{i=1}^N \delta\_{x\_{N, i}} \mu(\Omega\_{N, i})$. It is clear that $W\_\infty(\mu, \bar\mu^N) \leq \delta(N)$. The fact that the partitions are refinements implies, in particular, that $\bar\mu^{N\_2}(\Omega\_{N\_1, i}) = \bar\mu^{N\_1}(\Omega\_{N\_1, i})$ for $N\_2 \geq N\_1$.
Define $w(N) := \min\_{i=1, \dots, N} \mu(\Omega\_{N, i})$, which is positive by assumption on $\mu$.


Note that for any $N\_1, N\_2 \in \mathbb{N}$ with $N\_2 \geq N\_1$ and any $A \subset X$ open such that $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) \neq 1$, it holds
$$
\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) \geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + w(N\_1),
$$
the reason being as follows: $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) \neq 1$ implies that there exists some $i \in \{1, \dots, N\_1\}$ such that $\Omega\_{N\_1, i} \cap A^{2 \delta(N\_1)} = \emptyset$ (since otherwise $A^{3 \delta(N\_1)} = X$). In particular, by connectedness of $X$ there exists such an $i$ such that $\Omega\_{N\_1, i} \subset A^{4 \delta(N\_1)}$. I.e.,
\begin{align}
\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) &\geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + \bar\mu^{N\_2}(\Omega\_{N\_1, i})\\ &= \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + \bar\mu^{N\_1}(\Omega\_{N\_1, i})\\ &\geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + w(N\_1).
\end{align}


Now for $\varepsilon > 0$, we choose $N\_1$ large enough so that $\delta(N\_1) < \varepsilon$. Let $r(N\_1) := \min\{2 \delta(N\_1), w(N\_1)\}$. Choose $N\_2, k$ large enough so that $d\_P(\bar\mu^{N\_2}, \mu\_k) \leq r(N\_1)$.


Let $A \subset X$ open. Either $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) = 1$, and hence $\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) \geq \mu\_k(A)$ holds trivially. Or, $\bar\mu^{N\_2}(A^{3 \delta(N\_1)}) < 1$ and hence by the above
$$
\bar\mu^{N\_2}(A^{4 \delta(N\_1)}) \geq \bar\mu^{N\_2}(A^{2 \delta(N\_1)}) + w(N\_1) \geq \bar\mu^{N\_2}(A^{r(N\_1)}) + r(N\_1) \geq \mu\_k(A),
$$
where the last step follows since $d\_P(\bar\mu^{N\_2}, \mu\_k) \leq r(N\_1)$. Therefore, $W\_\infty(\bar\mu^{N\_2}, \mu\_k) \leq 4 \delta(N\_1) \leq 4 \varepsilon$. By the triangle inequality, $W\_\infty(\mu, \mu\_k) \leq W\_\infty(\bar\mu^{N\_2}, \mu\_k) + W\_\infty(\bar\mu^{N\_2}, \mu) \leq 5 \varepsilon$, which yields the claim.


