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This is a follow-up question to this question: Closure Wasserstein for pointmasses

Let $(X,d)$ be a metric space, and let $W_1(X)$ be the space of probability measures $\mu$ on $X$ having finite first moment with respect to $d$, i.e. $\int_X d(x_0,x)\,d\mu(x)<+\infty$ for some base-point $x_0\in X$. Endow $W_1(X)$ with the Earthmover, a.k.a. 1-Wasserstein distance. What is the closure of the set of finitely supported probability measures in $W_1(X)$? Are there sufficient conditions for the latter set to be dense in $W_1(X)$?

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    $\begingroup$ My first guess would be that this is true if and only if $X$ is separable. Do you know that this is not the correct condition? $\endgroup$
    – Anthony Quas
    Commented May 17, 2021 at 22:15
  • $\begingroup$ Maybe just a hint, but (sequential) convergence in $W_1$ is equivalent to (sequential) weak convergence (duality with bounded continuous functions) + convergence of the first moment. So the question somehow boils down to asking wether finitely supported measures are weakly closed. And then just as Anthony Quas I suspect that this is true iff $X$ is separable (well, clearly this is necessary, perhaps weird situations can arise when separability is not sufficient?) $\endgroup$
    – leo monsaingeon
    Commented May 17, 2021 at 22:42
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    $\begingroup$ @Anthony Quas: I convinced myself that density holds provided the metric space $X$ is proper, i.e. closed balls are compact... That's all I know basically... $\endgroup$
    – Alain Valette
    Commented May 18, 2021 at 22:09

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