Does complete and separable Wasserstein space imply a complete base space?

Question

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Let $(Z,d)$ be a metric space, and for $p\geq 1$, consider a metric space $(W_p,d_{W_p})$ defined by

The Wasserstein Space $\begin{align}W_p = \{\mu|\mu\textrm{ is a Borel probability measure on Z such that} \int_{Z}d(z_0,z)^p\mu(dz)<\infty \textrm{ for some }z_0 \} \end{align}$

and

The Wasserstein distance $\begin{align} d_{Wp}(\mu,\nu) = \inf_{\pi}\left(\int_{Z\times Z}d(z,z')^p\pi (dz\times dz')\right)^{1/p}\end{align}$

where $\pi$ is a coupling between $\mu$ and $\nu$ i.e. a probability measure on $Z\times Z$ such that $\pi(A\times Z)=\mu(A), \pi(Z\times B)=\nu(B)$ for any measurable $A,B\subset Z$.

Let us assume that $Z$ is separable (since otherwise, something like Nedoma's pathology can show up and make things complicated). It is well known that if $Z$ is complete in addition, the Wasserstein space is also complete. Is the converse true? That is, I want to prove the following:

Conjecture. Complete and separable Wasserstein distance implies $Z$ is complete (assuming $Z$ is separable).

My Attempt.

Google took me to Counter-example to the completeness of the Wasserstein metric, but this is about $Z$ complete $\Rightarrow$ W space is complete, so it is not really what I want.

I considered a simple case $Z=(0,1)$, and if the conjecture is true, the W space should be incomplete. To prove this, I took the sequence of Dirac measures $\{\delta_{1/n}\}$, which is a Cauchy sequence in the W distance, but I couldn't prove that it does not converge. Maybe it converges to some exotic probability measure? I don't know.

On the other hand, according to this thesis, it seems that Theorem A.1, p238 says a topological space is complete if and only if the weak convergence of probability measures is complete. However, this is not directly applicable since the fact that the W distance metrizes the weak convergence is only proven for complete & separable spaces.

I believe what we should do is to prove that the space of Dirac measures is sequentially closed, and this is proven for weak convergence in the thesis I mentioned above(Proposition A.4). But again we need to prove that for the W-distance and not weak covnergence.

Answer 1

$\newcommand\de\delta\newcommand\ep\varepsilon$The conjecture is true.

Indeed, let $(z_n)$ be a Cauchy-convergent sequence in $(Z,d)$. Then $$d_{W_p}(\de_{z_m},\de_{z_n})=d(z_m,z_n)\to0$$ (as $m,n\to\infty$). So, the sequence $(\de_{z_n})$ is Cauchy-convergent and hence converges in $(W_p,d_{W_p})$ to some probability measure $\mu$ over $Z$.

That is, $\int_Z \mu(dz)\,d(z,z_n)^p\to0$. So, by Markov's inequality, $1-\mu(B_\ep(z_n))=\int_Z \mu(dz)\,1(d(z,z_n)\ge\ep)\to0$ for each real $\ep>0$, where $B_\ep(z)$ denotes the open ball of radius $\ep$ centered at $z$. It follows that, if $u$ is any point in the support set $S_\mu$ of the probability measure $\mu$, then $z_n\to u$. So, any Cauchy-convergent sequence $(z_n)$ in $(Z,d)$ is convergent. $\quad\Box$

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This result holds for $p=\infty$ as well. Indeed, in this case the $\mu$-essential supremum of the function $d(\cdot,z_n)$ goes to $0$. So, again, if $u$ is any point in the support set $S_\mu$ of the probability measure $\mu$, then $z_n\to u$.