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$\newcommand{\P}{\mathcal{P}}$

Let $(E,d)$ be a complete metric space, let $\P(E)$ be the set of all probability measures on $(E,\mathcal{B}(E))$. Let $W_d$ be the $1$-Wasserstein (Kantorovich) metric on $\P(E)$.

It is well known that if additionally the space $E$ is separable, then the space $(\P(E),W_d)$ is complete. It is also known that in the general case (if $E$ is not separable) only the space $(\P_r(E),W_d)$ is complete, where $\P_r(E)$ is the set of all Radon probability measures.

My question is: can one construct an explicit example of a complete (and obviously non-separable) metric space$(E,d)$ such that the space $(\P(E),W_d)$ is not complete?

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    $\begingroup$ What about the discrete space $E$ with $\{0,1\}$-valued metric $d:E\times E\to\{0,1\}$, whose cardinality $|E|$ is a real-valued measurable cardinal (which means that $E$ admits a $\sigma$-additive probability measure $\mu:2^E\to[0,1]$ that vanishes on all singletons and is defined on the Boolean algebra of all subsets of $E$)? Is the space $(P(E),W_d)$ complete? $\endgroup$ – Taras Banakh Aug 15 '17 at 15:53
  • $\begingroup$ You mean all Borel probability measures, right? $\endgroup$ – Fedor Petrov Aug 15 '17 at 17:44
  • $\begingroup$ The equivalent definition of the Wasserstein metric as $W_d(\mu,\lambda)=\sup\{|\mu(f)-\lambda(f)|:f:X\to\mathbb R$ is non-expanding$\}$ implies that for a uniformly discrete metric space $(X,d)$ the metric space $(P(X),W_d)$ is complete. A metric space $(X,d)$ is uniformly discrete if $\inf\{d(x,y):x,y\in,\;x\ne y\}>0$. So, the example I suggested above does not fit the requirements. $\endgroup$ – Taras Banakh Aug 16 '17 at 4:46
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    $\begingroup$ It seems that the existence of a non-Radon $\sigma$-additive Borel measure on a complete metric space implies the existnce of a real-valued measurable cardinal. So, the required counterexample cannot exist in ZFC and should use real-valued measurable cardinals in some way. $\endgroup$ – Taras Banakh Aug 16 '17 at 5:07
  • $\begingroup$ The Vitali-Hahn-Saks Theorem implies that for any discrete topological space $X$ the space $P(X)$ with the weak$^*$ topology is sequentially complete. Then for any metric $d$ generating the discrete topology on $X$ the metric space $(P(X),W_d)$ is complete. $\endgroup$ – Taras Banakh Aug 16 '17 at 7:22

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