Let $(E,d)$ be a bounded polish space (separable, complete metric space satisfying $\sup_{x,y\in E} d(x,y) < \infty$). By $\mathcal{P}(E)$ we denote the space of Borel probability measures on $E$ endowed with the topology of weak convergence.

We fix $\varepsilon > 0$, $\theta \in \mathcal{P}(E)$ and a stochastic kernel $\pi$ on $E$ (I.e. $\pi: E\times Bor(E) \rightarrow [0,1]$ such that $\pi(x,\cdot)$ is a Borel measure for all $x\in E$ and $\pi(\cdot,A)$ is measurable for all Borel sets $A\subseteq E$).

We assume that $\pi$ satisfies the strong Feller property, i.e. $x_n \rightarrow x\in E \Rightarrow \pi(x_n,\cdot) \stackrel{w}{\rightarrow} \pi(x,\cdot)$, where $\stackrel{w}{\rightarrow}$ denotes weak convergence.

We define \begin{align*} M := \{ \mu \in \mathcal{P}(E^2) &: \text{If } \mu = \mu_1 \otimes K, \text{ where } \mu_1 \in \mathcal{P}(E) \text{ and } K \\ &\text{ is a stochastic kernel on } E, \text{ then } \\ &W_1(\mu_1,\theta) \leq \varepsilon \text{ and } \\&W_1(K(x,\cdot), \pi(x,\cdot)) \leq \varepsilon \text{ for } \mu_1\text{-almost all } x\in E.\}, \end{align*} where $W_1$ is the first Wasserstein distance on $\mathcal{P}(E)$ given by \begin{align*} W_1(\nu,\mu) &= \inf_{\pi \in \Pi(\nu,\mu)} \int_{E^2} d(x,y) \pi(dx,dy)\\ &= \sup_{\substack{f:E\rightarrow \mathbb{R},\\ |f(x)-f(y)| \leq d(x,y)}} \left( \int_E f d\nu - \int_E f d\mu\right) \end{align*} and $\Pi(\nu,\mu)$ is the set of measures on $E^2$ with first marginal $\nu$ and second marginal $\mu$.

Question: Is M closed?


By boundedness of $(E,d)$ weak convergence is compatible with the Wasserstein distance, i.e. $\mu_n \stackrel{w}{\rightarrow} \mu \in \mathcal{P}(E) \Leftrightarrow W_1(\mu_n,\mu) \rightarrow 0$.

If $\pi$ does not satisfy the strong Feller property, $M$ is in general not closed. Take for example $\varepsilon = 0.5, E = [0,1], \theta = \delta_0, \pi(0,\cdot) = \delta_1$ and $\pi(x,\cdot) = \delta_0$ for $x \neq 0$. Then $M\ni\delta_{1/n} \otimes \delta_0 \stackrel{w}{\rightarrow} \delta_0 \otimes \delta_0 \not\in M$.

Even if $\pi$ is constant, I don't know whether $M$ is closed and would be very interested in an answer for this case as well.

  • $\begingroup$ Are you willing to assume those Borel measures admit finit second moments? $\endgroup$ – Henry.L Mar 31 '17 at 15:44
  • $\begingroup$ This is certainly an important special case and a solution would be very interesting to me. Do you have a method in mind that requires finite second moments? And just to clarify, admitting finite second moments in the sense of restricting $M$ to measures that have finite second moments? $\endgroup$ – Steve Mar 31 '17 at 18:12
  • $\begingroup$ Yes, if you are willing to assume finite second moment, I will try to think of it and probably have a direction using duality. However, I have no idea beyond $L^2$. $\endgroup$ – Henry.L Mar 31 '17 at 19:03
  • $\begingroup$ Ok, as mentioned I would be very interested in this case. However, in the context I'm working in one doesn't usually require moment restrictions so the general case would still be preferable. But if it's just about accepting the answer then the $L^2$-case would be sufficient for me if nothing else turns up. $\endgroup$ – Steve Mar 31 '17 at 20:19
  • $\begingroup$ I think we can relax to finite first moment due to how I use the result. Correct the wrong solution I posted earlier. $\endgroup$ – Henry.L Mar 31 '17 at 22:57

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