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?**

Remarks:

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.