Wasserstein convergence of conditional measures Suppose $W_r(\mu_n,\mu)\to0$, where $\mu_n$ and $\mu$ are discrete probability measures on some metric space $\Omega$, and that all measures have the same number of atoms $d$ (but not the same atoms):
$$\mu_n = \sum_{i=1}^d p_{n,i}\delta_{\theta_{n,i}}, \quad \mu = \sum_{i=1}^d p_{i}\delta_{\theta_{i}}.$$
$W_r$ here is the usual $L_r$-Wasserstein metric (a.k.a. Kantorovich–Rubinstein, optimal transport distance). Assume further that the atoms are labeled such that $\theta_{n,i}\to\theta_i$ for each $i$.
Now let $S\subset\{1,\ldots,d\}$ be any subset and consider the restriction of $\mu_n$ and $\mu$ to $S$. Let $\widetilde{\mu}_n(S)$ and $\widetilde{\mu}(S)$  denote these measures after normalization (i.e. so that each has total mass 1), i.e.
$$\widetilde{\mu}_n(S) \propto \sum_{i\in S} p_{n,i}\delta_{\theta_{n,i}}, \quad \widetilde{\mu}(S) \propto \sum_{i\in S} p_{i}\delta_{\theta_{i}}.$$
Is it true that $W_r(\widetilde{\mu}_n(S), \widetilde{\mu}(S))\to 0$ for any $S$?
Note. These can be interpreted as "conditional probabilities" where we are conditioning on the atoms in $S$.
 A: The edited version of the question requires that $\langle\theta_{i,n}\rangle$ converges to $\theta_i$ for all $i$. In that case, the result is true. It suffices to show that we also have $\langle p_{i,n}\rangle$ converging to $p_i$. This is easy to see when we note that the Wasserstein-metric metrizes a topology stronger than the topology of weak convergence of measures under which $\langle\mu_n\rangle$ converges to $\mu$ if and only if $\lim_n\int g~\mathrm d\mu_n=\int g~\mathrm d\mu$ for every bounded continuous function $g$. Let $g_i$ be a bounded continuous function such that $g(\theta_i)=1$ and for each $j\neq i$ there is a neighborhood $U_j$ of $\theta_j$ such that $g$ vanishes on $U_j$. For example, we can take $g_i$ to be $\max\{1-k d(\theta_i),0\}$ for $k$ large enough. Then for $n$ large enough we have $g(\theta_{j,n})=0$ for all $j\neq i$. For such $n$, we have $\int g_i~\mathrm d\mu_n=p_{i,n}g(\theta_{i,n})$. Since $\lim_n g(\theta_{i,n})=g(\theta_i)$, we must have $\lim_np_{i,n}=p_i$.
Without the assumption that $\langle\theta_{i,n}\rangle$ converges to $\theta_i$, this fails for trivial reasons: You are not really conditioning on atoms but some arbitrary labeling of atoms. Let $\Omega=\{1,2\}$, $\theta_1=1$, $\theta_2=2$, $p_1=2/3$, $p_2=1/3$. For $n$ odd, let $\theta_{n,1}=1$, $\theta_{n,2}=2$, $p_{n,1}=2/3$, $p_{n,2}=1/3$. For $n$ even, let $\theta_{n,1}=2$, $\theta_{n,1}=2$, $p_{n,2}=2/3$, $p_{n,1}=1/3$. The sequence $\mu_n$ is constant and converges to $\mu$. But for $S=\{1\}$, the "conditional measures" are point masses switching between $1$ and $2$. 
