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](https://en.wikipedia.org/wiki/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 $S$.



Wasserstein convergence of conditional measures
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 $2$. 


