I am trying to understand the following:
Definition:
Map $t:X \to Y$ is called a transport map if it is measurable and $t_{\#}\mu=\nu$ where $\mu$ is a measure on $X$ and $\nu$ is a measure on $Y$.
Now, what is not clear to me is why the "class of all transportation maps is not closed with respect to any reasonable weak topology in the space of all measurable maps $X\to Y$"
I tried to construct a sequence of transport maps that converges to a map that is not a transport map i.e. does not satisfy $t_{\#}\mu=\nu$, but did not manage to do so. In particular, I was trying to work with atomic measures.
Can someone give me some hints?
The statement is correct and comes as a result in many papers, for example, https://link.springer.com/article/10.1007/s00229-021-01333-3 but I can not find the reason for this.
The class of transport maps is actually closed under the topology of convergence in probability. Indeed, if you take the class of Borel probability measures on $X\times Y$ with marginals $\nu$ and $\mu$, respectively, you get a closed set in the topology of weak convergence of measures when $X$ and $Y$ are, for example, Polish. If you identify transport maps with the induced measures supported on their graph, the trace topology of the weak topology is the usual topology of convergence in probability. One can adapt the proof of Proposition 1 in ["A Course on Young Measures" by Valadier] and use the fact that the weak topology coincides with the narrow topology for Young measures used there by the Scorza-Dragoni theorem.
However, we usually want to have a topology that is weak enough to be compact and strong enough that the total transport cost is continuous. That is why one usually works with the relaxed version given in terms of measures satisfying the marginal conditions given above. In the weak topology on that space, the set of transport maps is not closed. If $\nu$ is atomless, the space of support maps is actually a dense subspace, a fact that is proven in many introductions to optimal transport.