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The following question has come up while working on my senior thesis: Assume $\mu$ and $\nu$ are regular probability measures on $\mathbf{R}^n$. We are also given a coupling $\gamma$ of $\mu$ and $\nu$ such that if $(x,y)\in Supp(\gamma)$ then $|x-y| \leq 1$. So basically given random variable $X$ based on $\mu$ and another one $Y$ based on $\nu$ we know $|X-Y|\leq 1$ with probability $1$.

Now additionally assume $\mu$ is absolutely continuous with respect to Lebesgue measure. Show that there is a deterministic coupling of $\mu$ and $\nu$, i.e. a Borel mapping $s:\mathbf{R}^n \rightarrow \mathbf{R}^n$ such that $s$ pushes $\mu$ forward to $\nu$.

Note that this is true in dimension $1$ by taking monotone measure preserving map.

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    $\begingroup$ Did you try to apply optimal transport for a suitable cost function? $\endgroup$ Feb 27, 2011 at 22:23
  • $\begingroup$ Well, I tried the natural thing and I didn't see it working out for me. I tried approximating the costs by strictly convex costs converging to this cost and my attempt failed at multiple levels because of non-compactness of space of deterministic Borel couplings with respect to wk-* topology.(i.e. a sequence of mappings can go to a random coupling in the limit.) you can try approximate the cost in the other way, from left as opposed to right, and then your measures might not decompose in the way that is necessary for existence of mappings. $\endgroup$
    – Nick B.
    Feb 27, 2011 at 22:47
  • $\begingroup$ The other approach is just to look at extremal points of space of all such couplings of $\mu$ and $\nu$ and somehow show that the extremal points of this space, which exist I think by Krein_Milman, must be deterministic couplings. But I couldn't make that work either. $\endgroup$
    – Nick B.
    Feb 27, 2011 at 23:05
  • $\begingroup$ From what I remember, optimal transport solves your problem at least in case if your inequality is strict plus your measures are striclty positive on all open sets. [If you would ask a specialist, he would probably give you a solution immideately...] $\endgroup$ Feb 28, 2011 at 17:29

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