Hi,
I am reading the book "Topics in Optimal Transportation" by Cedric Villani.


The Brenier's theorem states (among other things) that there is a unique transport plan for the optimal transport with the quadratic cost if the measure $\mu$ (to be transported toward $\nu$) does not give mass to small sets.
(page 66)


A counter example, in the case of mu giving mass to small sets, is the case where $\mu$ and $\nu$ are measures on $\mathbb R^2$, concentrated on $\{(0,0),(1,1)\}$ and $\{(0,1),(1,0)\}$ respectively. In that case, there is no uniqueness. (page 67)


However, I am wondering why this is only the case when $\mu$ does not give mass to small sets. We can easily imagine a case where $\mu$ and $\nu$ are continuous but are made such that there is no uniqueness. For example, with an analogy with the example above, $\mu$ can spread its mass on a line between $(0,0)$ and $(1,1)$ with some thickness, and $\nu$ on a line between $(0,1)$ and $(1,0)$ with the same thickness (so the two lines are symmetric and don't have zero measure). In that case, I still see two possible transport plans...
Same thing if we consider 2 gaussians centered at $(0,0)$ and $(1,1)$ in the first measure and at $(0,1)$ and $(1,0)$ in the second one....


Any idea ?


Thanks 



Brenier's theorem
Because there's no reason to ship a whole gaussian en masse to the same target gaussian. It's cheaper to send all the mass that is to one side of the diagonal (line joining (0,0) to (1,1)) to one gaussian and the rest of the mass to the other side. The same is true for the thickened line.


