# Optimal transport plan induced by an optimal transport map

I am reading Santambrogio's book on optimal transport, remark 1.19. Let's consider an optimal transport problem between $$(X,\mu)$$ and $$(Y,\nu)$$.

(Remark 1.19) ... every time that we know that any optimal $$\gamma$$ must be induced by a map $$T$$, then we have uniqueness (of a Kantorovich solution). Indeed, suppose that two different plans $$\gamma_1=\gamma_{T_1}, \gamma_2=\gamma_{T_2}$$ are optimal: consider $$\frac{1}{2}\gamma_1+\frac{1}{2}\gamma_2$$, which is optimal as well by convexity. This last transport plan cannot be induced by a map unless $$T_1=T_2$$ $$\mu-a.e.$$, which gives a contradiction.

Here an optimal plan $$\gamma$$ refers to a solution of the Kantorovich problem and an optimal map $$T$$ refers to a solution of the Monge problem. The remark does not assume any regularity of the spaces.

I could prove the statement assuming that $$\gamma(\{(x,y)\in X\times Y:y=T(x)\})=1$$, but I could only show that $$\gamma$$ is concentrated on the closure of the set. From that I could deduce $$\gamma(\{(x,y)\in X\times Y:y=T(x)\})=1$$ when $$T$$ is continuous $$\mu$$-a.e. I am also not sure if optimality matters for the statement. So my questions are:

1. Can we prove or disprove that $$\gamma(\{(x,y)\in X\times Y:y=T(x)\})=1$$?
2. If the answer for 1. is no, how can we prove Remark 1.19?

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Edit: For people who are interested in the same question, I complete the proof for Remark 1.19 here. (Thanks to @gerw for the nice answer.)

Lemma If $$\gamma=\gamma_T:=(id, T)_{\#}\mu$$, then $$\gamma(\{(x,y)\}\in X\times Y:y=T(x))=1$$.

Proposition Let $$\gamma_1=\gamma_{T_1}$$ and $$\gamma_2=\gamma_{T_2}$$. Then $$\gamma=\frac{1}{2}\gamma_1+\frac{1}{2}\gamma_2$$ cannot be of the form $$\gamma=(id,T)_{\#}\mu$$ for any optimal transport map $$T$$ unless $$T_1=T_2$$ $$\mu-a.e.$$.

Proof Suppose $$\gamma=(id,T)_{\#}\mu$$ for some $$T$$. Then,

\begin{alignat*}{3} 1=\gamma(y=T(x)) &=&&\gamma(\{y=T(x)\}\cap\{T(x)=T_1(x)\ne T_2(x)\}) \\ & &&+\gamma(\{y=T(x)\}\cap\{T(x)=T_2(x)\ne T_1(x)\})\\ & &&+\gamma(\{y=T(x)\}\cap\{T(x)=T_1(x)= T_2(x)\})\\ & &&+\gamma(\{y=T(x)\}\cap\{T(x)\ne T_1(x), T(x)\ne T_2(x)\})\\ &=&& \frac{1}{2}\gamma_1(\{y=T(x)\}\cap\{T(x)=T_1(x)\ne T_2(x)\})\\ & &&+ \frac{1}{2}\gamma_2(\{y=T(x)\}\cap\{T(x)=T_2(x)\ne T_1(x)\})\\ & &&+ \frac{1}{2}\gamma_1(\{y=T(x)\}\cap\{T(x)=T_1(x)= T_2(x)\})\\ & &&+ \frac{1}{2}\gamma_2(\{y=T(x)\}\cap\{T(x)=T_2(x)= T_1(x)\})\\ &=&& \frac{1}{2}\gamma_1(\{y=T_1(x)\}\cap\{T(x)=T_1(x)\})\\ & &&+\frac{1}{2}\gamma_2(\{y=T_2(x)\}\cap\{T(x)=T_2(x)\})\\ &=&& \frac{1}{2}\gamma_1(T(x)=T_1(x))+\frac{1}{2}\gamma_2(T(x)=T_2(x))\\ &=&& \frac{1}{2}\mu(T(x)=T_1(x))+\frac{1}{2}\mu(T(x)=T_2(x)). \end{alignat*}

Thus $$\mu(T(x)=T_1(x))=\mu(T(x)=T_2(x))=1$$.

• What is the definition of "induced by a map"? Commented May 28 at 17:14
• Good point! The plan induced by a map $T$ is $\gamma_T=(id,T)_{\#}\mu$. Commented May 29 at 0:46

Set $$G := \{(x,y) \in X \times Y : y = T(x)\}$$. I think for $$\gamma = \gamma_T$$ we just have \begin{align*} \gamma(G) = \int_{X \times Y} \chi_G(x,y) \,\mathrm{d}\gamma(x,y) = \int_X \chi_G \mathbin\circ (id,T) (x) \, \mathrm{d} \mu(x) = 1. \end{align*}