Why is this probability $1$?

Question

Let's consider two metric spaces $X$ and $Y$.

Definition: Given $\phi: X \to \mathbb{R}\cup\{+\infty,-\infty\}$ and $c:X \times Y \to \mathbb{R}$ we define $\phi^c(y)=\inf\limits_{x \in X} c(x,y)-\phi(x)$.

Problem: Let's assume that $\pi$ is a probability measure on $X \times Y$ ($\pi \in P(X \times Y)$) with marginals $\mu$ on $X$ and $\nu$ on $Y$. Let's assume that $\Gamma= \{(x,y) \in X \times Y: \phi(x)+\phi^c(y) = c(x,y)\}$ is such that $\pi(\Gamma)=1$. Let's finally assume that $\max \{ \phi,0 \} \in L^1(\mu)$. How can I prove that $\mu(\{x : \phi(x) \in \mathbb{R}\})=1$?

Attempt: I tried in the following way. $1= \pi (\Gamma)= \mu (\{x : k_x(\Gamma_x)=1 \})$ where $k$ is the disintegration kernel of $\pi$ with respect to $\mu$ in the sense that $\pi(f)= \int f(x,y) k_x(dy) \mu(dx)=:(\mu \times k)(f)$ for every $f \in C_b(X \times Y)$. To conclude I would like to prove that $\{x : k_x(\Gamma_x)=1 \} \subset \{x : \phi(x) \in \mathbb{R}\}$ but I don't know how to do it.

Answer 1

Assume moreover that $\phi$ is not be constantly equal to $-\infty$ (otherwise all the other hypothesis are fulfilled but the desired result is clearly false).

We want to prove that $\mu(\{x\in X\mid\phi(x)=\pm\infty\})=0.$

• $\mu(\{x\in X\mid\phi(x)=+\infty\})=0$ because $\max \{ \phi,0 \} \in L^1(\mu).$
• $\mu(\{x\in X\mid\phi(x)=-\infty\})=\pi(\{(x,y)\in X\times Y\mid\phi(x)=-\infty\})=0$ because whenever $\phi(x)=-\infty,$ we have $c(x,y)-\phi(x)=+\infty>\phi^c(y)$ hence $(x,y)\notin\Gamma,$ and $\pi(\Gamma)=1.$