Let $\mu$ and $\nu$ two probability measures on $\mathbb{R}$. Let $\pi$ a probability measure in $\mathbb{R}^{2}$ such as $$ \pi (A\times \mathbb{R}) =\mu(A) \text{ and } \pi(\mathbb{R} \times B)=\nu(B) $$ and $$ \forall (x,y),(x',y')\in \Gamma, x<x' \Rightarrow y\le y' $$ with $\Gamma$ the support of the measure $\pi$.
Theorem : If $\mu$ is atomless then it exists a map $T : \mathbb{R} \rightarrow \mathbb{R}$ defined $\mu$-a.e. such that $\pi$ is concentrated on the graph of $T$.
Does somebody know how to prove it? It would be a relief for me because I don't see how.
And just to be sure the sentence "$\pi$ is concentrated on the graph of $T$" means $\Gamma = N \cup G(T)$ with $\pi(N)=0$ and $G(T)$ the graph of $T$ ?
Thank you so much for your help.
I made a progress, let $\mu$ and $\nu$ two probability measures on $\mathbb{R}$. Let $\pi$ a probability measure in $\mathbb{R}^{2}$ such as $$ \pi (A\times \mathbb{R}) =\mu(A) \text{ and } \pi(\mathbb{R} \times B)=\nu(B) $$
Let $\Gamma$ be the support of $\pi$. If $$ \Gamma \subseteq G_{T} $$ with $G_{T}$ the graph of the $\mu$ mesurable function $T$ then $$ T_{\text{#}} \mu = \nu $$
A question, for the beauty of mathematics, is the controverse true ?
Anyway, let's go back to our problem, this step is not important to answer to my question. But it's a important application for me, I hope you will understand why I'm really interested in this problem now.
Thank you for your help and regards.
PS : Maybe we can use convex range for atomless measure to solve it !