Let $X$ and $\bar X$ be two standard Borel spaces, and let $A\subseteq X\times\bar X$ be an analytic subset of the product space. Let $P$ be any probability measure such that $P(A) = 1$, and denote by $p$ and $\bar p$ marginals of $P$ on $X$ and $\bar X$ respectively. Does there exist a universally measurable map $f:X\to \bar X$ such that $\bar p = p\circ f^{1}$ and such that the graph of $f$ is contained in $A$?
$\begingroup$
$\endgroup$
4

1$\begingroup$ Are some conditions missing? Take $X = \{0\}$, $\bar{X} = [0,1]$, $A = X \times \bar{X}$, $p = \delta_0$ is a point mass, $\bar{p}$ is Lebesgue measure, and $P = p \times \bar{p}$. Then the marginals are as required but clearly $\bar{p}$ cannot be the pushforward of $p$. $\endgroup$– Nate EldredgeCommented Jan 14, 2014 at 2:09

$\begingroup$ @NateEldredge: Thanks. Nope, your example applies  you can post it as an answer (unless the question will be closed before that as a trivial one) $\endgroup$– SBFCommented Jan 14, 2014 at 2:15

1$\begingroup$ In optimal transport theory one minimzes an integral over probabilities on a product space with given marginals. If the solution is supported on the graph of a function, one has a Mongesolution. There is a very, very large and active literature on the topic that might provide useful sufficient conditions. $\endgroup$– Michael GreineckerCommented Jan 14, 2014 at 2:23

$\begingroup$ @MichaelGreinecker: thanks, but Nate's example (appied to $X$ being an arbitrary finite set and $\bar X = [0,1]$) is something I'm also dealing with, so surely in general in the situation I'm interested in $\bar p$ may not be a pushforward of $p$. Btw, you shall know that one I guess. $\endgroup$– SBFCommented Jan 14, 2014 at 2:25
Add a comment

1 Answer
$\begingroup$
$\endgroup$
No. The measures $p, \bar{p}$ could be almost anything; there's nothing in the given conditions that forces $\bar{p}$ to be a possible pushforward of $p$.
As an extreme example, take $X = \{0\}$ to be a singleton, $\bar{X} = [0,1]$, $A = X \times \bar{X}$, $P= \delta_0 \times m$ where $m$ is Lebesgue measure on $[0,1]$. Then $p = \delta_0$, $\bar{p} = m$, but for any $f : X \to \bar{X}$, $p \circ f^{1}$ can only be a point mass at $f(0)$.