Define a coupling $\pi\in \Pi(\mu,\nu)$ on the product space $(X\times X,\mathcal{F}\times\mathcal{F})$. let $\pi\_x$ be the disintegration of $\pi$ with respect to the $\mu$, i.e. there exists a Borel measurable function $x\mapsto\pi\_x$ such that $$\pi(dx,dy)=\mu(dx)K(x,dy)$$ where $K(\cdot,\cdot)$ is a probability transition kernel.


If $\pi\_x=\delta\_{f(x)}$ ($\delta$ is the Dirac measure), can we say that there exists a measurable function $f$ such that
$$
\nu=f\_{\#}\mu
$$ where $f\_{\#}\mu$ is the pushforward of measure $\mu$, i.e. $\mu(f^{-1}(A))=\nu(A)$ for any Borel measurable sets $A\subset X$.



Can we say that there exists a measurable function $f$ such that $ \nu=f_{\#}\mu$?
Yes: if $\pi\_x=\delta\_{f(x)}$, then
$$\nu(A)=\pi(X\times A)=\int\_X\mu(dx)\pi\_x(A) \\ 
=\int\_X\mu(dx)\,1(f(x)\in A)=\int\_X\mu(dx)\,1(x\in f^{-1}(A))
=\mu(f^{-1}(A))$$
for all Borel subsets $A$ of $X$, as desired.


