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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 $A\subset X$.

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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.

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  • $\begingroup$ Why is $f$ measurable? $\endgroup$
    – Dieter Kadelka
    Jul 17, 2022 at 19:37
  • $\begingroup$ @DieterKadelka : This follows because $(\pi_x)$ is a disintegration of $\pi$, which implies that all the equalities in my answer must make sense. If $f$ were not measurable, then the latter integral there, as well as $\mu(f^{-1}(A))$, would not make sense for some Borel $A$. $\endgroup$
    – Iosif Pinelis
    Jul 17, 2022 at 19:48

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