# Can we say that there exists a measurable function $f$ such that $\nu=f_{\#}\mu$?

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 $$A\subset X$$.

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.
• Why is $f$ measurable? Jul 17, 2022 at 19:37
• @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 $A$. Jul 17, 2022 at 19:48