Regular conditional probability and absolute continuity

Question

Let's say I have two measure spaces $(X,\mathcal{F},\mu)$ and $(Y,\mathcal{G},\nu)$. I can consider the measure $\mu\times\nu$ over the product measurable space $(X\times Y, \mathcal{F} \times \mathcal{G})$. Let us also say that there is a coupling $\pi$ of $\mu,\nu$ such that $\pi \ll \mu\times\nu$. I.e., another probability measure over the same measurable space and such that $\pi(X\times A) = \nu(A), \pi(B\times Y) = \mu(B$). From my understanding, if the spaces are Radon, I can define a regular conditional probability and disintegrate $\pi$ using $\mu$ and this conditional measure. I am trying to show that $\mu\times\nu \ll \pi$. Indeed, let us call $\omega$ this conditional measure. I have that for a given measurable set $E$, $$(\mu \times \nu)(E) = \int_Y \mu(E_y) d\nu(y) = \int_X \nu(E_x)d\mu(x),$$ where $E_y = \{ x \in X : (x,y)\in E\}$ and $E_x$ is similarly defined. Similarly, for $\pi$, $$ \pi(E) = \int_Y \mu(E_y) dw(y|E_y).$$ If $\mu(E_y)=0$ for every $y$ then we clearly satisfy the absolute continuity constraint. Assume then that there exists a $y$ s.t. $\mu(E_y)\neq 0$ and that $\pi(E)=0.$ Using the disintegration in the other direction we also have that $$\pi(E) = \int_X \nu(E_x)d\xi(x|E_x).$$ Similar reasoning for $\nu(E_x)$. If $\pi(E)=0$ then, denoting with $\hat{y}$ the element such that $\mu(E_\hat{y})\neq 0$ we must have that $\omega(\{\hat{y}\}|E_\hat{y})=0$. Is it possible that $\nu(\{\hat{y}\})\neq 0 $?

Answer 1

Correct me if I've misunderstood, but it seems to me like you're trying to prove that $\mu \times \nu \ll \pi$ for any coupling $\pi$ of $\mu$ and $\nu$. But this is definitely false in general. For example, let $\mu$ and $\nu$ both be Lebesgue measure on $[0,1]$, and $\pi$ be (normalized) Lebesgue measure on the diagonal $\{ (x,x) : 0 \leq x \leq 1 \}$. Then it's clear that $\mu \times \nu$, which is $2$-dimensional Lebesgue measure, is not absolutely continuous with respect to $\pi$.