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Suppose I have two separable measurable spaces $X$ and $Y$ with finite measures $\mu_X$ and $\mu_Y$. Then I build the product space $Z=X\times Y$ with the product measure $\mu_X\otimes\mu_Y$. Moreover we assume a joint measure $\mu(dx,dy)$ is also given.

I have some troubles seeing why the following two assumptions should be equivalent

$$\mu(dx,dy)\ll \mu_X\otimes\mu_Y(dx,dy) \iff \mu_{Y|X}(\cdot,dy)\ll \mu_Y(dy) \text{ $\mu_X$-a.s}$$

where $\mu_{Y|X}(x,dy)=f(x,y)\mu_Y(dy)$ is the conditional measure for the Randon-Nikodym derivative.

What always holds for any measure on $\mu$ on $Z$ is $\mu(dx,dy)=\mu_{Y|X}(x,dy)\mu_X(dx)$. The question arises from this paper, equation $(4)$ and $(8)$.

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  • $\begingroup$ Maybe you should start by rewriting the two statements more explicitly? For example by writing what $\mu(dx,dy) \ll \mu_X\otimes \mu_Y (dx,dy)$ is. It would also help if you write what you have tried so far. $\endgroup$ May 29, 2022 at 8:10

1 Answer 1

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Hint

You can translate these statements to:

  • $\mu(dx,dy)\ll \mu_X\otimes\mu_Y(dx,dy)$ means that every $G\in \mathcal{B}_X\otimes \mathcal{B}_Y$ which satisfies $\mu_X\otimes \mu_Y(G)=0$, implies also that $\mu(G)$.

  • $\mu_{Y|X}(\cdot,dy)\ll \mu_Y(dy) \text{ $\mu_X$-a.s}$ means that for $\mu_X$ almost every $x\in X$ and for every $A\in \mathcal{B}_Y$ satisfying $\mu_Y(A)=0$, implies also that $\mu_{Y\vert X}(x,A)=0$. $\mu_{Y\vert X}(x,A)$ also equals to $\int_Y f(x,y) \mu_Y(dy)$.

Using these more explicit formulations and what you said, can you use this and perhaps Fubini's theorem, to show the if and only if?

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  • $\begingroup$ thanks for the hint. Can we argue for $1\Rightarrow 2$. let $A\in\mathcal{B}_Y$ s.t. $\mu_Y(A)=0$. Then $\int_X\int_A\mu_{Y|X}(x,dy)\mu_X(dx) = \int_{X\times A}\mu_X\otimes\mu_Y(dx,dy)=\int_A\int_X\mu_{X|Y}(dx,y)\mu_Y(dy) = 0$ since $\mu_Y(A)=0$. Hence $\mu_{Y|X}(x,A)$ for a.e. $x$? Similar for the other direction? $\endgroup$
    – swissy
    May 29, 2022 at 14:31
  • $\begingroup$ I don't think you can say that exactly. From your phrasing $\mu$ and $\mu_X\otimes \mu_Y$ are not the same. I was thinking more along the lines that $\mu_X(X)\otimes \mu_Y(X\times A)= \mu_Y(A)\cdot \mu_X(X)=0$ and therefore $0=\mu(X\times A)=\int_X \int_A \mu_{Y\vert X}(x,dy)$. $\endgroup$ May 29, 2022 at 16:14

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