# equivalence of absolute continuity of measures

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 $$(8)$$.

• 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. May 29, 2022 at 8:10

• $$\mu(dx,dy)\ll \mu_X\otimes\mu_Y(dx,dy)$$ 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 $$\int_Y f(x,y) \mu_Y(dy)$$.
• 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? May 29, 2022 at 14:31
• 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 $0=\mu(X\times A)=\int_X \int_A \mu_{Y\vert X}(x,dy)$. May 29, 2022 at 16:14