Let $(X,\mathcal{M})$ and $(Y, \Sigma)$ be two measure spaces, and let $\mu_1$ and $\mu_2$ $\sigma$-finite measures defined respectively in $X$ and $Y$. Now let $\nu_1$ and $\nu_2$ finite measures in $X$ and $Y$ such that $\nu_1 \ll \mu_1$ and $\nu_2 \ll \mu_2$. We want to prove that $\nu_1 \otimes \nu_2 \ll \mu_1 \otimes \mu_2$, where $\otimes$ denote the product of measures.
My attempt is the following: take $A_1 \in \mathcal{M}$ and $A_2 \in \Sigma$ such that $\mu_1(A_1)\mu_2(A_2) = \mu_1\otimes \mu_2(A_1 \times A_2) = 0$. Then, using the absolute continuity, we deduce that $$ \nu_1\otimes\nu_2(A_1\times A_2) = \nu_1 (A_1) \nu_2(A_2) = 0.$$
I don't know if it's enough to prove what I want. I mean, $\nu_1 \otimes \nu_2$ and $\mu_1 \otimes \mu_2$ are defined in $(\mathcal{M}\times \Sigma)^\sigma$, id est, the $\sigma$-algebra generated by the product of the $\sigma$-algebras.
Is it enough to prove this property for the sets of $\mathcal{M}\times \Sigma$? I think we must ask them to be at least a semi-ring, but I'm not very sure.
Can anyone help me? Thank you very much.
Best Answer
Let $(\mu_1 \otimes \mu_2) (E)=0$. By Fubini's Theorem $\mu_1(E_x)=0$ for almost all $x$ w.r.t. $\mu_2$ (where $E_x=\{y:(x,y) \in E\}$. Since $\nu_2 << \mu_2$ we see that $\mu_1(E_x)=0$ for almost all $x$ w.r.t. $\nu_2$. Since $\nu_1 << \mu_1$ we see that $\nu_1(E_x)=0$ for almost all $x$ w.r.t. $\nu_2$. Another application of Fubini's Theorem gives $(\nu_1 \otimes \nu_2) (E)=0$.