# If $\nu_1 \ll \mu_1$ and $\nu_2 \ll \mu_2$ then $\nu_1 \otimes \nu_2 \ll \mu_1 \otimes \mu_2$

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.

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 $$(\nu_1 \otimes \nu_2) (E)=0$$.