Product measures and absolute continuity For $j=1,2$, let $\mu_j,\nu_j$ be $\sigma-$finite measures on $(X_j, M_j)$ such that $\nu_j<<\mu_j$. Prove that $\nu_1 \times \nu_2 << \mu_1 \times \mu_2$.
Here, does it suffices to prove that $\nu_1 \times \nu_2(A\times B)=0$ whenever $\mu_1 \times \mu_2(A\times B)=0$ for any box $A\times B$? 
Or should I prove that $\nu_1 \times \nu_2(E)=0$ whenever $\mu_1 \times \mu_2(E)=0$ for any $E\in M_1\otimes M_2$?
 A: Assume that 
$A \in M_1 \otimes M_2$ such that
$\mu_1\times \mu_2(A)=0$. We have to show that $\nu_1\times \nu_2(A)=0$.
By Fubini theorem we have
$$0=\mu_1\times \mu_2(A)=\int_{X_1}\mu_2(A \cap (x \times X_2))d\mu_1(x).$$
This means that
$$
\mu_{1}(\{ x : \mu_2(A \cap (x \times X_2))>0 \})=0,
$$
equivalently
$$
\mu_{1}( X_1  \setminus  \{ x : \mu_2(A \cap (x \times X_2))=0 \})=0.
$$
Since $\nu_2 \ll \mu_2$ we have
$$
 \{ x : \mu_2(A \cap (x \times X_2))=0 \} \subseteq \{ x : \nu_2(A \cap (x \times X_2))=0 \}.
$$
Since $\nu_1 \ll \mu_1$ and 
$$
\mu_{1}( X_1  \setminus  \{ x : \nu_2(A \cap (x \times X_2))=0\})=0,
$$
we have 
$$
\nu_{1}( X_1  \setminus  \{ x : \nu_2(A \cap (x \times X_2))=0\})=\nu_{1}(\{ x : \nu_2(A \cap (x \times X_2))>0\})=0.
$$
Finally, we get
$$ \nu_1\times \nu_2(A)=\int_{X_1}\nu_2(A \cap \{x \times X_2\})d\nu_1(x)=$$
$$\int_{\{ x : \nu_2(A \cap (x \times X_2))>0\} }\nu_2(A \cap (x \times X_2))d\nu_1(x)+$$
$$\int_{\{ x : \nu_2(A \cap (x \times X_2))=0\}}\nu_2(A \cap (x \times X_2))d\nu_1(x)=0.
$$
