Product of Absolutely Continuous Measures is Absolutely Continuous I am stuck on this problem from Folland's Real Analysis, Second Edition:
For $j = 1, 2$, let $\mu_j, \nu_j$ be $\sigma$-finite measures on $(X_j, \mathcal{M}_j)$ such that $\nu_j <\!\!< \mu_j$.  Then $\nu_1 \times \nu_2 <\!\!< \mu_1 \times \mu_2$.
Here is about where I am at.
(1)  It is immediate that if $\mu_1 \times \mu_2(A \times B) = 0$, then $\nu_1 \times \nu_2(A \times B) = 0$.  
(2)  Because $\nu_1 \times \nu_2$ is $\sigma$-finite, it is enough to prove the result for when $\nu_1 \times \nu_2$ is finite.  Since the result is quickly verified if either $\nu_1$ or $\nu_2$ is the zero measure, we may assume that both $\nu_1$ and $\nu_2$ are finite.  We then have the following equivalent formulation for $\nu_1 <\!\!< \mu_1$:
For every $\epsilon > 0$, there exists $\delta > 0$ such that $\mu_1(E) < \delta$ implies $\nu_1(E) < \epsilon$.
And similarly for $\nu_2 <\!\!< \mu_2$.  
I thought I might be able to prove the same condition for $\nu_1 \times \nu_2$ with respect to $\mu_1 \times \mu_2$, which would then imply $\nu_1 \times \nu_2 <\!\!< \mu_1 \times \mu_2$ since $\nu_1 \times \nu_2$ has been reduced to being finite.  As a suggestion, following Folland's technique, it might be easier to argue by contradiction, assuming first the $\epsilon-\delta$ condition is false.  
(3)  If $\mu_1 \times \mu_2(E) = 0$, then, by definition,
$$0 = \inf \bigg\{ \sum_n \mu_1(A_n)\mu_2(B_n) \colon A_n \times B_n \text{ are rectangles such that } E \subset \bigcup_n A_n \times B_n\bigg\}.$$
To show $\nu_1 \times \nu_2(E) = 0$, we want to show that the analogous equation holds for $\nu_1 \times \nu_2$.  
I can't seem to put the pieces together, despite some effort.  
Any help would be greatly appreciated.  Thanks.  
 A: Assume that 
$E \in M_1 \times M_2$ such that
$\mu_1\times \mu_2(E)=0$. We have to show that $\nu_1\times \nu_2(E)=0$.
By Fubini theorem we have
$$0=\mu_1\times \mu_2(E)=\int_{X_1}\mu_2(E \cap (\{x\} \times X_2))d\mu_1(x).$$
This means that
$$
\mu_{1}(\{ x : \mu_2(E \cap (\{x\} \times X_2))>0 \})=0,
$$
equivalently
$$
\mu_{1}( X_1  \setminus  \{ x : \mu_2(E \cap (\{x\} \times X_2))=0 \})=0.
$$
Since $\nu_2 \ll \mu_2$ we have
$$
 \{ x : \mu_2(E \cap (\{x\} \times X_2))=0 \} \subseteq \{ x : \nu_2(E \cap (\{x\} \times X_2))=0 \}.
$$
Since $\nu_1 \ll \mu_1$ and 
$$
\mu_{1}( X_1  \setminus  \{ x : \nu_2(E \cap (\{x\} \times X_2))=0\})=0,
$$
we have 
$$
\nu_{1}( X_1  \setminus  \{ x : \nu_2(E \cap (\{x\} \times X_2))=0\})=\nu_{1}(\{ x : \nu_2(E\cap (\{x\} \times X_2))>0\})=0.
$$
Finally, we get
$$\nu_1\times \nu_2(E)=\int_{X_1}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=
$$
$$
\int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))>0\} }\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)+
$$
$$
\int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))=0\}}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=0.
$$
