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.

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

$$\nu_1\times \nu_2(E)=\int_{X_1}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=$$ $$\int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))>0\} }\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)+$$ $$\int_{\{ x : \nu_2(E \cap (\{x\} \times X_2))=0\}}\nu_2(E \cap (\{x\} \times X_2))d\nu_1(x)=0.$$