This is exercise 3.12 from Folland's Real Analysis. It took me a long times to come up with a solution to this problem, and I'd appreciate it if anyone could verify if my answer is correct.
For $j=1,2,$ let $\mu_j, \nu_j$ be $\sigma$-finite measures on $(X_j,\mathcal{M}_j)$ such that $\nu_j \ll \nu_j$. Then $\nu_1 \times \nu_2 \ll \mu_1 \times \mu_2$ and
$$\frac{d(\nu_1\times \nu_2)}{d(\mu_1 \times \mu_2)}(x_1,x_2)=\frac{d\nu_1}{d\mu_1}(x_1)\frac{d\nu_2}{d\mu_2}(x_2).$$
The theorems I used from the text are captured at the bottom.
Proof:
First, assume that $\mu_1 \times \mu_2(E)=0.$ Then $\mu_1 \times \mu_2 (E)=\int \mu_1(E^{x_2})d\mu_2=0$. So $\mu_1(E^{x_2})=0$ $\mu_2$-a.e. Since $\nu_1 \ll \mu_1$, for a.e. on $X_2$, $\nu_1(E^{x_2})=0$, and hence $\nu_1 \times \nu_2 (E)=\int \nu_1(E^{x_2})d\nu_2=0$, which shows that $\nu_1 \times \nu_2 \ll \mu_1 \times \mu_2$.
To prove the second part, by definition of the Radon-Nikodym derivative and its a.e.-uniqueness, it will suffice to show that for all $E \in \mathcal{M}\otimes \mathcal{N}$, we have
$$\int_E \frac{d(\nu_1\times \nu_2)}{d(\mu_1 \times \mu_2)}d(\mu_1 \times \mu_2)=\nu_1 \times \nu_2 (E)=\int_E \frac{d\nu_1}{d\mu_1}(x_1)\frac{d\nu_2}{d\mu_2}(x_2)d(\mu_1 \times \mu_2).$$
To show this I need to use the Tonelli's theorem restricted on a measurable set. Namely, $\int_E f d(\mu \times \nu)=\int f\chi_E d(\mu \times \nu)=\int [\int f_x \chi_{E_x}d\nu]d\mu=\int[\int_{E_x} f_x d\nu] d\mu$. (I believe this is correct and my proof hinges on it, so tell me if it's wrong).
Finally, we have
$\nu_1 \times \nu_2(E)=\int \nu_2(E_{x_1})d\nu_1=\int \nu_2(E_{x_1})\frac{d\nu_1}{d\mu_1} d\mu_1=\int \int_{E_{x_1}}\frac{d\nu_2}{d\mu_2}d\mu_2 \frac{d\nu_1}{d\mu_1} d\mu_1 $. where the third equality follows from Prop 3.9(a) and the last equality follows from the definition of $d\nu_2/d\mu_2$.
Also,
$\int_E \frac{d\nu_1}{d\mu_1}(x_1)\frac{d\nu_2}{d\mu_2}(x_2) d(\mu_1 \times \mu_2)=\int \int_{E_{x_1}}\frac{d\nu_1}{d\mu_1}(x_1)\frac{d\nu_2}{d\mu_2}(x_2) d\mu_2(x_2) d\mu_1(x_1) $,
from the Tonelli's theorem I've derived above. Now we have the desired equality and so the two derivatives are equal a.e. The problem doesn't state the equality in terms of a.e., but I believe it is just omitted. I'd appreciate any comments on this.
