Show that $\frac{d(\alpha\times \beta)}{d( \mu\times \nu)}(x,y)=\frac{d\alpha}{d\mu}(x)\frac{d\beta}{d\nu}(y)$ Suppose $(X,\mathcal M,\alpha)$ and $(Y,\mathcal N,\beta)$ are positive finite measure spaces and $(X,\mathcal M,\mu)$ and $(Y,\mathcal N,\nu)$ are positive $\sigma$-finite measure spaces with $\alpha\ll \mu$ and $\beta\ll \nu$.  Prove that $\alpha\times \beta \ll \mu\times \nu$ and
$$\frac{d(\alpha\times \beta)}{d( \mu\times \nu)}(x,y)=\frac{d\alpha}{d\mu}(x)\frac{d\beta}{d\nu}(y)$$
for a.e. $(x,y)\in X\times Y$.
I think showing that $\alpha\times \beta \ll \mu\times \nu$ will be an application of the radon-nikodym theorem, but I don't see how exactly. I've been struggling with this problem for over a week, if somebody could hold me hand on this one and walk me through it you'd be a lifesaver!!
 A: Showing $(\alpha \times \beta) \ll (\mu \times \nu)$ looks difficult.
After all, how would you show something like $$(\alpha \times \beta)A = \int_{A} \frac{d \alpha}{d \mu} (x)\frac{d \beta}{d \nu}(y)(\mu \times \nu)(x,y) \tag{1}$$ for every set $A$ in $M \otimes N$, the product sigma-algebra? It seems difficult because the nature of $A$ is not ascertainable from its membership in the product sigma-algebra, so we can't really simplify both sides of the equation.
The idea is to show that $\mathcal S = \{A \in M \otimes N : (1) \text{ holds for } A\}$ is a sigma-algebra which contains a generating set of $M \otimes N$. 
What would the generating set be? Quite naturally, it would be the set of all rectangles in $M \times N$. That is:
$$
\mathcal T = \left\{A \times B : A \in \mathcal M, B \in \mathcal N \right\}
$$
therefore, it is enough to show that for any $A \in \mathcal T$, the identity $1$ holds, and that $\mathcal S$ is a sigma-algebra.

Why does $(1)$ hold for $A \in \mathcal S$? Use the definition of the product measure. On any set of the form $A \times B$,
$$
(\alpha \times \beta)(A \times B) = \alpha(A) \times \beta(B) = \int_{A} \frac{d \alpha}{d \mu}(x)dx \int_B \frac{d \beta}{d \nu}(y)dy
$$
Now, by definition of the product measure, the integral on the RHS of $(1)$ also simplifies. Recall that the integral of a function over a set in a product sigma-algebra is given by the integral of the slice functions (can take either variable first by Fubini's theorem). That is, if $f :  X \times Y \to \mathbb R$ is integrable, then :
$$
\int_{X \times Y} fd(\alpha \times \beta)(x,y) = \int_X \left(\int_Y f_x d\beta(y)\right) d \alpha(x) = \int_{Y} \left(\int_X f_x d \alpha(x)\right)d\beta(y) 
$$
where $f_x(y) = f(x,y) : Y \to \mathbb R$, similarly $f_x$.
With that, observe that if a function $h(x,y)$ is of the form $h(x,y) = f(x)g(y)$, then :
$$
\int_Y h_x d \beta(y) = f(x)\int_Y g(y) d \beta(y) 
$$
where $\int_Y g(y) d \beta(y)$ is a constant independent of $x$, so it can be slipped out of any integral involving $x$.
Therefore, using this to simplify the RHS of $1$ :
$$
\int_{A \times B} \frac{d \alpha}{d \mu}(x) \frac{d \beta}{d \nu}(y) (\mu \times \nu)(x,y) = \int_A \frac{d \alpha}{d \mu}(x)d \mu(x) \int_B \frac{d \beta}{d \nu} d \nu(y) 
$$
which proves $1$ for $A \times B$.

Next, we show that $\mathcal S$ is a sigma-algebra. For this, we employ the Dynkin $\pi-\lambda$ theorem. To show this , we need a $\pi$ system which generates $M \otimes N$, and need to show that $\mathcal S$ is a $\lambda$ system.
The $\pi$ system is straightforward enough : the set of all finite disjoint unions of sets from $\mathcal T$ is a $\pi$ system (one can easily check this). That $\mathcal S$ contains this set follows from the fact that the product is integral is finitely additive.
Now, $\mathcal S$ is a $\lambda$ system needs to be checked. Check the conditions on Wikipedia for yourself( the second set of conditions) and see that they are satisfied (it is fairly clear).
$(1)$ clearly implies that $\alpha \times \beta \ll \mu \times \nu$.
Therefore, $(1)$ holds on $M \otimes N$. Because the RHS should also be $\int_A \frac{d(\alpha \times \beta)}{d (\nu \times \mu)}(x,y) (\mu \times \nu)(x,y)$, from $(1)$ holding for all such subsets we get by uniqueness of the RN derivative , the equality a.e. of the required functions.
