The following problem is proving stubborn. I humbly request assistance.
If $f$ and $g$ are integrable functions on $\mathbb R$ and $F(x,y) = f(x)g(y)$, then $F$ is measurable, integrable on $\mathbb R\times \mathbb R$ and $\int_ {\mathbb R\times \mathbb R}F~d(\mu\times \mu)=\int_{\mathbb R}f~d\mu \int_{\mathbb R}g~d\mu.$
Can I do this for the first two parts of the problem?
If I let $A$ and $B$ be measurable subsets of $\mathbb R$. Set $f = 1_A, g=1_B$. Then $f = 1_{A\times B}$, $A\times B$ is measurable, so $f$ is measurable. $1_{A\times B}$ is integrable, so $f$ is integrable on $\mathbb R \times \mathbb R$. Furthermore $\int_{\mathbb R} F~d(\mu \times \mu) = (\mu\times \mu)(A\times B) = \mu(A)\cdot \mu(B) = \int_{\mathbb R} f ~d\mu \int_{\mathbb R }g~d\mu .$
This is all I'm able to do now. How about the second part?