Showing that a set is Lebesgue Measurable in Higher Dimensions and Applying Fubini's Theorem I have an idea of how to proceed, but I'm suspicious that my efforts were of no use. Let $A\in\mathcal{M}$ be Lebesgue measurable, and let $g,h:A \rightarrow \bar{\mathbb{R}}$ be Lebesgue measurable functions with $g(x)\leq h(x) \ \forall x \in A$. With this, we define $E=\{(x,y) \in \mathbb{R^2} : g(x) \leq y \leq h(x)\}$. Prove that $E\in \mathcal{M^2}=\mathcal{M} \times \mathcal{M}$.
And following this, show that if $f:E \rightarrow [0,\infty]$ is $\mathcal{M^2}$ measurable, then
$\displaystyle\int_Ef dm^2=\int_A(\int_{g(x)}^{h(x)}f(x,y) \ dy) \ dx$, where $m^2=m \times m$.
For the first part, may we write $E=\{(x,y) \in \mathbb{R^2} : g(x) \leq y\} \cap \{(x,y) \in \mathbb{R^2} : y \leq h(x)\}$
$\Rightarrow E=\bigcup\limits_{q \in \mathbb{Q}} \{(x,y) \in \mathbb{R^2} : g(x) \leq q \leq y\} \cap \{(x,y) \in \mathbb{R^2} : y \leq q \leq h(x)\}$
and since $h$ and $g$ are measurable, this is a countable union of measurable sets? I don't feel so good about this reasoning, as I'm pretty sure that I would have to find a way to rewrite $E$ as cartesian product of 2 Lebesgue measurable sets (ideally, one with respect to x and the other with respect to y).
So perhaps $E=(g^{-1}[-\infty,y]\cap h^{-1}[y,\infty]) \times \{y\}$ works instead?
As for the second part, I'm not particularly sure how to use Fubini's in this scenario. I'd be very gracious for any help and guidance.
 A: It is easier to use that $H \colon \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $H(x,y) = -x+y$ is continuous and thus measurable. Therefore also $$h_1(x,y)=H(g(x),y) = -g(x)+y \quad \text{and} \quad h_2(x,y)=H(h(x),y) = -h(x)+y$$ are measurable too - as functions from $A \times \mathbb{R}$ to $\mathbb{R}$.
Note that we have used that the composition of measurable functions are measurable and that $$(x,y) \rightarrow (g(x),y), \quad \text{resp.} \quad (x,y) \rightarrow (h(x),y),$$ is measurable. This can checked for product sets easily. Since we only need to show measurability on a generator, this already proves the measurability according to the product-$\sigma$-algebra.
Coming back to the first  step:  We see that the set $$h^{-1}_1([0,\infty) \cap h_2^{-1}((-\infty,0]) = \{ (x,y) \in A \times \mathbb{R} : g(x) \le y \le h(x)\} $$
is measurable. 
The second part can be shown by Fubini's theorem as follows: We have
\begin{align}
\int_{\mathbb{R}^2} f(x,y) 1_{E}(x,y) \, \lambda^2(x,y) &= \int_{\mathbb{R}} \int_{\mathbb{R}} f(x,y) 1_{E}(x,y) \,dy \, dx \\
&=\int_{A} \int_{[g(x),h(x)]} f(x,y) 1_{E}(x,y) \,dy \, dx. 
\end{align}
In the last line we have used that $g(x) \le h(x)$ for all $x \in A$.
