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.
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 $x \in A$.