Let ($\Omega$, $\mathcal{A}$, $\mu$) be a $\sigma$-finite measure space and $f: \Omega \rightarrow \mathbb{R}$ be measurable and nonnegative. Define \begin{equation} G_f = \left\{(\omega, y) \in \Omega \times [0, \infty ) : y ≤ f(\omega)\right\}. \end{equation} Show that $G_f \in \mathcal{A} ⊗ \mathcal{B}$ and $(\mu ⊗ m)(G_f) = \int_Ω f d\mu$.
The first part was pretty doable just by proving that $\omega \in \mathcal{A}$ and $y \in \mathcal{B}$ separately by the definitions of the $\sigma$-finite algebra and the Borel $\sigma$-algebra.
But I don't really understand how to show the second part. I see that it measures the area under $f$, but I don't know how to formally show that.
$$\begin{aligned}\left(\mu\otimes m\right)\left(G_{f}\right) & =\int1_{G_{f}}d\left(\mu\otimes m\right)\\ & =\int\int1_{G_{f}}\left(\omega,y\right)m\left(dy\right)\mu\left(d\omega\right)\\ & =\int\int1_{\left[0,f\left(\omega\right)\right]}\left(y\right)m\left(dy\right)\mu\left(d\omega\right)\\ & =\int m\left(\left[0,f\left(\omega\right)\right]\right)\mu\left(d\omega\right) \end{aligned} $$
If $m$ denotes the Lebesgue measure then we end up with: $$\left(\mu\otimes m\right)\left(G_{f}\right)=\int m\left(\left[0,f\left(\omega\right)\right]\right)\mu\left(d\omega\right)=\int f\left(\omega\right)\mu\left(d\omega\right)=\int fd\mu$$