Let $(X,\Sigma,\mu)$ be a $\sigma$-finte measure space and $f \in L^+(X,\Sigma)$. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$.
Theorem:
Define the area under the graph of $f$ to be $S_f = \{(x,y) \in X \times [0,\infty]\mid y \leq f(x)\}$. Then $S_f \in \Sigma \otimes \mathcal{B}(\mathbb{R})$, where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$.
Moreover, $\mu \times \lambda(S_f) = \int f\,d\mu$
My proof: $(x,y) \to f(x) - y$ is the composition of $\psi(x,y) = (f(x),y)$ and the subtraction $\phi(z,y) = z-y: [0,\infty] \times [0,\infty[ \to [-\infty,\infty[$. The former is automatically measurable since $f \in L^+(X)$, and $$ \phi^{-1}(z,y) = z + (-1)y $$ is the composite of measurable functions, and is measurable. Therefore $S_f = (\psi \circ \phi)^{-1} [0,\infty[ \cup (f^{-1}[\{\infty\}] \times \infty) \in \Sigma\otimes\mathcal{B}(\mathbb{R})$.
For the second part, extend $\lambda$ to $\mathbb{R} \cup \{\infty\}$ by setting $\lambda\{\infty\} = 0$. By Tonelli's Theorem, $$\begin{align*} \mu \times \lambda(S_f) &= \int_{X \times [0,\infty]}\chi_{S_f}\,d\mu\times\lambda \\ &= \int_X \int_{[0,\infty]} [y \leq f(x)]\,d\lambda(y)\,d\mu(x) \\ &= \int_X f(x) \,d\mu(x) \end{align*}$$ Is my proof correct?