If the area under graph of $f$ is measurable then $f$ is measurable

Question

Let $(X,\mathcal{M},\mu)$ be a $\sigma$-finite measure space and $f \colon X \to [0,\infty]$ such that \begin{equation} \{ (x,y) \in X \times \mathbb{R} \colon 0<y<f(x) \} \end{equation} is measurable respect to $\mu \otimes \lambda$ where $\lambda$ is the Lebesgue measure on $\mathbb{R}$. How can I prove that $f$ is measurable?

Answer 1

If $E$ is a measurable set in $X \times \mathbb R$ then its sections are measurable. (This is part of Fubini's Theorem). Let $E$ be the graph of $f$. Fix $y$ consider the section $E^{y}\equiv \{x:(x,y)\in E\}$. This is nothing but $\{x: f(x) >y\}$. Hence $\{x: f(x) >y\}$ is measurable for each $y$ which proves that $f$ is measurable.