I have the following problem:
Let $f$ be a non-negative function defined on a Lebesgue measurable subset $ E \subset [0,1]$. Show that $f$ is Lebesgue measurable if the region $$A=\{\ (x,y): x \in E, \ f(x) \geq y\}$$is a Lebesgue measurable subset of $\mathbb{R}^2$.
My attempt on a solution:
$f: E \to [0, +\infty)$ is Lebesgue measurable $\iff$ for any $y \geq 0$ we have $f^{-1} ( [y, +\infty) ) \in \mathcal{L}(\mathbb{R}) \cap E.$
Now notice that $ f^{-1} ( [y, +\infty) ) = \{ x \in E \ | \ f(x) \geq y \} = A^y$ - the $y$-section of the set $A$.
We can apply Cavaliery's principle for complete measures (consequence of Fubini-Tonelli's theorem for complete measures):
Since $A \in \mathcal{L}(\mathbb{R}^2) = \overline{\mathcal{L}(\mathbb{R}) \otimes \mathcal{L}(\mathbb{R})}$, we conclude that $A^y \in \mathcal{L}(\mathbb{R})$ for almost any $y \in \mathbb{R}$ (or $y \in [0, + \infty)$ for that matter).
This is close, but how to get rid of the "almost any $y$" part?
Update:
I finally found a slightly different approach which allows to take care of this nuisance (essentially that is what William hinted at in the comments):
Define the set $F=A \cap ( E \times [0, +\infty) )=\{ (x,y) \ | \ x \in E, f(x) \geq y \geq 0 \} \in \mathcal{L}(\mathbb{R}^2)$.
Take the indicator $\chi_F: \mathbb{R}^2 \to \mathbb{R}$ which is therefore measurable.
Applying Fubini-Tonelli's theorem for complete measures, we get that almost everywhere defined function $g(x) = \int \chi_{F}^{x} (y) d \lambda(y) = \lambda (\{y \in \mathbb{R} \ | \ f(x) \geq y \geq 0 \} ) = f(x)$ is measurable.
Going back to the proof of Fubini-Tonelli's thm for complete measures, we see that $g(x)$ can be extended to a measurable function (that is what people mean when they use the term "almost everywhere defined"). Anyway, we get that $f(x)$ is almost everywhere equal to a measurable function, and since we are on a complete measure space, the result follows.