Show that $E=\{(x,\alpha )\mid|f(x)|>\alpha\geq 0 \}$ is measurable, where $f$ is integrable.
Is my proof correct?
Let $A$ measurable and $f=1_A$. Then $E=A\times [0,1]$ which is measurable.
If $f=\sum \limits _{i=1}^n a_i 1_{A_i}$ is simple, then $E=\bigcup \limits _{i=1}^n A_i\times [0,a_i]$ which is measurable. If $f\geq 0$, there is an increasing sequence $(f_n)$ of simple function s.t. $f_n\to f$.
If I denote $E^n=\{x\mid |f_n(x)|>\alpha \geq 0\}$ then $E^n\subset E^{n+1}$ and $E=\bigcup \limits _{n=1}^\infty E^n$. Since $E$ are measurable and the union is countable, then $E$ is measurable. If $f$ is measurable, then we can take a sequence $|f_n|\to |f|$ of simple function that is increasing and do as we did previously.
If $f=1_A$, then $E=\{(x,\alpha )\in A\times \mathbb R \mid 0\leq \alpha <1\}=A\times [0,1)$. Same with $f$ simple. Otherwise your proof is perfectly correct.