Let $f : \mathbb{R} \to \mathbb{R}^+$ be a positive, Lebesgue measurable function.
Denote the set $A := \{(x,y)\in \mathbb{R}^2 \mid 0 < y < f(x)\}$ .
Show that $A$ is Lebesgue measurable.
Is the following correct?
$\textbf{My attempt:}$
we can write that there exists a rational $r$ that $f(x)-y >r>0$ so $f(x)>r+y>0$
$$A = \{(x,y)\in \mathbb{R}^2 \mid f(x) – y>0 \} = \bigcup_{r\in\mathbb{Q} \\ r >0} \bigg[ \bigg( \{(x,y)\in\mathbb{R}^2 \mid f(x)>r\}\times \mathbb{R}\bigg)\cap \bigg( \{(x,y)\in\mathbb{R}^2 \mid y>-r\}\times \mathbb{R}\bigg) \bigg]$$
Best Answer
You can write $$\{ 0 < y < f(x)\} = \{y>0\}\cap\{f(x)-y>0\} = \{y>0\}\cap P^{-1}(0,\infty)$$ where $$P:\mathbb R^2 \to \mathbb R , \quad P(x,y) = f(x) - y.$$ $P$ is measurable, so the result follows.
Closer to your attempt (along the lines of Jake's comments)- For $q\in\mathbb Q_+$, let $$A_q = B_q \times (0,q), \quad B_q = \{x\in\mathbb R : 0 < q < f(x)\}.$$ For $x\in B_q$, we have $f(x)>q$, i.e. $(x,q)\in A$. Thus $A_q \subset A$. Therefore $\bigcup_q A_q\subset A$.
For $(x,y)\in A$, we have $(x,r)\in A$ for some rational $0<y<r<f(x)$, and $x\in B_r$. Thus $(x,y)\in A_r \subset \bigcup_q A_q, $ and therefore $A \subset \bigcup_q A_q$. In conclusion: $$A = \bigcup_{q\in\mathbb Q_+} A_q.$$