if $f \geq 0$ Lebesgue measurable the set $A := \{(x,y)\in \mathbb{R}^2 \mid 0 < y < f(x)\}$ is measurable - proof verification 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 $f(x)>r+y>0$ 
$$A = \{(x,y)\in \mathbb{R}^2 \mid f(x) - y>0 \} = \bigcup_{r\in\mathbb{Q} \\ r >0} \bigg[ \bigg( \{(x,y)\in\mathbb{R}^2 \mid f(x)>r\}\times \mathbb{R}\bigg)\cap  \bigg( \{(x,y)\in\mathbb{R}^2 \mid y>-r\}\times \mathbb{R}\bigg) \bigg]$$

I think you can try the hint of Folland's textbook which is considering
the map $(x,y) \to f(x)-y$ is the composition of $(x,y) \to (f(x),y)$ and $(z,y) \to z -y$. It is suffices to show all maps are measurable. 
