Product measure, measurable set and integral

Question

I come up with is $m{ (x,y) \in. But I am not sure if it is useful. Please give me some hint.

I am really thankful for your help

Answer 1

$G=\{(x,y):y>0, (x,f(x)) \in A\}$ where $A=\{(a,b): b>a\}$. $A$ is an open set and the function $x \to (x,f(x))$ is measurable: $\{x: (x,f(x)) \in C\times D\}=C\cap f^{-1}(D)\}$ is measurable for any Borel sets $C$ and $D$ in $\mathbb R$. [ Sets of the type $C\times D$ generate the Borel sigma algebra of $\mathbb R^{2}$]. Now $(m\times m) (G)=\int \int _0^{f(x)} dy dx=\int f(x) dx$ by Fubini's Theorem.

Answer 2

Hint: The function $g(x,y) = f(x) - y$ is measurable (why?). Can your set G be realized as the pre-image of a measurable set under $g$?

Hint 2: Once you prove that the set $G$ is measurable, then the indicator function on $G$ will be a measurable function. Note that the measure of the $x$-section $G$ is $f(x)$. Apply Tonelli's theorem to $m \ x \ m(G)$ to conclude.