Let us have $f,g$ measurable functions in a measure space. We have the following set:
{$x: f(x) < y < g(x)$}. Prove, that if the measure of this set is $0$ for every $y \in \mathbb{R}$, then the measure of {$ x: f(x) < g(x)$} set is $0$ too.
This question seems somehow trivial for me. If $f(x) < g(x)$, we can always find such $y$, that $f(x) < y < g(x)$, so the two sets must be equivalent, therefore the statement is proven.
Am I thinking something wrong? Any help appreciated.
Hint: If $x$ is such that $f(x) < g(x)$, then there exists a rational $y$ such that $f(x) < y < g(x)$. Now, consider the appropriate countable union.