$f$ measurable iff $f^2$ measurable and $\{f > 0\}$ measurable

Question

I know for sure that if $f^2$ is measurable, that doesn't imply that $f$ is measurable, but how does the condition:

$$ \{f > 0\} \text{ measurable }$$

play into making $f$ automatically measurable?

Answer 1

Hint: For $a>0$, we have $\{f>a\} =\{f^2 >a^2\} \cap\{f>0\}$.

Answer 2

Another Hint: It holds $$f =\sqrt{f^2}1_{\{f>0\}}-\sqrt{f^2}1_{\{f>0\}^c}. $$