# $f:[a,b] \rightarrow \Bbb R$ measurable function, $f\geq0$ a.e. If $f^{-1}((0,\infty))$ has measure $>0$ , $f^{-1}((1/n,\infty))$ does for some n.

This is exercise #8, p.103 if Gail S. Nelson's A User-Friendly Guide to Lebesgue Measure and Integration. Here is my attempt: For each $$n$$ let $$E_n:=f^{-1}((1/n,\infty))$$. Note that $$f$$ is measurable, so $$E_n$$ is measurable for all $$n$$. Then $$f^{-1}((0,\infty))=\bigcup_{n=1}^{\infty}E_n$$ (the reverse inclusion is clear, the forward inclusion follows from the Archimedean Property.) Then $$m(f^{-1}((0,\infty))) = m(\bigcup_{n=1}^{\infty}E_n) \leq \sum_{n=1}^{\infty} m(E_n)$$ by subadditivity. So if every $$E_n$$ had zero measure, so would $$f^{-1}((0,\infty))$$, which proves the contrapositive. I'm fairly sure this is wrong, since I haven't used that $$f$$ is nonnegative almost everywhere; I'm just not sure where I went wrong. Any help would be greatly appreciated!

In general, subadditivity shows that if $$E_n$$ has measure zero each, then so does $$\cup E_n.$$ Hence the exercise. Q.E.D.
• Either $f^{-1}((0, \infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e. – Will M. Nov 18 '18 at 6:19
Prove by contradiction. If $$f^{-1}((\frac 1 n , \infty)$$ has measure $$0$$ for each $$n$$ then the union of these sets has measure $$0$$. The union is exactly $$f^{-1}((0 , \infty)$$