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

Question

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!

Answer 1

In general, subadditivity shows that if $E_n$ has measure zero each, then so does $\cup E_n.$ Hence the exercise. Q.E.D.

Answer 2

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 $f^{-1}((0 , \infty)$