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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!

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

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  • $\begingroup$ So is the given hypothesis that f is nonnegative almost everywhere not even necessary for this problem? $\endgroup$ – Alex Sanger Nov 16 '18 at 23:46
  • $\begingroup$ Either $f^{-1}((0, \infty))$ has measure $> 0$ is necessary or else, $f > 0$ a.e. $\endgroup$ – Will M. Nov 18 '18 at 6:19
  • $\begingroup$ Ah, I see! Thanks for your replies! $\endgroup$ – Alex Sanger Nov 18 '18 at 22:43
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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 $f^{-1}((0 , \infty)$

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