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Let $f$ be a measurable nonnegative function with domain $\mathbb{R}$, such that \begin{equation} \int_a^b f(x) \ dx > 0 \end{equation} when $a < b$. Can we say that $f>0$ almost everywhere? (or $f\neq 0$ almost everywhere, because is a nonnegative function)

We work with Lebesgue measure and the integral above is a Lebesgue integral.

I'm thinking that the answer is yes, but I can't prove it.

I tried to get a contradiction if I suppose that $m(\{ f=0 \}) > 0$, or maybe a subset of $\{ f=0 \}$ with measure greater than $0$, but I got stuck.

I already showed that for any open set $G$ we have \begin{equation} \int_G f(x) \ dx > 0 \end{equation} So maybe I could find an open set $G$ such that $m(G) = m(\{ f=0 \})$, but I can't.

And that's it, I have no more ideas :(

Thanks

itnasx
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    The answer is of course, negative. $f = 1_{[a,b-r]}$ – Andrew Nov 12 '23 at 02:01
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    One extreme counterexample is the indicator function of a set which has "positive but not full measure in every interval", in which case there is not even a nontrivial interval in which $f > 0$ a.e. Here is my favourite construction of such a set. – Izaak van Dongen Nov 12 '23 at 02:04

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No: let $X \subseteq \mathbb{R}$ be a fat Cantor set and let $f$ be a function that is $0$ on $X$ and $1$ elsewhere. Since $X$ is nowhere dense, $\int_a^b f\, dx>0$ for all $a<b$ since $[a,b] \setminus X$ will contain a nontrivial interval. At the same time, by definition the set where $f$ vanishes has positive measure.

Inbo Gottlieb-Fenves
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    This also shows why our intuition can be misleading. The (wrong) proofs I were imaging involved finding a contradiction with $f$ vanishing on an interval. But "not measure zero" does not mean "contains an interval." My brain still wants to believe that even after knowing about the Cantor set. – pancini Nov 12 '23 at 02:12
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    Thanks! I didn't know about the fat Cantor set (don't judge me) – itnasx Nov 12 '23 at 02:37