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I am working through a Real Analysis textbook and found the following exercise.

Let $f\,:\,[a,b]\rightarrow\mathbb R$ be Riemann integrable and nonnegative. Suppose $$\int_a^bf(x)\,dx=0.$$ Find $f$.

I know that this should come out to be that $f(x)=0$ logically, but how could I show this with some rigor?

Alexander
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  • Well... it's not actually true that $f(x)=0$ is the only such function. It may be zero except at countably many discontinuities where the function is nonzero. Usually such a problem assumes $f$ is continuous instead of merely Riemann integrable, and under that assumption, only $f(x)=0$ works. – Brian Moehring Feb 06 '21 at 02:54
  • @BrianMoehring Ah, I see. I'm still not completely sure how to approach the proof for that though. – Alexander Feb 06 '21 at 02:57
  • Do you know that Riemann integrable functions are discontinuous at countably many points? – Brian Moehring Feb 06 '21 at 03:02
  • @BrianMoehring Yes, my understanding is that this is the Lebesgue Criteria. – Alexander Feb 06 '21 at 03:20
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    Then I would show that if such an $f$ is continuous at $x=c$, then $f(c)=0$ (by contradiction). – Brian Moehring Feb 06 '21 at 03:34

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