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Suppose that f : [0, 1] −→ (0, ∞) is a Riemann-integrable function. Prove that the integral is strictly positive.

Here is my thought process:

The function is riemann integral implies that the set of discontiuties has measure zero. Assume that f is contiuous at some c in [0,1] This implies that there exists some r,z such that f>=r on [c-z,c+z] which implies (????) that the integral on [0,1] is bigger than the integral on [c-z,c+z] is bigger than or equal to 2zr, which is positive.

I am not sure if I am making any assumptions here that I can't make, especially where I marked with (????)

23523rff
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  • I would think that you could appeal to partitions, and take a simple one or two element partition and show that the lower bound must be positive and therefore the integral must be positive. – abiessu Mar 31 '16 at 00:05
  • I find your reasoning correct. You need to pick $r>0$, and you need to note that the integral on the rest of $[0,1]$ besides $[c-z,c+z]$ is nonnegative and so can't cancel the positive value of the integral on $[c-z,c+z]$. – ForgotALot Mar 31 '16 at 00:07
  • Related: http://math.stackexchange.com/questions/351157/is-the-riemann-integral-of-a-strictly-positive-function-positive –  Mar 31 '16 at 00:09
  • Welcome to math.stackexchange, I'd recommend you the page http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference for future reference. It will make your question much more easier to read. – BigbearZzz Mar 31 '16 at 01:16

1 Answers1

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If $f$ is nonnegative and has $0$ integral, it must be $0$ a.e., but it isn't, so the integral must be greater than $0$.

Henricus V.
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