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Suppose $f:[0,1] \to (0,\infty)$ is a Riemann integrable function. Prove that the integral of the function from $0$ to $1$ is strictly positive.

I have been trying to do this for awhile but I can't seem to get it. Here is my thought process: If the function is Riemann integrable, then its set of discontinuities has measure zero (Lebesgue). I am not sure how to connect this to the integral being positive

user26857
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JD162
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1 Answers1

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Say $f$ is continuous at $x$. Then there exist $r>0$ and $\delta>0$ so that $f\ge r$ on $[x-\delta,x+\delta]$. This shows easily that $$\int_0^1 f\ge\int_{x-\delta}^{x+\delta}f\ge2\delta r>0.$$

user26857
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David C. Ullrich
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  • Aren't you implicitly using the fact that the integral of the positive function is positive in the step $$\int _0^1 f\geq \int _{x-\delta}^{x+\delta }f$$ ? – eyedropper Mar 11 '16 at 16:22
  • @eyedropper: $f$ must be continuous at some point $x\in I$. As $f(x)>0$, this means that there is an $r>0$ and an interval $I_1\subseteq I$ such that $f>r$ on $I_1$ so $f$ is being compared to $r$ and obviously then $\int f\geq r\vert I_1\vert>0$ – Matematleta Mar 11 '16 at 16:31
  • That's all fine, I was just remarking upon the first step (the one I mentioned in the comment). By linearity we have $$\int _0^1f-\int _{x-\delta }^{x+\delta }f=\int _0^{x-\delta }f+\int _{x+\delta }^1$$ and the condition $$\int _0^1f\geq\int _{x-\delta }^{x+\delta }f$$ seems implied by $$\int _0^{x-\delta }f+\int _{x+\delta }^1f\geq 0$$ which is what should be proved. – eyedropper Mar 11 '16 at 16:41
  • @eyedropper No, that's not what should be proved. That $f\ge 0$ implies $\int f \ge 0$ is easy. – zhw. Mar 11 '16 at 17:47
  • @eyedropper No. I'm implicitly using the fact that $f\ge0$ implies $\int f\ge0$. That's very easy from the definition. – David C. Ullrich Mar 11 '16 at 19:31
  • @DavidC.Ullrich but isn't that what we're asked to prove? That f≥0 implies ∫f≥0 – JD162 Mar 11 '16 at 20:01
  • @eyedropper No! The problem is to show that $f>0$ implies $\int f>0$. That's much harder. Look. Suppose $f\ge0$. Then every Riemann sum is $\ge0$, so $\int f\ge0$, nothing at all. Now suppose that $f>0$. So every Riemann sum is $>0$. But that does not imply immediately that $\int f>0$. Because a limit of strictly positive numbers need not be strictly positive, – David C. Ullrich Mar 11 '16 at 20:41
  • @DavidC.Ullrich the OP asked that :) either way, thanks for clarification – eyedropper Mar 11 '16 at 21:26
  • @JD162 No. See the reply I addressed to eyedropper... – David C. Ullrich Mar 11 '16 at 21:46
  • @DavidC.Ullrich You should explain why assume that there is an $x$ such that $f$ is continuous at $x$. – user26857 Mar 03 '18 at 06:54
  • @user26857 Assume $f$ is discontinuous everywhere then $f$ is not Riemann integrable. A contradiction. So $f$ must be continuous at some $x$. – Shashi Mar 03 '18 at 08:15