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Here we're talking about Riemann integrable real functions defined on $[a,b]$. Even though this is a simple question I didn't find it here, if it's duplicated I'm sorry...

I could prove this if $f$ was continuous, or $$ f\geq0 \rightarrow \int f\geq0. $$

Trying to prove this one I stucked in a point that if I could prove the following

if for every partition of $[a,b]$, $$ \inf \{f(x)|x\in[x_{t_i},x_{t_{i+1}}]\}=0 $$ for all intervals of the partition then $f=0$,

then I could prove the initial thing... But I couldn't do this neither. Anyone can help me with that?

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  • $\begingroup$ $f>0 \iff f(x)>0$ for all $x\in[a,b]$ $\endgroup$ – Robson May 5 '18 at 23:57
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    $\begingroup$ Another way: $f$ is Riemann integrable implies $f$ is a.e. continuous. Pick $x_0$ where $f$ is continuous at $x_0$. Choose $\delta>0$ such that $f(x)> \frac{1}{2} f(x_0)$ whenever $x\in[x_0-\delta, x_0+\delta]$. Then $\int f \geq \int_{x_0-\delta}^{x_0+\delta} f \geq \frac{1}{2}f(x_0) \cdot 2\delta$. $\endgroup$ – Danny Pak-Keung Chan May 6 '18 at 0:08
  • $\begingroup$ math.stackexchange.com/questions/2731800/… $\endgroup$ – user284331 May 6 '18 at 0:11
  • $\begingroup$ This is not easy; actually quite a difficult problem, but is certain to be on MathSE already. $\endgroup$ – T_M Nov 7 '18 at 14:56
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    $\begingroup$ @T_M: the proof begins by showing that a Riemann integrable function must be continuous somewhere. Proving this without measure theory is not easy but luckily it is available here: math.stackexchange.com/a/519921/72031 $\endgroup$ – Paramanand Singh Nov 8 '18 at 13:25
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This is also true in Lebesgue sense. We can prove it as follows: Let $A=\{x\mid f(x)>0\}$. For each $n\in\mathbb{N}$, let $A_n=\{x\mid f(x)\geq \frac{1}{n}\}.$ Note that $A=\cup_n A_n$. Since $A$ has non-zero measure, there exists $n$ such that $A_n$ has non-zero measure. Now $\int f \geq \int_{A_n} f \geq \frac{1}{n}m(A_n)>0$, where $m(A_n)$ denotes the Lebesgue measure of $A_n$.

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  • $\begingroup$ Yes, yes, one uses Lebesgue theory to prove that, but actually one can use Riemann partition to prove that as well. By the way, your name reminiscent me of the Hong Kong singer 陳百強. $\endgroup$ – user284331 May 6 '18 at 0:17
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    $\begingroup$ @user284331 Ah! You also know that. He was a singer in 80's in Hong Kong. $\endgroup$ – Danny Pak-Keung Chan May 6 '18 at 0:19
  • $\begingroup$ Typo in the definition of the set $A$. I just fix it. Thank you. $\endgroup$ – Danny Pak-Keung Chan May 6 '18 at 2:03
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    $\begingroup$ I think it's fair to down vote, sorry, but this isn't what is asked so it's confusing that its currently here as the most upvoted answer. $\endgroup$ – T_M Nov 7 '18 at 14:55
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Since $f(x) \gt 0$, then $\exists m \in \Re$ s.t. $f(x) \ge m \gt 0$.

Compose a constant function $\phi$ s.t.

$\forall x \in [a,b]$, $\phi(x) = m$.

Then

$\int_a^b f(x) \ge \int_a^b \phi(x) = (b-a)*m \gt 0$.

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  • $\begingroup$ Welcome to math.SE. This is very hard to read, please learn some MathJax and edit your post to use it. $\endgroup$ – Henrik Nov 7 '18 at 14:56

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