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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?

Marcelo
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  • $f>0 \iff f(x)>0$ for all $x\in[a,b]$ – Marcelo May 05 '18 at 23:57
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    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 $\int f \geq \int_{x_0-\delta}^{x_0+\delta} f \geq \frac{1}{2}f(x_0) \cdot 2\delta$. – Danny Pak-Keung Chan May 06 '18 at 00:08
  • https://math.stackexchange.com/questions/2731800/strictly-positive-riemann-integrable-function – user284331 May 06 '18 at 00:11
  • This is not easy; actually quite a difficult problem, but is certain to be on MathSE already. – SBK Nov 07 '18 at 14:56
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    @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: https://math.stackexchange.com/a/519921/72031 – Paramanand Singh Nov 08 '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 $A_n$.

Danny Pak-Keung Chan
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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

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

Kate Zhu
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