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Let $f: [a,b] \longrightarrow \mathbb{R}$ is a integrable function and $X \subset [a,b]$ dense in $[a,b]$. Suppose that $f(x) = 0$ for all $x \in X$. Prove that $\displaystyle \int\limits_{a}^{b}f(x)dx = 0$.

$\textbf{My idea:}$ Let $P = \lbrace a=t_{0}, t_{1}, ..., t_{n} = b \rbrace$ an arbitrary partition of $[a,b]$. Since $X$ in dense, in any partitions intervals $[t_{i-1}, t_{i}]$ of $[a,b]$, $\inf([t_{i-1}, t_{i}]) = 0$. Therefore, $s(|f|, P) = 0$, than $\displaystyle \int\limits_{a}^{b}f(x)dx = 0$, because $\sup s(|f|, P) =\inf S(|f|, P)$

$-$ $s(f, P)$ is the left riemann sum and $S(f, P)$ is the right riemann sum

Is the correct?

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$f$ is not nonnegative, it does not follow that $\inf f[t_{i-1},t_{i}]=0$. But this can be amended if we consider the integrable function $|f|$ and the fact that $\left|\displaystyle\int_{a}^{b}f(x)dx\right|\leq\displaystyle\int_{a}^{b}|f(x)|dx$.

Further, actually it is not necessarily that $s(f,P)=S(f,P)$ for an integrable function $f$.

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  • $\begingroup$ Yeah. I forgot that sup and inf hahaha and was considering $|f|$ too (missed placed in here). Thanks for the help! $\endgroup$ – Corrêa Feb 11 '18 at 20:06
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For $n>1$ and $k=0,1,2...n, $

let $$x_k=a+k\frac {b-a}{n}. $$

$f $ is integrable implies $$\lim_{n\to+\infty}\frac {b-a}{n}\sum_{k=0}^{n-1}f (c_k)=\int_a^bf $$

where $c_k\in [x_k,x_{k+1}] .$

As $X $ is dense in $[x_k,x_{k+1}] $, we can always find $c_k\in X\cap [x_k,x_{k+1}] $ with $f (c_k)=0$.

the Riemann sum and its limit $\int_a^bf $ will be both zero.

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  • $\begingroup$ The approach via Riemann sums is the easiest here. +1 $\endgroup$ – Paramanand Singh Feb 12 '18 at 11:00

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