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Suppose $f(x)$ is defined as $$f(x) =\begin{cases} 1 & \text{ if } x= (a+b)/2 \\ 0 & \text{ otherwise.} \end{cases} $$ where $f$ is real valued function on $I = [a,b]$. I want to compute the Riemann Integral for this function on the interval $I$. To compute the Riemann Integral we need to compute upper and Lower Integral. $$\underline{\int_{a}^{b}} f dx = \sup_P \left\{\sum_{k=1}^n \mid I_k\mid \inf_{x\in I_k} f(x)\right\} $$ $ \forall x \in [a,b] \quad \inf_{x} =0 $ $$\underline{\int_{a}^{b}} f dx = 0$$ Let's compute Upper Reimann Integral $$\overline{\int_{a}^{b}} f dx = \inf_P \left\{\sum_{k=1}^n \mid I_k\mid \sup_{x\in I_k} f(x)\right\}$$ $if \quad x= (a+b)/2 $ , $\exists I_j $ for which $sup_{x\in I_j} =1$ , $x= (a+b)/2 \in I_j$ $$\overline{\int_{a}^{b}} f dx = \mid I_j \mid $$ Let there are in $n$ partition, $$\mid I_j \mid = \frac {\mid b-a \mid} n $$ as $n \rightarrow \infty$ , $\mid I_j \mid = 0$ $$\overline{\int_{a}^{b}} f dx = 0$$ $$\overline{\int_{a}^{b}} f dx = \underline{\int_{a}^{b}} f dx = \int_{a}^{b} f dx =0$$ If there are $m$ values where $f(x) =1$ then $m \frac {\mid b-a \mid} n$ also tends to zero. Therefore, Riemann integral is equal to zero even if there are more than one different value of function at a single point. Is this correct reasoning to compute Riemann integral of $f(x)$?

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It is a bit overcomplicated, but yes. Basically it is like this: A finite amount of points can be enclosed by a finite amount of intervals with arbitrarily small length. Thus (since $f$ is bounded) follows that they do not affect the riemann integral at all. Even countably many points can be enclosed in such a way, that the sum of the lengths of the intervals is arbitrarily small.

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  • $\begingroup$ But changing the function at countably many points may affect the Riemann integral. For example let $f(x) $ equal $1$ at the rational points of $[a, b] $ and equal $0$ at irrational points of $[a, b] $. Then $f$ is not Riemann integrable on $[a, b] $. $\endgroup$
    – Paramanand Singh
    Aug 28 at 12:29
  • $\begingroup$ Yes, but it does only change the integrability, not the value. Generally we may only change points in such a way that the cluster points of these points are nowhere dense (and thus finite). Because then we can still trap these clusters in arbitrarily small intervals. $\endgroup$
    – Lazy
    Aug 28 at 16:17
  • $\begingroup$ See math.stackexchange.com/q/3694790/72031 $\endgroup$
    – Paramanand Singh
    Aug 28 at 16:20

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