Evaluating Riemann Integral with finitely many different values of function at different points 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 $f(x)$?

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.
