Let
$$f(x)=\begin{cases}0 , \text{ if } x \in [0,1]- \left\{\frac{\sqrt{2}}{2} \right\} \\ 1 \text{ if } x= \frac{\sqrt{2}}{2}\end{cases}$$
Then are the right-hand Riemann sum and the left hand Riemann sum are both zero? i.e. $R_n = \sum_{i=1}f (\frac{i}{n})= 0 = \sum_{i=1}f \left(\frac{i-1}{n}\right) =L_n$
If we are talking about an arbitrary partition $P = \{ 0=t_0 <t_1 < \cdots < t_n=1\}$, would the right hand and left hand Riemann sum associated with $P$ still be zero? i.e.$$R= \sum_{j=1} f(t_j) \cdot \Delta t_j=0 = \sum_{j=1} f(t_{j-1}) \cdot \Delta t_j=L.$$
I think it's true, because the integral of this "almost continuous function" would be zero. Could anyone let me know if this is correct?
Thanks!
Your right and left Riemann sum is always $0$ because there is not a partition where ${\sqrt 2\over 2}=\sup t_j \text{ or } \inf t_j$ so you are always taking the $0$. If you use the Darboux sum, the upper and lower sum always are the same value because the range of $f(x)$ is the singleton $\{1\}$ and so taking the limit:
$$\lim_{n\to\infty} {1\over n} =0 $$