Integral of discontinuous function

Question

Let $$f(x) = \begin{cases} 1 & \text{if } x = 1/n \text{ for some } n \in \mathbb{N} \\ 0 & \text{otherwise} \end{cases}$$ Show that $f$ is Riemann-integrable on $[0,1]$ and compute $\int_{0}^{1}f$.

My thought is to define partitions $P_n= \{0, 1/n, 2/n, ..., (n-1)/n, 1\}$ and use the sequential criterion for integrability. For any $P_n$ it must be that $L(f, P_n)= 0$ so $\lim_{n\to\infty} L(f, P_n) = 0$. The tricky part is showing that $\lim U(f, P_n) = 0$ as well. Even so, it doesn't seem right to me that the value of the integral would be $0$. Does anyone have a better idea?

Answer 1

For each $n > 1$ we can choose a partition

$$P_n: 0< a_n< 1/n< b_n < a_{n-1}< 1/(n-1) < b_{n-1} < \cdots < a_2 < 1/2< b_2 <1$$

such that $(b_n-a_n) + (b_{n-1}-a_{n-1}) + \cdots + (b_{2}-a_2) < 1/n.$ For this partition, we have

$$U(f,P_n) - L(f,P_n) \le (1/n)\cdot a_n + (1/n)\cdot(b_n-a_n) + (1/(n-1))\cdot(b_{n-1}-a_{n-1})\,+$$ $$\cdots + (1/2)\cdot (b_{2}-a_2) < 1/n^2 + 1/n.$$

This implies the desired result.

Answer 2

HINT:

Consider the partitions

$$P_k = \{0, \frac1k, \cdots, \frac14, \frac13, \frac12, 1\} $$

When $k \to \infty $.