calculus - How to prove that this function is integrable on $[0,1]$

Here I tried to find two step functions, one of them is less than $f$ on $[0,1]$ whereas one of them is greater than $f$ on the same closed interval, to prove this function is Riemann-integrable on this interval, however, I could not find such two step functions to help me show this function is integrable. May I get some hint about the solution to this problem?

Answer

Hint:

$$\int_0^1 (-1)^{[1/x]} \, dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^{[1/x]} dx = \lim_{n \to \infty}\sum_{k=1}^n\int_{1/(k+1)}^{1/k} (-1)^k dx \\ = \sum_{k=1}^{\infty} (-1)^k \left[\frac1{k}-\frac1{k+1} \right]=\sum_{k=1}^{\infty} \frac{(-1)^k}{k(k+1)}$$

Alternatively consider a partition $P = (0, 1/n, \ldots, 1/2, 1)$.

The difference between upper and lower Riemann sums is $1/n$ since

$$\sup_{(1/(k+1),1/k]} f(x) - \inf_{(1/(k+1),1/k]} f(x) = (-1)^k - (-1)^k = 0,$$

and

$$\sup_{[0,1/n]} f(x) - \inf_{[0,1/n]} f(x) = 1.$$

By choosing $n$ sufficiently large, this difference can be made smaller than any $\epsilon > 0$ and $f$ is Riemann integrable.