Is this Riemann-Integrable?

Question

Let $\lfloor x\rfloor$ be the integer part function. Let $$f(x)= \sum_{n=0}^\infty \frac{(-1)^n}{2^n} \lfloor nx\rfloor$$

Is $f(x)$ Riemann-Integrable ? Thanks in advance

Answer 1

Let $[a, b]$ be a closed interval. Let's consider the Riemann integrability of $f$ on $[a, b]$.

The function $g_n(x) = \left(-\frac{1}{2}\right)^n \lfloor n x \rfloor$ is discontinuous at $\frac{1}{n} \Bbb Z$ and continuous everywhere else. It has a finite number of discontinuities on $[a, b]$. Thus, it's Riemann integrable on $[a, b]$.

Put $M = \lceil \max(|a|, |b|) \rceil$. We have:

$$ |g_n(x)| = \left| \left(-\frac{1}{2}\right)^n \lfloor n x \rfloor\right| \le M \left(\frac{1}{2}\right)^n n $$

Since $\displaystyle \sum_{n=0}^\infty \left(\frac{1}{2}\right)^n n$ converges, $f(x) = \displaystyle \sum_{n=0}^\infty g_n(x)$ converges uniformly by the Weierstrass M-test. Hence $f$ is Riemann integrable.

Answer 2

Note that $$ f(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2^n} \lfloor nx\rfloor=-\frac{2x}{9}-\sum_{n=0}^\infty \frac{(-1)^n}{2^n}\{nx\} $$ So it is remains to show that the function $$ g(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2^n}\{nx\} $$ is Riemann integrable. Now follow the same steps presented in this answer to conclude that $g$ is indeed Riemann integrable.