Is $f(x)=\left.\begin{cases}x\,\text{sgn}(\sin\frac{1}{x})&\text{if $x\neq0$}\\0&\text{if $x=0$}\end{cases}\right\}$ Riemann integrable?

Question

For $x\in[-1,1]$, let $$ f(x)= \begin{cases} x\,\operatorname{sgn}(\sin\frac{1}{x}), &\text{if $x\neq0$} \\ 0, &\text{if $x=0$} \end{cases} $$ where $\text{sgn}$ denotes the signum function. Then:

1. $f$ is continuous on $[-1,1]$
2. $f$ is not differentiable at any point of $[-1,1]$
3. $f$ is Riemann integrable on $[-1,1]$.
4. The set of points of discontinuity of $f$ in $[-1,1]$ is finite.

I found this question in a previous year entrance paper. The answer given is option $3$. Now, I know that $f$ is discontinuous at $x=0$, and that $f$ is differentiable at many points in $[-1,1]$. But what about options $3$ and $4$? Aren't they both essentially the same?

Answer 1

For each $0 < \epsilon <1$, $f$ is piecewise differentiable on $[-1,-\epsilon] \cup [\epsilon , 1]$ and therefore Riemann integrable on those intervals. Also $f$ is discontinuous at all points of the infinite set $S=\{1/k\pi \mid k \in \mathbb Z\}$.

This makes 1., 2. and 4. claims false.

Finally $f$ is bounded on $[-1,1]$. So given the above, $f$ is Riemann integrable on $[-1,1]$ and 3. is correct.

Recall that a map $f$ that is bounded on $[a,b]$ and Riemann integrable on all $[c,b]$ with $a<c<b$ is Riemann integrable on $[a,b]$.