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?

  • 2
    $\begingroup$ I always cringe when I see ”Reimann” integrable on this site. It is "Riemann”!!! $\endgroup$
    – Martin R
    Jul 27, 2020 at 8:12
  • $\begingroup$ I wonder why this happens so often: math.stackexchange.com/search?q=Reimann $\endgroup$
    – Martin R
    Jul 27, 2020 at 8:13
  • $\begingroup$ @MartinR According to "How common is your last name" the last name "Reimann" is more common than "Riemann". But I digress $\endgroup$
    – Graviton
    Jul 27, 2020 at 8:16
  • 1
    $\begingroup$ The set of points of discontinuites is a countably infiinite set so 3) is true and 4) is false. $\endgroup$ Jul 27, 2020 at 8:16
  • $\begingroup$ Wonder why is $f$ discontinuous at $x=0$. $\endgroup$ Jul 27, 2020 at 8:20

1 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 $[a,b]$.

  • $\begingroup$ A bounded function is not necessarily Riemann integrable. $\endgroup$
    – Martin R
    Jul 27, 2020 at 8:23
  • $\begingroup$ @MartinR Right. But a function that is bounded on $[a, b]$ and Riemann integrable on all $[\epsilon, b]$ for $a< \epsilon <b$ is Riemann integrable on $[a,b]$. $\endgroup$ Jul 27, 2020 at 8:25

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