Consider the function

$$f(x) = \begin{cases} x^2\sin(1/x) & \text { if } x \neq 0 \\ 0 & \text{ otherwise} \end{cases} $$

has integrable derivative on $(-1, 1)$.

I found

$$f'(x) = \begin{cases} 2x\sin(1/x) - \cos(1/x) & \text { if } x \neq 0 \\ 0 & \text{ otherwise,} \end{cases} $$

but I have no clue how to show that it is integrable on $(-1, 1)$. Can someone please help me?


Your computations are correct and they show that $f'$ is bounded and it has a single discontinuity point. Therefore, it Riemann-integrable (and this would still be true if it had a countable set of discontinuity points).

  • $\begingroup$ Is that because the set of discontinuities is measure zero? $\endgroup$ – joseph Dec 11 '18 at 18:37
  • $\begingroup$ Indeed: a bounded function is Riemann-integrable if and only if its set of discontinuity points has measure zero. $\endgroup$ – José Carlos Santos Dec 11 '18 at 18:40
  • 1
    $\begingroup$ @joseph it's much easier than that $\endgroup$ – zhw. Dec 11 '18 at 18:54
  • $\begingroup$ @zhw. I agree. It's easy to prove directly from the definion (together with the fact that continuous $\implies$ Riemann-integrable) that if a bounded function only has finitely many points of discontinuity, then it is Riemann-integrable. $\endgroup$ – José Carlos Santos Dec 11 '18 at 18:56
  • $\begingroup$ How do we know it is bounded? $\endgroup$ – joseph Dec 11 '18 at 19:07

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