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If we have a real valued function $f$ continuous at some point $a$, is it necessarily true that $f$ is Riemann integrable on the interval $[a - \delta, a + \delta]$ if a $\delta > 0$ exists?

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No: The function

$$f(x) = \left\{\begin{array}{c} x : x \in \mathbb{Q} \\ 0 : x \notin \mathbb{Q} \end{array}\right.$$

is continuous only at the point $0$, and its set of discontinuities on $[-\delta, \delta]$ has positive measure.

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  • $\begingroup$ ... and therefore $f$ is not Riemann integrable on $[-\delta,\delta]$. $\endgroup$ – GEdgar Feb 2 '14 at 18:03
  • $\begingroup$ And why is this function non integrable? $\endgroup$ – user2850514 Feb 3 '14 at 17:26
  • $\begingroup$ @user2850514 Its set of discontinuities has positive measure, so by the Lebesgue criterion, it's not integrable. $\endgroup$ – user61527 Feb 3 '14 at 18:49
  • $\begingroup$ I am not familiar with the set of discontinuities and the Lebesgue criterion, is there another way to explain this? $\endgroup$ – user2850514 Feb 3 '14 at 19:22
  • $\begingroup$ @user2850514 If you set up a Riemann sum to try and compute the integral of $f$ over some small interval, you'll find that the lower sums are always $0$ and the upper sums are always $2\delta$; since these don't get any closer as you refine a partition, the Riemann integral won't exist. $\endgroup$ – user61527 Feb 3 '14 at 23:04

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