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Question: Is the set of points of continuity of any Riemann integrable function uncountable?

There's a question in my Analysis assignment asking us to prove if $f$ is integrable then it has infinitely many points of continuity (and the set is dense). I wonder if the set is also uncountable. I've seen examples of Riemann integrable functions that have uncountably many discontinuities but I haven't seen an example with only countably many points of continuity.

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  • $\begingroup$ The indicator function of the 1/3-Cantor set is Riemann integrable in $0,1]$. It is discontinue at every point in the cantor set, which is uncountable. The indicator function of $\mathbb{Q}$ which is dens in $\mathbb{R}$ is on the other hand not Riemann integrable in $[0,1]$ since it is discontinue at any point (uncountable st of discontinuities); The indicator function of the sequence $\{1/n:n\in\mathbb{N}\}$ is Riemann integrable and its set of discontinuities is countable. $\endgroup$
    – Mittens
    Oct 19, 2022 at 16:47
  • $\begingroup$ Using elementary arguments one can prove that if $f$ is Riemann integrable on $[a, b] $ it is continuous at some point in $[a, b] $. And then since it is integrable on every subinterval, every subinterval of $[a, b] $ contains points of continuity of $f$. $\endgroup$
    – Paramanand Singh
    Nov 5, 2022 at 9:25
  • $\begingroup$ @ParamanandSingh Thanks for your reply! But wouldn't that just prove there are infinitely many points of continuity? Every subinterval of [0,1] contains a rational number but there are only countably many of them. $\endgroup$
    – Eric Chan
    Nov 5, 2022 at 14:14
  • $\begingroup$ Yeah that argument only proves that points of continuity are dense in that interval. It doesn't guarantee uncountability. $\endgroup$
    – Paramanand Singh
    Nov 6, 2022 at 6:25
  • $\begingroup$ By the way density as well being $G_\delta$ proves it as uncountable. See math.stackexchange.com/q/708681/72031 $\endgroup$
    – Paramanand Singh
    Nov 6, 2022 at 7:36

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The set of points of discontinuity has Lebesgue measure $0$ (by to a theorem due to Lebesgue). So, the set of points of continuity has Lebesgue measure greater than $0$, and therefore it is uncountable.

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    $\begingroup$ the set of points of continuity has Lebesgue measure greater than $0$, and therefore it is uncountable --- In fact, the set is "uncountably dense" (i.e. uncountable intersection with every nonempty open interval). With measurability we can strengthen this to $c$-dense (i.e. cardinality continuum intersection with every nonempty open interval) using regularity and the fact that uncountable $F_{\sigma}$ sets have cardinality continuum (by Cantor-Bendixson theorem have this for closed sets, and easy to extend this to $F_{\sigma}$ sets). $\endgroup$ Oct 19, 2022 at 17:42

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