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I'm trying to construct a nowhere continuous function while it is Riemann integrable. There's a nowhere differentiable while continuous function, but I don't know about the latter one?

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    No such function exists. For bounded functions defined on a closed interval, they are Riemann integrable iff they're continuous a.e. – MathematicsStudent1122 May 18 '17 at 23:51
  • It is possible to prove (without using measure theory) that a Riemann integrable function must be continuous at some point in the interval and hence continuous at some point in every subinterval. See this answer https://math.stackexchange.com/a/519921/72031 – Paramanand Singh May 19 '17 at 05:53

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This is not possible. Lebesgue's criterion for Riemann integrability says that a bounded function is Riemann-integrable if and only if it is continuous except on a null set.

Chappers
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  • Is there a simple counterexample to the same statement but with Lebesgue integrable functions? Ie. is there a function that is Lebesgue integrable but not continuous on any set of co-measure 0? (Feel free to assume the function is bounded and the domain is compact, etc) – Faraz Masroor Jul 01 '22 at 18:01
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Riemann integration is too weak for this, there is a theorem which states that it is Riemann integrable when the discontinuities are at most countable or in a set of measure zero, so nowhere continuous is too strong.

For Lebesgue integral, $1_{\mathbb R\setminus \mathbb Q}$ is nowhere continuous but integrable (of value $b-a$ on $[a,b]$).

zwim
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