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Is it true that for a function defined on a compact interval of the real line, if it is Riemann integrable it is possible for it to still have a countably infinite number of discontinuities, so long as they are not dense in the interval of integration?

Riccardo Caiulo
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    A real function on a closed interval is Riemann integrable if and only if its set of discontinuities is a zero-measure set. See this link https://math.stackexchange.com/questions/1039740/example-of-riemann-integrable-f-0-1-to-mathbb-r-whose-set-of-discontinui – Alessandro Fenu Nov 18 '24 at 17:05
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    The set of discontinuities certainly can be dense. There is a well-known Riemann-integrable function that is continuous at precisely the irrational points in any interval $[a,b]$. – Ted Shifrin Nov 18 '24 at 19:03

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