When I was studying calculus (I used Purcell's book), it is stated that the bounded function $f$ on closed bounded interval is Riemann integrable if and only if $f$ has countable discontinuity points, which means if $f$ is discontinuous on Cantor set, then it is not Riemann integrable.
However, when I was learning real analysis (I used Bartle's book Introduction to Real Analysis 3rd ed.), in the section 7.3, the Lebesgue's Integrability Criterion said that the bounded function $f$ on closed bounded interval is Riemann integrable if and only if the set of discontinuity points is a null set (the measure of such set is $0$), which means if $f$ is discontinuous on the Cantor set, then it is still Riemann integrable.
Which one is the truth?
I remember (a long time ago) I asked this question to my folks to discuss, and he gave me an example when $f$ is discontinuous on the Cantor set and it is not Riemann integrable, and I didn't find any fault in his argument.
I am sorry I cannot give you that example guys, I lost the notes.
Best Answer
Having only countably many discontinuities implies a bounded function on a closed bounded interval is Riemann integrable, but it is certainly possible to have a function that is Riemann integrable that has uncountable many discontinuities.
One example would be the characteristic function of the usual Cantor set in $[0,1]$.
See these related questions:
Characteristic function of Cantor set is Riemann integrable
Example of Riemann integrable $f: [0,1] \to \mathbb R $ whose set of discontinuity points is an uncountable and dense set in $[0,1]$