Ah, thanks for the clarification! Would it have been accurate then to have said:
"iff it is bounded and has countable discontinuities"?
Or, are there some uncountable sets which also have Lebesgue measure 0?
Ah, thanks for the clarification! Would it have been accurate then to have said:
"iff it is bounded and has countable discontinuities"?
Or, are there some uncountable sets which also have Lebesgue measure 0?
No, it's really sets of measure zero. The Cantor set is an example of an uncountable set of measure 0: https://en.wikipedia.org/wiki/Cantor_set
The indicator function of the Cantor set is Riemann integrable. Like you said, though, the Dirichlet function (which is the indicator function of the rationals) is not Riemann integrable.
The reason is because the Dirchlet function is discontinuous everywhere on [0,1], so the set of discontinuities has measure 1. The Cantor function is discontinuous only on the Cantor set.
Likewise, the indicator function of a "fat Cantor set" (a way of constructing a Cantor-like set w/ positive measure) is not Riemann integrable: https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%9...
No that's not true either. A quick Google will reveal many examples, in particular the "Cantor set".
The Cantor set is uncountable and has Lebesgue measure 0.
Great example