> FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities".

It is not: for example, the piece-wise constant function f: [0,1] -> [0,1] which starts at f(0) = 0, stays constant until suddenly f(1/2) = 1, until f(3/4) = 0, until f(7/8) = 1, etc. is Riemann integrable.

"Continuous almost everywhere" means that the set of its discontinuities has Lebesgue measure 0. Many infinite sets have Lebesgue measure 0, including all countable sets.

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emacdona 6 hours ago | parent [-]

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?

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jfarmer 6 hours ago | parent | next [-]

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...

	▲	5 hours ago \| parent [-]
		[deleted]

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ironSkillet 6 hours ago | parent | prev | next [-]

No that's not true either. A quick Google will reveal many examples, in particular the "Cantor set".

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thaumasiotes 6 hours ago | parent | prev [-]

The Cantor set is uncountable and has Lebesgue measure 0.

	▲	dalvrosa 5 hours ago \| parent [-]
		Great example