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> f is Riemann integrable iff it is bounded and continuous almost everywhere. FWIW, I think this is the same as saying "iff it is bounded and has finite discon

> f is Riemann integrable iff it is bounded and continuous almost everywhere. FWIW, I think this is the same as saying "iff it is bounded and has finite discontinuities". I like that characterization b/c it seems more precise than "almost everywhere", but I've heard both. I mention that because when I read the first footnote, I thought this was a mistake: > boundedness alone ensures the subinterval infima and suprema are finite. But it wasn't. It does, in fact, insure that infima and suprema are finite. It just does NOT ensure that it is Riemann integrable (which, of course the last paragraph in the first section mentions). Thanks for posting. This was a fun diversion down memory lane whilst having my morning coffee. If anyone wants a rabbit hole to go down: Think about why the Dirichlet function [1], which is bounded -- and therefore has upper and lower sums -- is not Riemann integrable (hint: its upper and lower sums don't converge. why?) Then, if you want to keep going down the rabbit hole, learn how you _can_ integrate it (ie: how you _can_ assign a number to the area it bounds) [2] [1] One of my favorite functions. It seems its purpose in life is to serve as a counter example. https://en.wikipedia.org/wiki/Dirichlet_function https://en.wikipedia.org/wiki/Dirichlet_function [2] https://en.wikipedia.org/wiki/Lebesgue_integral https://en.wikipedia.org/wiki/Lebesgue_integral

mjdv 5mo ago
> 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.
- emacdona 5mo ago
  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?
  - thaumasiotes 5mo ago
    The Cantor set is uncountable and has Lebesgue measure 0.
    - dalvrosa 5mo ago
      Great example
  - ironSkillet 5mo ago
    No that's not true either. A quick Google will reveal many examples, in particular the "Cantor set".
  - jfarmer 5mo ago
    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 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%93Cantor_set https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%9...
    - deleted 5mo ago
      [deleted]
jfarmer 5mo ago
"Almost everywhere" means "everywhere except on a set of measure 0", in the Lebesgue measure sense. Here's an example of a Riemann integrable function w/ infinitely many discontinuities: https://en.wikipedia.org/wiki/Thomae%27s_function https://en.wikipedia.org/wiki/Thomae%27s_function Anyone interested in this should check out the Prologue to Lebesgue's 1901 paper: http://scratchpost.dreamhosters.com/math/Lebesgue_Integral.pdf http://scratchpost.dreamhosters.com/math/Lebesgue_Integral.p... It gives several reasons why we "knew" the Riemann integral wasn't capturing the full notion of integral / antiderivative
Jaxan 5mo ago
“Almost everywhere” is a mathematical term and can mean two things (I think): - except finitely many, or - except a set of measure zero.
- dalvrosa 5mo ago
  Here is used in the Lebesgue measure theory sense
bandrami 5mo ago
"Almost everywhere" is precisely defined, and it is broader than that. E.g. the real numbers are almost everywhere normal, but there are uncountably many non-normal numbers between any two normal reals.
sambapa 5mo ago
"almost everywhere" can mean the curve has countably infinite number of discontinuities

mjdv 5mo ago

> 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.

emacdona 5mo ago
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?
- thaumasiotes 5mo ago
  The Cantor set is uncountable and has Lebesgue measure 0.
  - dalvrosa 5mo ago
    Great example
- ironSkillet 5mo ago
  No that's not true either. A quick Google will reveal many examples, in particular the "Cantor set".
- jfarmer 5mo ago
  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 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%93Cantor_set https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%9...
  - deleted 5mo ago
    [deleted]

jfarmer 5mo ago

"Almost everywhere" means "everywhere except on a set of measure 0", in the Lebesgue measure sense. Here's an example of a Riemann integrable function w/ infinitely many discontinuities: https://en.wikipedia.org/wiki/Thomae%27s_function https://en.wikipedia.org/wiki/Thomae%27s_function Anyone interested in this should check out the Prologue to Lebesgue's 1901 paper: http://scratchpost.dreamhosters.com/math/Lebesgue_Integral.pdf http://scratchpost.dreamhosters.com/math/Lebesgue_Integral.p... It gives several reasons why we "knew" the Riemann integral wasn't capturing the full notion of integral / antiderivative

Jaxan 5mo ago

“Almost everywhere” is a mathematical term and can mean two things (I think): - except finitely many, or - except a set of measure zero.

dalvrosa 5mo ago
Here is used in the Lebesgue measure theory sense

bandrami 5mo ago

"Almost everywhere" is precisely defined, and it is broader than that. E.g. the real numbers are almost everywhere normal, but there are uncountably many non-normal numbers between any two normal reals.

sambapa 5mo ago

"almost everywhere" can mean the curve has countably infinite number of discontinuities