I've studied the proofs before but there's still something mystical and unintuitive for me about the area under an entire curve being related to the derivative at only two points, especially for wobbly non monotonic functions.
I feel similar about the trace of a matrix being equal to the sum of eigenvalues.
Probably this means I should sit with it more until it is obvious, but I also kind of like this feeling.
It is not determined by the derivative, it's the antiderivative, as someone else mentioned. The derivative is the rate of change of a function. The "area under a curve" of the graph of a function measures how much the function is "accumulating", which is intuitively a sum of rates of change (taken to an infinitesimal limit).
> 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]
> 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.
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.
"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.
I feel similar about the trace of a matrix being equal to the sum of eigenvalues.
Probably this means I should sit with it more until it is obvious, but I also kind of like this feeling.
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
[2] https://en.wikipedia.org/wiki/Lebesgue_integral
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.
"iff it is bounded and has countable discontinuities"?
Or, are there some uncountable sets which also have Lebesgue measure 0?
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...
Here's an example of a Riemann integrable function w/ infinitely many discontinuities: 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.p...
It gives several reasons why we "knew" the Riemann integral wasn't capturing the full notion of integral / antiderivative
- except finitely many, or
- except a set of measure zero.
--edit: The font used for those initials is called Goudy Initialen: https://www.dafont.com/goudy-initialen.font