The easy and less generic integral. The harder one is the Lebesgue integral.
"More complex and general" integral. Matches the Riemann integral for "simple functions", but also works for some "funkier" functions that Riemann does not work for.
Ciro Santilli sometimes wonders how much someone can gain from learning this besides the beauty of mathematics, since we can hand-wave a Lebesgue integral on almost anything that is of practical use. The beauty is good reason enough though.
Advantages over Riemann:
Video 1.
Riemann integral vs. Lebesgue integral by The Bright Side Of Mathematics (2018)
. Source. shows how Lebesgue can be visualized as a partition of the function range instead of domain, and then you just have to be able to measure the size of pre-images.
One advantage of that is that the range is always one dimensional.
But the main advantage is that having infinitely many discontinuities does not matter.
Infinitely many discontinuities can make the Riemann partitioning diverge.
But in Lebesgue, you are instead measuring the size of preimage, and to fit infinitely many discontinuities in a finite domain, the size of this preimage is going to be zero.
So then the question becomes more of "how to define the measure of a subset of the domain".
Which is why we then fall into measure theory!
In "practice" it is likely "useless", because the functions that it can integrate that Riemann can't are just too funky to appear in practice :-)
Its value is much more indirect and subtle, as in "it serves as a solid basis of quantum mechanics" due to the definition of Hilbert spaces.
And then this is why quantum mechanics basically lives in : not being complete makes no sense physically, it would mean that you can get closer and closer to states that don't exist!
TODO intuition
A measurable function defined on a closed interval is square integrable (and therefore in ) if and only if Fourier series converges in norm the function:
Riesz-Fischer theorem is a norm version of it, and Carleson's theorem is stronger pointwise almost everywhere version.
Note that the Riesz-Fischer theorem is weaker because the pointwise limit could not exist just according to it: norm sequence convergence does not imply pointwise convergence.
There are explicit examples of this. We can have ever thinner disturbances to convergence that keep getting less and less area, but never cease to move around.
If it does converge pointwise to something, then it must match of course.
The Fourier series of an function (i.e. the function generated from the infinite sum of weighted sines) converges to the function pointwise almost everywhere.
The theorem also seems to hold (maybe trivially given the transform result) for the Fourier series (TODO if trivially, why trivially).
Only proved in 1966, and known to be a hard result without any known simple proof.
This theorem of course implies that Fourier basis is complete for , as it explicitly constructs a decomposition into the Fourier basis for every single function.
TODO vs Riesz-Fischer theorem. Is this just a stronger pointwise result, while Riesz-Fischer is about norms only?
Integrable functions to the power , usually and in this text assumed under the Lebesgue integral because: Lebesgue integral of is complete but Riemann isn't
for .
is by far the most important of because it is quantum mechanics states live, because the total probability of being in any state has to be 1!
has some crucially important properties that other don't (TODO confirm and make those more precise):
Some sources say that this is just the part that says that the norm of a function is the same as the norm of its Fourier transform.
Others say that this theorem actually says that the Fourier transform is bijective.
The comment at may be of interest, it says that the bijection statement is an easy consequence from the norm one, thus the confusion.
As mentioned at Section "Plancherel theorem", some people call this part of Plancherel theorem, while others say it is just a corollary.
This is an important fact in quantum mechanics, since it is because of this that it makes sense to talk about position and momentum space as two dual representations of the wave function that contain the exact same amount of information.

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