Real Analysis 51 | Riemann Integral - Definition

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The Bright Side of Mathematics

The Bright Side of Mathematics

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@fabriziocabrera6376
@fabriziocabrera6376 2 жыл бұрын
Thank you for your amazing work! Does the playlist of real analysis end here? How many more topics are you going to cover?
@brightsideofmaths
@brightsideofmaths 2 жыл бұрын
A lot of videos are planned to come. I cover everything about the Riemann integral here :)
@mastershooter64
@mastershooter64 2 жыл бұрын
@@brightsideofmaths what about multivariable analysis?
@brightsideofmaths
@brightsideofmaths 2 жыл бұрын
@@mastershooter64 This will be a separate course.
@mastershooter64
@mastershooter64 2 жыл бұрын
@@brightsideofmaths awesome!
@fabriziocabrera6376
@fabriziocabrera6376 2 жыл бұрын
​@@brightsideofmaths Great!
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
Technically, what is defined here in this video is not the Riemann integral, but the Darboux integral. Also, for those who are wondering how this generalizes to the Lebesgue integral, consider the mathematical space ([a, b], Π([a, b]), J). Here, Π([a, b]) is the set of closed intervals which almost partition [a, b]: this is to say, closed intervals which only intersect at a point, and whose union is [a, b]. J is a function Π([a, b]) -> [0, +♾) satisfying the axioms of a content: specifically, this is called the Jordan content in 1 dimension. Consider x = {[x(i), x(i + 1)]: i in [0, n], i is an integer} a partition in Π([a, b]). The Jordan content is defined such that x(i + 1) - x(i) = J([x(i), x(i + 1)]). The lower Darboux sum L(f, x) is defined as the sum of inf({f[t(i)] : t in [x(i), x(i + 1)]})·J([x(i), x(i + 1)]), with i in [0, n], i being an integer. The upper Darboux sum U(f, x) is defined as the sum of sup({f[t(i)] : t in [x(i), x(i + 1)]})·J([x(i), x(i + 1)]). The lower Darboux integral is defined as sup({L(f, x) : x in Π([a, b])}). The upper Darboux integral is defined analogously. The Darboux integral exists if and only if the upper Darboux integral and the lower Darboux integral are equal, and if it does exist, then the Riemann integral (which is defined differently) also exists and is equal to the Darboux integral. To generalize the Darboux integral to the Lebesgue integral, one does the following: rather than focusing on the set Π([a, b]), one allows the partitions of [a, b] to come from any elements of the Borel sigma algebra of [a, b], denoted β([a, b]), and the Jordan content is extended to the Lebesgue measure λ, a function β([a, b]) -> [0, +♾] which satisfies the axioms of a measure, rather than just a content. This gives the Lebesgue measure of a bounded function f whose domain is [a, b]. However, the Lebesgue integral is more powerful than this: the domain need not be compact, and it could be any subset U of R, with the sigma algebra being appropriately chosen as β(U), and the Lebesgue measure simply being defined on β(U) (in general, it is defined on β(R)). To ensure this extension of the Lebesgue integral is well-defined, one considers the lower Lebesgue integral of max(0, f) and the upper Lebesgue integral of -min(0, f), and if their minimum is finite, then the Lebesgue integral of f is equal to their sum.
@brightsideofmaths
@brightsideofmaths 2 жыл бұрын
True, not the historical Riemann integral but it is a equivalent definition, therefore, I wouldn't change the name :)
@hyperduality2838
@hyperduality2838 2 жыл бұрын
@@brightsideofmaths Equivalence, similarity = duality! Supremum is dual to infimum synthesizes the Riemann integral. Thesis is dual to anti-thesis creates the converging thesis or synthesis -- the time independent Hegelian dialectic. Injective is dual to surjective synthesizes bijective or isomorphism. "Always two there are" -- Yoda. Integration (convergence, syntropy) is dual to differentiation (divergence, entropy) -- the 4th law of thermodynamics.
@rafaelschipiura9865
@rafaelschipiura9865 Жыл бұрын
@@brightsideofmaths Doesn't one need the definition of the Riemann integral to be the historic one, so that any point in the function can be chosen, even at random, so that one can have path integrals? Talking about "bigger" and "smaller" is difficult when leaving lower dimensions... I know this is Real analysis, but I just wonder if there are implications for later.
@brightsideofmaths
@brightsideofmaths Жыл бұрын
@@rafaelschipiura9865 I am not sure which path integrals you mean but choosing points at random makes it not simpler in my opinion. For higher dimensions, the Lebesgue integral is the right approach.
@rafaelschipiura9865
@rafaelschipiura9865 Жыл бұрын
@@brightsideofmaths Right, thanks.
@tamimaziz3631
@tamimaziz3631 2 жыл бұрын
Sir, Should we take the supremum of {integration a to b phi(x) such that 0
@PrzemyslawSliwinski
@PrzemyslawSliwinski 2 жыл бұрын
4:32 - I remember this definition from the Rudin's book - but wonder how it encompasses e.g. beta probability density functions (which can be unbounded)?
@brightsideofmaths
@brightsideofmaths 2 жыл бұрын
Unbounded functions cannot be Riemann integrable with this definition. One can retrieve this by using limits, also known as improper Riemann integrals. We talk about them later.
@obviamorbisterrarum
@obviamorbisterrarum 2 жыл бұрын
Awesome, thank you...
@Infinium
@Infinium 2 жыл бұрын
This is a really nice video! I produced something similar not too long ago!
@brightsideofmaths
@brightsideofmaths 2 жыл бұрын
Thanks. I will check that out :)
@mengyangli6412
@mengyangli6412 Жыл бұрын
Did you mean the infimum for the lower sums and the supremum for the upper sums? That's what you said you "Riemann integral vs Lebesgue integral,".
@mengyangli6412
@mengyangli6412 Жыл бұрын
It was at 8:20 in that video.
@brightsideofmaths
@brightsideofmaths Жыл бұрын
No, here it's correct :)
@brightsideofmaths
@brightsideofmaths Жыл бұрын
@@mengyangli6412 I watched the time stamp. There I talk about something different: the values of the function we take inside the rectangles.
@mengyangli6412
@mengyangli6412 Жыл бұрын
@@brightsideofmaths Okay. Thank you.
@hyperduality2838
@hyperduality2838 2 жыл бұрын
Supremum is dual to infimum synthesizes the Riemann integral. Thesis is dual to anti-thesis creates the converging thesis or synthesis -- the time independent Hegelian dialectic. Injective is dual to surjective synthesizes bijective or isomorphism. "Always two there are" -- Yoda.
@thomasjefferson6225
@thomasjefferson6225 Жыл бұрын
I'm sure you're German, but isn't this darbouxs integral? 😂 Just kidding, this is one of the few times the German and French agree on something right!
@brightsideofmaths
@brightsideofmaths Жыл бұрын
Yes, you are right: this is the Riemann integral with the approach developed by Jean Gaston Darboux.
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