Could you suggest a simple book about measure theory that follows the same steps you took in explaining this course? Maybe with some exercises , thanks!
I would like to thank the professor who explained the Lebesgue Integral in a so clear intuitive way to make undertand the logic behind. I have bad memories when at school the issue of Lebesgue vs Riemann was done by tens of demostrations and theorem and level functions and at the end, by magic, Lebesgue integral was more robust than Riemann. Finally, after 30 years now I know, at least I have an idea. Thank you.
Lebesgue integrability is more robust than Riemann integrability, because Lebesgue integrability is constructed to be a direct generalization of Riemann integrability, even though this is not at all obvious from the way it is most often taught in courses.
Just found this by accident. I did a maths degree several decades ago and sadly have forgotten 99.9% of what I learned! I really enjoyed watching this - it was very clearly explained.
The anxiety, fear and panic caused by having to learn and to remember is only exceeded by the anxiety , fear and panic caused at having unlearned and forgotten it. I wonder if this true for all professions, and for spheres of learning. Most probably it is, which is a shame.
This presenter gets it: a solid concept at the beginning is way more powerful than a collection of very good theorems. Great job sir, the internet and I thank you ☺️ 🙏🏽💯🙌🏽
Finally, I found someone who can beautifully explain Real Analysis. Once, I hated this course because I couldn't understand it. You're doing a great job 👍🙏
Thanks so much for this! I FINALLY understand the concept of the Lebesgue integral! Every time I would try to look up a definition of the Lebesgue integral, it always came across as very abstract but that the idea was that we use arbitrarily sized Δx's. I never got a sense of how you parameterize those Δx's so you could use them to systematically calculate the integral. Thanks to you, I now see that the division is based on the output of the function and parameterized based on that. Hence the need to study Measure Theory to know the proper way to systematically "measure" these new types of divisions. Great video!!
The Δx(i) are not themselves what we parametrize to integrate. Δx(i) is an output family of a function applied to sets partitioning the domain of the function being integrated. The partitions are what we parametrize over.
This series is quite straightforward after watching your real analysis series, and surprisingly, it looks like functional analysis is not required to study before this series. Well done!
Very nice video. After reading the comment section, I’m really grateful for how good my professor explains and emphasizes these matters. Riemann’s notation does have advantages as well, like its conceptually easier to understand (despite those technical difficulties), and the calculation is often more easier to be done by human and computer. Lebesgue’s notation is more desirable and suitable for theoretical purpose. Another advantage I would like to add is, actually the most important one for me, Lebesgue integral is closed under the convergence that we consider (point-wise, in L^p, in measure), that is, sequence of Lebesgue integrable functions has again a Lebesgue integrable limit, which is very crucial for analysis as we no longer consider function as a correlation but an object in a vector space. For Riemann integrable functions, the limit is almost always not Riemann integrable again.
This is amazing!!! After countless courses in math, I never understood what the Lebesgue integral was even about. Thank you for this video. Every math teacher should introduce courses like this !!!
I am studying Stochastic modelling, and we need to use Lebesgue integral for the Renewal Theory. Your video is an excellent material to begin with Lebesgue integral. Thanks for the time for making this video!
Technically, this video did not work with the definition of the Riemann integral, but the definition of the Darboux integral. However, we know that a function is Riemann integrable if and only if it is Darboux integrable, and since the definition of the Darboux integral is simpler, this abuse of language is justified in context. Personally, I prefer to motivate measure theory and measure-theoretic integration by starting with the definition of the Darboux integral, and thinking of ways to directly loosening the definition, in order to produce a generalized notion of integration, along with reasons with why we should create such a generalization, by explaining each loosening adequately. The upper Darboux integral of f : [a, b] -> R is defined as the infimum of the upper Darboux sums, which are sums of sup{f[t(i)] : t(i) in [x(i), x(i + 1)]}·[x(i + 1) - x(i)], with x(i) < x(i + 1), and the union of every [x(i), x(i + 1)] is [a, b]. The lower Darboux integral is the supremum of the lower Darboux sums, which are sums of inf{f[t(i)] : t(i) in [x(i), x(i + 1)]}·[x(i + 1) - x(i)]. A function is Darboux integrable if the upper Darboux integral and the lower Darboux integral both exist, and are equal. For the purposes of motivating the Lebesgue integral, I will focus only on the lower Darboux integral, since the Lebesgue integral, as covered in this channel, is exactly a direct generalization of the lower Darboux integral. First, we acknowledge that the reason the Darboux sums are defined as such is because the partitions of [a, b] into [x(i), x(i + 1)] are done so that the length of [x(i), x(i + 1)] is the width of the rectangle under f, and inf{f[t(i)] : t(i) in [x(i), x(i + 1)]} is the lower bound to the height of the rectangle, while sup{f[t(i)] : t(i) in [x(i), x(i + 1)]} is the upper bound to the height of the rectangle. An idea here is to make the relationship between the partition [x(i), x(i + 1)] and the widths more natural by having a length function λ with the property that λ([x(i), x(i + 1)]) = x(i + 1) - x(i). Thus, we can write the lower Darboux sums as sums of inf{f[t(i)] : t(i) in [x(i), x(i + 1)]}·λ([x(i), x(i + 1)]). Here, the visual and abstract connection between the individual lower Darboux sums and the individual partitions is made natural and intuitive. A more concise notation can be adopted, by letting S(i) = [x(i), x(i + 1)], so that we can simply write the lower Darboux sums as sums of inf{f[t(i)] : t(i) in S(i)}·λ[S(i)], where each S(i) is an adjacent closed interval, and the set of S(i) partitions [a, b]. Then λ[S(i)] should be interpreted as the length of S(i). This adequately explains how to motivate the definition of the lower Darboux integral from our intuitive idea of "area enclosed by the graph of f". As the video explains, there are problems with this definition. For instance, there are functions whose graph is such that it cannot enclose enclose rectangles in its area, so partitioning [a, b] into S(i), where each S(i) is a closed interval, is inadequate for capturing the idea of width formally. Rather than partitioning [a, b] into closed intervals, the elements of the partition S(i) should be allowed to be arbitrary elements of a collection A of subsets of [a, b]. Accordingly, we need to generalize what our length function is, so that λ[S(i)] can be well-defined even if S(i) is not a closed interval, but rather, λ should admit any element of A as an input, and the output should be a quantity that aptly captures the idea of "length of A". On that note, this generalization should still be applicable if the domain is some arbitrary non-empty set X, rather than specifically [a, b]. The codomain may also be a Banach space over R, rather than just R, although this is not strictly speaking a generalization. Here, we have three concepts that have been loosened: the domain of f has been loosened from being a closed interval [a, b] of R into being simply an arbitrary non-empty set X; the collection of subsets of [a, b] has been loosened from only containing closed subintervals of [a, b] into simply containing arbitrary subsets of X, and this collection of subsets is A; the function λ has been loosened from only acting on closed subintervals of [a, b] to now acting on arbitrary elements of A, being now a function μ with domain A, and we may even allow for μ([x(i), x(i + 1)]) = x(i + 1) - x(i) to not be satisfied in general for the sake of broader applicability. This gives us a space (X, A, μ) with respect which our notion of generalized or loosened lower Darboux sums is defined. These sums may be called lower weighted sums, and the supremum of the set of such sums will be our new notion of integral, or our new notion of lower integral, depending on how powerful this notion is. The lower Darboux integral is the special case where X = [a, b], A is the set of closed subintervals of [a, b], and μ is the restriction of the Lebesgue measure to the set A as domain. Measure-theoretic integration is what you get when you restrict A to being a σ-algebra of X, and μ to be a measure with domain A. Of course, alternative formulations of integration exist, where μ is even more restricted than a measure, or is more loose than a measure, and A is even more restricted than a σ-algebra, or more loose than a σ-algebra. Measure theory is the systematic, axiomatic study of these different types of spaces, their properties, and restrictions or loosenings thereof, along with the study of the properties of the morphisms between these spaces, which are the functions we want to integrate. Of course, understanding how these spaces can be restricted into being measure spaces, and how can they be generalized, opens up the door for a theory that allows for integrals that are, so to speak, in between the Darboux integral and the Lebesgue integral, or notions of integration that are even stronger and more applicable than the Lebesgue integral, such as the gauge integral or the Khinchin integral.
Shout out from S.Korea from about 14:00 It was a moment of reckoning for me I have been learning measure theory and Lebesgue intergal in class but I didnt exactly know why we need these quite clearly Your video just gave whole new meaning to my Real Analysis study Thank you very much sir NOW I AM ENLIGHTENED!!!
Good job, I was never taught to think of Lebesgues integrals this way but I quickly learned what you showed thanks to a friend back when I was learning it. I was expecting exactly what you you showed in the video, and I am happy to see that any student could see that on TH-cam because sadly it is easy to be lost in the theory (I have myself some bad memories about mesure theory) without realizing the motivation behind it.
Perfect work. What I didn’t like in my university education was that professors often skipped motivation and history. And it makes math _too_ abstract. I have a book with biographies of greatest mathematicians and the book is written not only about theirs lives but also about theirs discoveries. And it helps!
@@kaiwalpanchal5872 Actually I had several but it's been a long, long time. Right now I have found only *"Tales of Mathematicians and Physicists"* by _Simon Gindikin_ on my shelves. It's good though.
thank you! Now I finally understood what I "learned" ten years ago in my math lecture :) Good work embedding the important statements concisely in twenty minutes!
You are partly right. Infinitely many discontinuity points CAN destroy the Riemann integrability. That is what I said and meant there. For example, all monotonically increasing functions are Riemann integrable. However, having measure zero for this set of discontinuities is not sufficient since the function has also to be bounded (Lebesgue Criterion).
Other major problems of the Riemann integral include the case when the integration inteval [a,b] is infinite and/or when the function has unbounded singularity points. The so called improper Riemann integral used to deal with these is a mess. Riemann integration only works well when functions fall inside a box.
Both have their problems, if I recall correctly from university (its been a long time ...). The integral of sin(x) on ]-inf, inf[ is 0, isn't it? But you cannot use the Lebesgue integral to integrate that, AFAIK. If I recall correctly, it's due to the construction of Lebesgue: 1. Define the integral for non-negative functions 2. Define the integral for purely negative functions f as the negative of the integral of -f 3. For an arbitrary functions, separate its definition set into 2 parts: the part where f is negative, and the part where it is positive, integrate both parts and then take the sum If you applied this to sin(x) on ]-inf;inf[ is, that you had to go with the 3rd option. Thus you had to construct the integral of sin(x) for the subset where sin(x) is positive, but the integral of sin(x) over {e el ]-t,t [ | sin(e) >= 0} for t -> inf is not defined as it would become infinitely large (and the same for the negative part...) But on the other hand, its been a long time, and I was but a normal student, thus perhaps there is a fix on higher levels?
@@w1darr No, the integral of sin(x) on ]-inf, inf[ is undefined, as what you actually do is integrate sin(x) from a to b then take the limit of (a, b) -> (-inf, inf). This depends on your actual limit 2D, while int sin(x) from -n to n is 0, int sin(x) from -2pi * n to pi * (2n + 1) alternates between -2 and 2 and is hence divergent. So the improper integral does not exist.
I love it when people make me feel dumb for not understanding things that once seemed hard. It shows me that more that I imagine is possible. Thank you very much, I will study Lebesgue Integration in detail now ☺️🙌🏽❤️
Thank you for this Lebesgue vs Riemann video! I'm currently auditing MIT's online Functional Analysis course for my personal enrichment (currently covering sigma-algebras, outer measure, & Lebesgue measurable sets), and these clear illustrations are invaluable.....I'd venture to say I learned more in 19 min than in nearly 3.5 hrs worth of proofs lecturing!
I don't know if I haven't been listening to my professor but this is the first time that I finally understand the overlap between Lebesgue measure and the Lebesgue integral. Thanks for the help.
The idea of integration, in the context of measure theory, is that you integrate a (measurable) function with respect to an interval. Riemann integration can be taken to be integration with respect to the Lebesgue measure, together with many restrictions. Getting rid of those restrictions leaves with just Lebesgue integration, although the phrase Lebesgue integration is sometimes used to refer to all measure-theoretic integration in general, rather than integration with respect to the Lebesgue measure specifically.
So 'Everyone wants to work with the lebesgue integral' is clearly bullshit. People don't use it. They don't even teach it at universities for stuff like physics
@@peterbonnema8913 This is a math topic, applied physicists don't need to learn about or work with intricate mathematical definitions. I don't know what applications you are looking for with a topic from analysis
Summary: Riemann integral: maps R to R, related to area under a graph which is approximated by a lower sum and higher sum of squares. problems: 1) Doesn't scale to higher dimensions easily. As we increase dimensions the shape of the partition we need to specify the range of the integral increases exponentially. 2) Functions with infinitely many discontinuities cannot be integrated with the Riemann integral. 3) We can only pull the limit inside the function when the uniform convergence property holds. We want an integral that works well in every dimension. Lebesgue Integral: Instead of partioning the X axis (which may be abstract or high dimensional), we decompose the Y axis. We want to find all the parts of the function that lie between small intervals ci. This causes discontinuities in the parts we get when are intervals are still big. To measure the lengths/areas/volumes of these discontinuous sets ( Which we call A). The total measure space is called mu. The area is thus the sum over the whole partition of the "rectangles", where Ci is its height and mu(A) its width
No, your list has many mistakes and inaccuracies. 0. The Riemann integral is not defined strictly for elements of R^R, only for elements R^[a, b]. Meanwhile, the Lebesgue integral is defined for functions B^X, where X is the carrier set of an arbitrary measure space, and B is a Banach space over R. 1. Some functions with infinitely many discontinuities are Riemann integrable, but most are not. 2. The Lebesgue integral still partitions the x-axis, not the y-axis. The difference is that it partitions it with measurable sets, rather than closed intervals of R, and rather than multiplying each infima by the length of such intervals, one multiplies by the measure of the sets in the partition. 3. The total measure space is (X, Σ, μ), not μ itself. μ is simple called the measure, and it is a function from Σ to [0, +♾].
Nice vid. I did a bachelor in maths many years ago so I've pretty much forgotten all of it but I could follow this very easily while eating my breakfast.
@@TheJProducti0ns Nothing I'm sorry to say. I tried to be hired as a trainee actuary but I think they probably thought I was too old. I also tried to get into the central statistics office where I live but although I did well on the language and maths test I didn't do well on the managerial test so they didn't hire me. I was working as a software developer at the time so I kept doing that and still do. Around 2008 I worked for a year as a secondary-school maths teacher (substituting for a woman who had a nervous breakdown) but when the year ended she was ready to come back and they wanted me to pay for a teachers education yet again for less hours so I didn't continue that and worked again as a software developer. Which is what I still do.
@@delberry8777 damn. do you ever regret not doing anything with your maths degree? I'm sure software developing pays quite well but do you ever wonder what could have been?
@@delberry8777 You can refresh a lot, I guess. Just use a little time every week to read, to watch some video and to solve some problems. Math can be fun :)
The nice thing about not knowing something is the hit of dopamine you get when you see "the big picture" ☺️❤️🙌🏽🔥 Thank you for such an eye-opening lecture ☺️
THANK YOU THANK YOU THANK YOU. I need to visualise stuff to actually comprehend it, especially stuff that is, at first sight, more complicated for me. I was really struggling to find a nice, simple explanation of the Lebesgue integrals and I found this. Thank you so much!!
Wow, good video! I actually have learned in calculus how to generalize the Riemann integral to higher dimension, but I hadn't seen the Lebesgue integral yet. So... Thanks for the extra knowledge!
I first watched this video around when it was posted and I was still in secondary school. At the time I was unsure about studying maths much further. Now I am at one of the top universities in the world studying maths. Crazy.
great video. i wish i had this vid on the first day of measure theory as it starts with the motivation and a direct comparison to the riemann integral which make every definition that follows more reasonable.
Same reason here. I'm Engineer and I had no Measure Theory on the undergraduate course. Although I don't use Lebesgue Integral, it always came across here and there. But every single explanation I found assumed I knew more math than the necessary to just say: "Ok, I understood de diference, I now know the basic concept, it has become clear the limits of Riemann Integral and how Lebesgue expands those limits".
This is in my opinion the best way to introduce Lesbegue integration. Maybe I was lucky, but when I came in touch the first time with Lesbegue integration it was exactly the same explanation: instead of partitioning the domain into small pieces and doing the limit process, the range of the function was partitioned and the corresponding parts of the domain had to be measured. For this the domain needs to be "measurable" to get a notion of "volume". This concept seemed to be very intuitive to me. The best part of it was the possibility to integrate functions that are not Riemann integrable like the Dirichlet function.
Nobody can really understand a graduate probability course without understanding this. There will no piece of mind if someone tries, as most including me did.Thanks.
The problem to be solved in the riemann function was how to partition abstract higher dimension spaces. Lebesgue found a way and changed the problem to , how to measure the volumes of abstract spaces
Amazing video man, really helped me understand the whole motivation behind Measure Theory. Also juast as I was studying Functional Analysis in my AI Bachelor you released a playlist about it. God bless u 😂
Thanks for the video, I am now wanting to learn about the Lebesgue-Stieltjes integral. I was wondering if you would recommend any videos or resources to learn this?
@@brightsideofmaths Hallo, ihre Videos sind sehr hilfreich, dass ich mich frage, was sie hauptberuflich machen. Arbeiten sie zufälligerweise an einer Universität oder machen sie das nur nebenbei?
The point you said about extending Riemann Integration to higher dimension: Why would have to have partition the domain that we're integrating over? I thought the standard approach is just to cover it with a box and then integrate the function "f" times the "inticator function" of your domain, over the box.
The question is still: how do you define the integral for a function g? You still need to partition the domain even if you choose as a box. Of course, the box makes it simpler but the other problems remain.
OMG! Why couldn't someone have shown me this years ago when I muddled through real analysis? That makes so much more sense. Now I want to read Royden again.
thanks a lot for that amazing explanation, i watched it after learning the lebesgue integral chapter and it really helped me to imagine its' dynamism. I guess this is going to help me solve futur "hard" problems where we've to pose functions,... thanks again
What software did you use to record it? I've seen it in action before (at least I think it was the same one) but I forgot its name - I was taught physics with it and it was great :D
A major pet peeve of mine is the typical drawing of Lebesgue integration as horizontal rectangles, when this is simply not how the integral is defined or calculated. Of course, you can define an equivalent integral by calculating those horizontal rectangles, but that's not what's taught, and so that particular graphic confuses so many students, my past self included. The fact that the thumbnail and the video itself don't use this graphic, and instead use one with vertical bars makes me very happy.
Very interesting explanation of a difference between Riemann and Lebesgue integral, motivation and why the later is preferred. I was only missing one piece at the end. You started your discussion with Riemann integral over an interval [a, b]. It would be nice to see how one calculate Lebesgue integral on [a,b] especially for non injective functions, as the one in you example with Lebesgue integral.
Thank you! I have a whole series about measure theory where you can find some examples. Just calculating a Lebesgue integral for a function f: [a,b] -> R is not very interesting in this regard. In the video above, I wanted to show the motivation and why one should study measure theory :)
Danke für die tollen Videos :) Sie haben mir echt geholfen das Lebesgue Integral also vor allem die Motivation und hier den Unterschied zB besser zu verstehen, aber irgendwie schaffe ich es noch nicht tatsächlich ein Integral einer nicht Treppenfunktion über den Lebesgue Weg zu berechnen/bestimmen. Darum fände ich eine Empfehlung, wo ich Beispielrechnungen finden kann oder auch ein kurzes Video zu Beispielen super hilfreich
Gerne! Ich habe eine ganze Video-Reihe über Maßtheorie: th-cam.com/video/4DHP8cBcg_o/w-d-xo.html (Deutsch oder Englisch). Vielleicht hilft das schon mal. Es ist ja auch nicht so, dass man das Lebesgue-Integral für explizite Berechnungen benutzen muss. Dafür ist der gewählte Integralbegriff ja oft ziemlich egal. Es geht um die gesamte Theorie, die man mit dem Lebesgue-Integral aufziehen kann.
Great video! May I ask what equipment and software you are using? I've been looking into digital note taking for mathematics but haven't decided upon anything wholeheartedly as of yet.
Thanks. I use Xournal and a Wacom for writing. For digital note taking, I am now a fan of Boox Note Ebook reader. However, I advise you to test some things. For me, for example, Microsoft Surface is just not accurate enough but other colleagues love it.
Please would you kindly add this to your measure theory playlist. It really filled in a lot of gaps while I was going through the series. Specifically why Lebague integration was helpful.
Suppose µ is counting measure on Z + and b1, b2, . . . is a sequence of nonnegative numbers. Think of b as the function from Z + to [0, ∞) defined by b(k) = bk . Then Z b dµ = ∞ ∑ k=1 bk , How we solve it
hello I think what you describe at around 12min might be a bit inaccurate. the Riemann integral is not defined via upper and lower sums. in the Riemann integral the Riemann sum converges to the integral of a function as the partition norm / mesh -> 0. What you describe at min 12 are technically the Darboux sums. however due to equivalence of the two it doesn't really matter. great video.
I typically think of the Riemann integral as being more naturally related to adding a large but finite number of discrete terms represented within a sigma term. Is that just a prejudice of mine? If not, is there any finite analogue to the Lebesgue integral, (I guess being one where we index over terms with respect to the output of a function rather than the arguments fed into it?), and does that analogue have any unique niceness to it? I can see that this matches the notation that you use near the end of the video, but what I mean is, this seems like maybe an idea that has utility even outside the context of calculus?
Yes, that is just a prejudice of yours. At no point in the definition of the Lebesgue integral are you ever adding infinitely many terms over the σ-algebra, so the Lebesgue integral is also a sum of an arbitrarily large but finite number of terms. To be precise, both the Riemann and Lebesgue integral can be defined as suprema of finite sums.
Hi, this μ( { x∣f(x) > y } ) is the way to choose partition A in video 16:29, right? Which one would it be for another partition? I get confused because the next partition would not meet the condition of f(x) > y, it would be mixed with the previous one, partition A meets it because it is the highest partition with the highest c_i but another one would not meet it, I don't know if I made myself clear, Thanks anyway.
![Riemann integral vs. Lebesgue integral [dark version]](http://i.ytimg.com/vi/AczklPvD9Ug/mqdefault.jpg)
PDF version here: tbsom.de/s/ov
Could you suggest a simple book about measure theory that follows the same steps you took in explaining this course? Maybe with some exercises , thanks!
Thank you for clearly explaining the *motivation* behind the Lebesque integral.
The mathematician's name was Henri Léon Lebesgue!
Common, but awful mistake ;).
@@active285 Thank you for correcting me!
Und jetzt es ist auch hier: @mathupdate
I would like to thank the professor who explained the Lebesgue Integral in a so clear intuitive way to make undertand the logic behind. I have bad memories when at school the issue of Lebesgue vs Riemann was done by tens of demostrations and theorem and level functions and at the end, by magic, Lebesgue integral was more robust than Riemann. Finally, after 30 years now I know, at least I have an idea. Thank you.
MCQ on Riemann integral
pathaksir2.blogspot.com/2020/11/mcq-on-riemann-integral.html?m=1.
Yes, but be careful to Dirichlet function, where only L integral can be applied.
Lebesgue integrability is more robust than Riemann integrability, because Lebesgue integrability is constructed to be a direct generalization of Riemann integrability, even though this is not at all obvious from the way it is most often taught in courses.
You cut the abstract clutter and clearly explained the concepts. Brilliant !
Just found this by accident. I did a maths degree several decades ago and sadly have forgotten 99.9% of what I learned! I really enjoyed watching this - it was very clearly explained.
Thank you very much! Please use my videos to refresh your Math memories :)
The anxiety, fear and panic caused by having to learn and to remember is only exceeded by the anxiety , fear and panic caused at having unlearned and forgotten it. I wonder if this true for all professions, and for spheres of learning. Most probably it is, which is a shame.
Brilliant! The shortest and simplest introduction to the Lebesque integral. You’re a good teacher.
I like how you used a yellow background; easier on my eyes when watching this in the dark.
Thank you. That is my style :)
MCQ on Riemann integral
pathaksir2.blogspot.com/2020/11/mcq-on-riemann-integral.html?m=1.
This presenter gets it: a solid concept at the beginning is way more powerful than a collection of very good theorems. Great job sir, the internet and I thank you ☺️ 🙏🏽💯🙌🏽
Thank you :)
Finally, I found someone who can beautifully explain Real Analysis. Once, I hated this course because I couldn't understand it.
You're doing a great job 👍🙏
Me too, real analysis was one of my nightmares
ahh, your channel is a gem....
Thanks so much for this! I FINALLY understand the concept of the Lebesgue integral! Every time I would try to look up a definition of the Lebesgue integral, it always came across as very abstract but that the idea was that we use arbitrarily sized Δx's. I never got a sense of how you parameterize those Δx's so you could use them to systematically calculate the integral. Thanks to you, I now see that the division is based on the output of the function and parameterized based on that. Hence the need to study Measure Theory to know the proper way to systematically "measure" these new types of divisions. Great video!!
The Δx(i) are not themselves what we parametrize to integrate. Δx(i) is an output family of a function applied to sets partitioning the domain of the function being integrated. The partitions are what we parametrize over.
This series is quite straightforward after watching your real analysis series, and surprisingly, it looks like functional analysis is not required to study before this series. Well done!
Thank you very much. You can check the map here to see what is required :)
tbsom.de/startpage
@@brightsideofmaths Ohh man you do have such a map! I just found a rough one from your "start learning maths" series lol. Thanks very much!
Very nice video. After reading the comment section, I’m really grateful for how good my professor explains and emphasizes these matters.
Riemann’s notation does have advantages as well, like its conceptually easier to understand (despite those technical difficulties), and the calculation is often more easier to be done by human and computer. Lebesgue’s notation is more desirable and suitable for theoretical purpose. Another advantage I would like to add is, actually the most important one for me, Lebesgue integral is closed under the convergence that we consider (point-wise, in L^p, in measure), that is, sequence of Lebesgue integrable functions has again a Lebesgue integrable limit, which is very crucial for analysis as we no longer consider function as a correlation but an object in a vector space. For Riemann integrable functions, the limit is almost always not Riemann integrable again.
This is amazing!!! After countless courses in math, I never understood what the Lebesgue integral was even about. Thank you for this video. Every math teacher should introduce courses like this !!!
I am studying Stochastic modelling, and we need to use Lebesgue integral for the Renewal Theory. Your video is an excellent material to begin with Lebesgue integral. Thanks for the time for making this video!
I wish every teacher would be as good as you at explaining things. Thank you very much!
Technically, this video did not work with the definition of the Riemann integral, but the definition of the Darboux integral. However, we know that a function is Riemann integrable if and only if it is Darboux integrable, and since the definition of the Darboux integral is simpler, this abuse of language is justified in context.
Personally, I prefer to motivate measure theory and measure-theoretic integration by starting with the definition of the Darboux integral, and thinking of ways to directly loosening the definition, in order to produce a generalized notion of integration, along with reasons with why we should create such a generalization, by explaining each loosening adequately. The upper Darboux integral of f : [a, b] -> R is defined as the infimum of the upper Darboux sums, which are sums of sup{f[t(i)] : t(i) in [x(i), x(i + 1)]}·[x(i + 1) - x(i)], with x(i) < x(i + 1), and the union of every [x(i), x(i + 1)] is [a, b]. The lower Darboux integral is the supremum of the lower Darboux sums, which are sums of inf{f[t(i)] : t(i) in [x(i), x(i + 1)]}·[x(i + 1) - x(i)]. A function is Darboux integrable if the upper Darboux integral and the lower Darboux integral both exist, and are equal. For the purposes of motivating the Lebesgue integral, I will focus only on the lower Darboux integral, since the Lebesgue integral, as covered in this channel, is exactly a direct generalization of the lower Darboux integral.
First, we acknowledge that the reason the Darboux sums are defined as such is because the partitions of [a, b] into [x(i), x(i + 1)] are done so that the length of [x(i), x(i + 1)] is the width of the rectangle under f, and inf{f[t(i)] : t(i) in [x(i), x(i + 1)]} is the lower bound to the height of the rectangle, while sup{f[t(i)] : t(i) in [x(i), x(i + 1)]} is the upper bound to the height of the rectangle. An idea here is to make the relationship between the partition [x(i), x(i + 1)] and the widths more natural by having a length function λ with the property that λ([x(i), x(i + 1)]) = x(i + 1) - x(i). Thus, we can write the lower Darboux sums as sums of inf{f[t(i)] : t(i) in [x(i), x(i + 1)]}·λ([x(i), x(i + 1)]). Here, the visual and abstract connection between the individual lower Darboux sums and the individual partitions is made natural and intuitive. A more concise notation can be adopted, by letting S(i) = [x(i), x(i + 1)], so that we can simply write the lower Darboux sums as sums of inf{f[t(i)] : t(i) in S(i)}·λ[S(i)], where each S(i) is an adjacent closed interval, and the set of S(i) partitions [a, b]. Then λ[S(i)] should be interpreted as the length of S(i). This adequately explains how to motivate the definition of the lower Darboux integral from our intuitive idea of "area enclosed by the graph of f".
As the video explains, there are problems with this definition. For instance, there are functions whose graph is such that it cannot enclose enclose rectangles in its area, so partitioning [a, b] into S(i), where each S(i) is a closed interval, is inadequate for capturing the idea of width formally. Rather than partitioning [a, b] into closed intervals, the elements of the partition S(i) should be allowed to be arbitrary elements of a collection A of subsets of [a, b]. Accordingly, we need to generalize what our length function is, so that λ[S(i)] can be well-defined even if S(i) is not a closed interval, but rather, λ should admit any element of A as an input, and the output should be a quantity that aptly captures the idea of "length of A". On that note, this generalization should still be applicable if the domain is some arbitrary non-empty set X, rather than specifically [a, b]. The codomain may also be a Banach space over R, rather than just R, although this is not strictly speaking a generalization.
Here, we have three concepts that have been loosened: the domain of f has been loosened from being a closed interval [a, b] of R into being simply an arbitrary non-empty set X; the collection of subsets of [a, b] has been loosened from only containing closed subintervals of [a, b] into simply containing arbitrary subsets of X, and this collection of subsets is A; the function λ has been loosened from only acting on closed subintervals of [a, b] to now acting on arbitrary elements of A, being now a function μ with domain A, and we may even allow for μ([x(i), x(i + 1)]) = x(i + 1) - x(i) to not be satisfied in general for the sake of broader applicability. This gives us a space (X, A, μ) with respect which our notion of generalized or loosened lower Darboux sums is defined. These sums may be called lower weighted sums, and the supremum of the set of such sums will be our new notion of integral, or our new notion of lower integral, depending on how powerful this notion is. The lower Darboux integral is the special case where X = [a, b], A is the set of closed subintervals of [a, b], and μ is the restriction of the Lebesgue measure to the set A as domain.
Measure-theoretic integration is what you get when you restrict A to being a σ-algebra of X, and μ to be a measure with domain A. Of course, alternative formulations of integration exist, where μ is even more restricted than a measure, or is more loose than a measure, and A is even more restricted than a σ-algebra, or more loose than a σ-algebra. Measure theory is the systematic, axiomatic study of these different types of spaces, their properties, and restrictions or loosenings thereof, along with the study of the properties of the morphisms between these spaces, which are the functions we want to integrate. Of course, understanding how these spaces can be restricted into being measure spaces, and how can they be generalized, opens up the door for a theory that allows for integrals that are, so to speak, in between the Darboux integral and the Lebesgue integral, or notions of integration that are even stronger and more applicable than the Lebesgue integral, such as the gauge integral or the Khinchin integral.
Shout out from S.Korea from about 14:00 It was a moment of reckoning for me I have been learning measure theory and Lebesgue intergal in class but I didnt exactly know why we need these quite clearly Your video just gave whole new meaning to my Real Analysis study Thank you very much sir NOW I AM ENLIGHTENED!!!
SUPER CLEAR AND INSIGHTFUL. LOVE!
Thanks a lot :)
Thank you sir. Very very good empressive explanation
Good job, I was never taught to think of Lebesgues integrals this way but I quickly learned what you showed thanks to a friend back when I was learning it.
I was expecting exactly what you you showed in the video, and I am happy to see that any student could see that on TH-cam because sadly it is easy to be lost in the theory (I have myself some bad memories about mesure theory) without realizing the motivation behind it.
Even though I had already taken a Measure Theory course, I didn't come across this interpretation until now. I loved it. Very beautiful ideas.
Perfect work. What I didn’t like in my university education was that professors often skipped motivation and history. And it makes math _too_ abstract.
I have a book with biographies of greatest mathematicians and the book is written not only about theirs lives but also about theirs discoveries. And it helps!
what is the name of that book?
@@kaiwalpanchal5872 Actually I had several but it's been a long, long time. Right now I have found only *"Tales of Mathematicians and Physicists"* by _Simon Gindikin_ on my shelves. It's good though.
thank you! Now I finally understood what I "learned" ten years ago in my math lecture :) Good work embedding the important statements concisely in twenty minutes!
A function having infinitely many discontinuities is also Riemann integrable(5:05) provided the measure of the set of discontinuities is zero.
You are partly right. Infinitely many discontinuity points CAN destroy the Riemann integrability. That is what I said and meant there. For example, all monotonically increasing functions are Riemann integrable. However, having measure zero for this set of discontinuities is not sufficient since the function has also to be bounded (Lebesgue Criterion).
Other major problems of the Riemann integral include the case when the integration inteval [a,b] is infinite and/or when the function has unbounded singularity points. The so called improper Riemann integral used to deal with these is a mess. Riemann integration only works well when functions fall inside a box.
Why do you think it is a mess?
@@Pklrs perhaps because how we define the thickness of the rectangles
Both have their problems, if I recall correctly from university (its been a long time ...).
The integral of sin(x) on ]-inf, inf[ is 0, isn't it?
But you cannot use the Lebesgue integral to integrate that, AFAIK.
If I recall correctly, it's due to the construction of Lebesgue:
1. Define the integral for non-negative functions
2. Define the integral for purely negative functions f as the negative of the integral of -f
3. For an arbitrary functions, separate its definition set into 2 parts: the part where f is negative, and the part where it is positive, integrate both parts and then take the sum
If you applied this to sin(x) on ]-inf;inf[ is, that you had to go with the 3rd option. Thus you had to construct the integral of sin(x) for the subset where sin(x) is positive, but the integral of sin(x) over {e el ]-t,t [ | sin(e) >= 0} for t -> inf is not defined as it would become infinitely large (and the same for the negative part...)
But on the other hand, its been a long time, and I was but a normal student, thus perhaps there is a fix on higher levels?
@@w1darr No, the integral of sin(x) on ]-inf, inf[ is undefined, as what you actually do is integrate sin(x) from a to b then take the limit of (a, b) -> (-inf, inf). This depends on your actual limit 2D, while int sin(x) from -n to n is 0, int sin(x) from -2pi * n to pi * (2n + 1) alternates between -2 and 2 and is hence divergent. So the improper integral does not exist.
Kartik Nair i guess the improper integral is like a limit with two variables going to infinity.
I love it when people make me feel dumb for not understanding things that once seemed hard. It shows me that more that I imagine is possible. Thank you very much, I will study Lebesgue Integration in detail now ☺️🙌🏽❤️
Thank you for this Lebesgue vs Riemann video! I'm currently auditing MIT's online Functional Analysis course for my personal enrichment (currently covering sigma-algebras, outer measure, & Lebesgue measurable sets), and these clear illustrations are invaluable.....I'd venture to say I learned more in 19 min than in nearly 3.5 hrs worth of proofs lecturing!
Thanks a lot! I also have a whole series about measure theory for more details :)
I don't know if I haven't been listening to my professor but this is the first time that I finally understand the overlap between Lebesgue measure and the Lebesgue integral. Thanks for the help.
The idea of integration, in the context of measure theory, is that you integrate a (measurable) function with respect to an interval. Riemann integration can be taken to be integration with respect to the Lebesgue measure, together with many restrictions. Getting rid of those restrictions leaves with just Lebesgue integration, although the phrase Lebesgue integration is sometimes used to refer to all measure-theoretic integration in general, rather than integration with respect to the Lebesgue measure specifically.
This is a lucid yet insightful introduction to the idea behind Lebesgue integral. Amazing! Keep up the great work!
Thank you!
Wow, i graduated in applied physics and I've never heard of Lebesgue.
I'm constantly baffled @ how much there is to learn.
Thanks!
So 'Everyone wants to work with the lebesgue integral' is clearly bullshit. People don't use it. They don't even teach it at universities for stuff like physics
@@peterbonnema8913 This is a math topic, applied physicists don't need to learn about or work with intricate mathematical definitions. I don't know what applications you are looking for with a topic from analysis
@@peterbonnema8913 dude you need a lebesgue integral to define a hilbert space which is important in theoretical physics and math it's used a lot
master shooter64 It’s easy. You just say: assuming the necessary assumptions, let H be a Hilbert space
@@jesiryt8583 You watch Andrew Dotson and Flammable maths too!
Summary:
Riemann integral:
maps R to R, related to area under a graph which is approximated by a lower sum and higher sum of squares.
problems:
1) Doesn't scale to higher dimensions easily. As we increase dimensions the shape of the partition we need to specify the range of the integral increases exponentially.
2) Functions with infinitely many discontinuities cannot be integrated with the Riemann integral.
3) We can only pull the limit inside the function when the uniform convergence property holds.
We want an integral that works well in every dimension.
Lebesgue Integral:
Instead of partioning the X axis (which may be abstract or high dimensional), we decompose the Y axis. We want to find all the parts of the function that lie between small intervals ci. This causes discontinuities in the parts we get when are intervals are still big. To measure the lengths/areas/volumes of these discontinuous sets ( Which we call A). The total measure space is called mu.
The area is thus the sum over the whole partition of the "rectangles", where Ci is its height and mu(A) its width
No, your list has many mistakes and inaccuracies.
0. The Riemann integral is not defined strictly for elements of R^R, only for elements R^[a, b]. Meanwhile, the Lebesgue integral is defined for functions B^X, where X is the carrier set of an arbitrary measure space, and B is a Banach space over R.
1. Some functions with infinitely many discontinuities are Riemann integrable, but most are not.
2. The Lebesgue integral still partitions the x-axis, not the y-axis. The difference is that it partitions it with measurable sets, rather than closed intervals of R, and rather than multiplying each infima by the length of such intervals, one multiplies by the measure of the sets in the partition.
3. The total measure space is (X, Σ, μ), not μ itself. μ is simple called the measure, and it is a function from Σ to [0, +♾].
Nice vid. I did a bachelor in maths many years ago so I've pretty much forgotten all of it but I could follow this very easily while eating my breakfast.
What did you end up doing with your degree?
@@TheJProducti0ns Nothing I'm sorry to say. I tried to be hired as a trainee actuary but I think they probably thought I was too old. I also tried to get into the central statistics office where I live but although I did well on the language and maths test I didn't do well on the managerial test so they didn't hire me. I was working as a software developer at the time so I kept doing that and still do. Around 2008 I worked for a year as a secondary-school maths teacher (substituting for a woman who had a nervous breakdown) but when the year ended she was ready to come back and they wanted me to pay for a teachers education yet again for less hours so I didn't continue that and worked again as a software developer. Which is what I still do.
@@delberry8777 damn. do you ever regret not doing anything with your maths degree? I'm sure software developing pays quite well but do you ever wonder what could have been?
@@icontrolthespice Yes I do but it might also have been very hard. I regret mostly not having kept it up and forgotten mostly everything I learned.
@@delberry8777 You can refresh a lot, I guess. Just use a little time every week to read, to watch some video and to solve some problems. Math can be fun :)
Thank you! People like you are the only ones restoring this world
The nice thing about not knowing something is the hit of dopamine you get when you see "the big picture" ☺️❤️🙌🏽🔥 Thank you for such an eye-opening lecture ☺️
👍
THANK YOU THANK YOU THANK YOU. I need to visualise stuff to actually comprehend it, especially stuff that is, at first sight, more complicated for me. I was really struggling to find a nice, simple explanation of the Lebesgue integrals and I found this. Thank you so much!!
I would love to thank you for your efforts to explain this important mathematics topic.
All my respect.
Thank you very much :)
Wow, good video! I actually have learned in calculus how to generalize the Riemann integral to higher dimension, but I hadn't seen the Lebesgue integral yet. So... Thanks for the extra knowledge!
I first watched this video around when it was posted and I was still in secondary school. At the time I was unsure about studying maths much further. Now I am at one of the top universities in the world studying maths. Crazy.
Wow. I am happy to help on your journey :)
great video. i wish i had this vid on the first day of measure theory as it starts with the motivation and a direct comparison to the riemann integral which make every definition that follows more reasonable.
19:05 "okay to sum it up" haha I see what you did there
Wow! Thanks for the video, it was really helpful for understanding the basic idea behind the Lebesgue integral.
Very instructive, especially for the consequences, which are sometimes hidden by the hardness of theorems' demonstrations
Thank u for all the course. I just finished. You are a great professor. Thank u so much and I wish you a happy life ❤
You are very welcome :) And thanks for your support!
All time classic. Just rewatched it :)
Thanks :)
Finally! A well - motivated explanation of the Lesbesgue integral.
Thank you for this clear explanation! I was struggling a bit through wikipedia definitions, but you gave very clear motivation behind this concept!
Great to hear!
I was surfing privately and found your channel, but yourcontent made me login to subscribe. Gr8 content man
Thanks :)
Fantastic video! I finally understand the lebesgue integral in one go, when looking in books and on Wikipedia just wasnt clear
Same reason here. I'm Engineer and I had no Measure Theory on the undergraduate course.
Although I don't use Lebesgue Integral, it always came across here and there. But every single explanation I found assumed I knew more math than the necessary to just say: "Ok, I understood de diference, I now know the basic concept, it has become clear the limits of Riemann Integral and how Lebesgue expands those limits".
way better than my teacher's explain!
This is in my opinion the best way to introduce Lesbegue integration.
Maybe I was lucky, but when I came in touch the first time with Lesbegue integration it was exactly the same explanation: instead of partitioning the domain into small pieces and doing the limit process, the range of the function was partitioned and the corresponding parts of the domain had to be measured. For this the domain needs to be "measurable" to get a notion of "volume".
This concept seemed to be very intuitive to me. The best part of it was the possibility to integrate functions that are not Riemann integrable like the Dirichlet function.
Thanks a lot :)
Thank you so much for this explanation.
You're very welcome! :)
Amazing video - I wish I could have seen this during my undergraduate studies!!
Nobody can really understand a graduate probability course without understanding this. There will no piece of mind if someone tries, as most including me did.Thanks.
Correct. Probability theory is a special case of measure theory.
the most comprehensible explanation I've ever heard
Extremely clear explanation. Thank you!
I wish you were my maths teacher.....such nice intuitive explanation
Wow, thanks
Thanks. This also gives a motivation for Measure Theory, which is very nice!
The problem to be solved in the riemann function was how to partition abstract higher dimension spaces. Lebesgue found a way and changed the problem to , how to measure the volumes of abstract spaces
Amazing video man, really helped me understand the whole motivation behind Measure Theory. Also juast as I was studying Functional Analysis in my AI Bachelor you released a playlist about it. God bless u 😂
Thanks for the video, I am now wanting to learn about the Lebesgue-Stieltjes integral. I was wondering if you would recommend any videos or resources to learn this?
How about my videos about it? tbsom.de/s/mt
Thank you for this channel, it’s amazing! Quick question, are you German? The accent sounds a bit like my German grandma and that’s cool!
Thank you very much! Yeah, my German accent will never vanish :D
@@brightsideofmaths Hallo, ihre Videos sind sehr hilfreich, dass ich mich frage, was sie hauptberuflich machen. Arbeiten sie zufälligerweise an einer Universität oder machen sie das nur nebenbei?
The point you said about extending Riemann Integration to higher dimension:
Why would have to have partition the domain that we're integrating over? I thought the standard approach is just to cover it with a box and then integrate the function "f" times the "inticator function" of your domain, over the box.
The question is still: how do you define the integral for a function g? You still need to partition the domain even if you choose as a box. Of course, the box makes it simpler but the other problems remain.
Thankyou. I enjoyed the colorings with it.
OMG! Why couldn't someone have shown me this years ago when I muddled through real analysis? That makes so much more sense. Now I want to read Royden again.
Very nice explanation 🎉
Very clear ! Thank you so much from France !
The one thing I learned from this video is that in Lebesque 's' is silent
Now learn it's spelled with a g rather than a q
I just found your channel, but this is awesome! Subscribed!
thanks a lot for that amazing explanation, i watched it after learning the lebesgue integral chapter and it really helped me to imagine its' dynamism. I guess this is going to help me solve futur "hard" problems where we've to pose functions,... thanks again
What software did you use to record it? I've seen it in action before (at least I think it was the same one) but I forgot its name - I was taught physics with it and it was great :D
I use Xournal :)
A major pet peeve of mine is the typical drawing of Lebesgue integration as horizontal rectangles, when this is simply not how the integral is defined or calculated. Of course, you can define an equivalent integral by calculating those horizontal rectangles, but that's not what's taught, and so that particular graphic confuses so many students, my past self included.
The fact that the thumbnail and the video itself don't use this graphic, and instead use one with vertical bars makes me very happy.
Thank you very much. I always try not confuse students :)
Great explanation sir...Now I have to subscribe your channel to learn more nice mathematics topics
Very interesting explanation of a difference between Riemann and Lebesgue integral, motivation and why the later is preferred. I was only missing one piece at the end. You started your discussion with Riemann integral over an interval [a, b]. It would be nice to see how one calculate Lebesgue integral on [a,b] especially for non injective functions, as the one in you example with Lebesgue integral.
Thank you! I have a whole series about measure theory where you can find some examples. Just calculating a Lebesgue integral for a function f: [a,b] -> R is not very interesting in this regard. In the video above, I wanted to show the motivation and why one should study measure theory :)
great explanation ...its very easy to understood more than studying definition. make more video in real and algebra tooo
thank you
Thank you for the nicest introduction for lebesgue integration. A lesson series about measure theoretical probability would be great btw :)
Excellent work man, thank you
Glad you liked it!
Thank you so much for this awesome video
You are welcome :) Thanks for watching!
Danke für die tollen Videos :)
Sie haben mir echt geholfen das Lebesgue Integral also vor allem die Motivation und hier den Unterschied zB besser zu verstehen, aber irgendwie schaffe ich es noch nicht tatsächlich ein Integral einer nicht Treppenfunktion über den Lebesgue Weg zu berechnen/bestimmen. Darum fände ich eine Empfehlung, wo ich Beispielrechnungen finden kann oder auch ein kurzes Video zu Beispielen super hilfreich
Gerne! Ich habe eine ganze Video-Reihe über Maßtheorie: th-cam.com/video/4DHP8cBcg_o/w-d-xo.html
(Deutsch oder Englisch).
Vielleicht hilft das schon mal. Es ist ja auch nicht so, dass man das Lebesgue-Integral für explizite Berechnungen benutzen muss. Dafür ist der gewählte Integralbegriff ja oft ziemlich egal. Es geht um die gesamte Theorie, die man mit dem Lebesgue-Integral aufziehen kann.
Most integral in partitions can be created as function of a function recursive. Ordinary partitions are just like quantised elementary partitions.
Not really. That is just a very specific type of high-dimensional integral in which you pre-assume Fubini's theorem.
Wow awesome...you deserve 1M subscribers keep it up
Great explanation thank youu so much ,please keep doing such videos in English 😊
Thanks. Yes, I will translate all my other videos at some point :)
Great video!
May I ask what equipment and software you are using? I've been looking into digital note taking for mathematics but haven't decided upon anything wholeheartedly as of yet.
Thanks. I use Xournal and a Wacom for writing. For digital note taking, I am now a fan of Boox Note Ebook reader.
However, I advise you to test some things. For me, for example, Microsoft Surface is just not accurate enough but other colleagues love it.
@@brightsideofmaths Thank you! I hope you continue to make great videos in English :)
Please would you kindly add this to your measure theory playlist. It really filled in a lot of gaps while I was going through the series. Specifically why Lebague integration was helpful.
Thank you. However, I don't know where this would fit in :) Suggestions?
@@brightsideofmaths I was thinking just after Part 6 perhaps as a part 2. Thank you for such a wonderful series by the way :)
Nice presentation! To complete it, it would be nice to add an example on how to calculate the Lebesgue integral. Thank you.
Thank you. You can see examples and more in my measure theory series: th-cam.com/play/PLBh2i93oe2qvMVqAzsX1Kuv6-4fjazZ8j.html
I appreciate you
A lot of thanks for good representation and explanation.
Is this playlist finished? I would like to watch a full playlist from this channel that is already finished
See here: tbsom.de/s/mt
Nice explanation! Thanks so much!!!
really awesome work, definitely a hidden gem, thanks for sharing!
Well, that is very nice and clear as far as it goes. But I am left wondering how to obtain mu(A_i) for a given c_i.
For that question, I have a whole lecture series prepared :)
th-cam.com/play/PLBh2i93oe2qvMVqAzsX1Kuv6-4fjazZ8j.html
Danke sehr! Dieses Video ist wunderbar!
MCQ on Riemann integral
pathaksir2.blogspot.com/2020/11/mcq-on-riemann-integral.html?m=1.
Great video! Thank you!! What is the whiteboard software that you use?
Thanks! It is Xournal.
Is the classical example of a non integrable function with an infinite number of discontinuities the indicator function of rational numbers ?
Yes, for a non-Riemann-integrable function.
Suppose µ is counting measure on Z
+ and b1, b2, . . . is a sequence of nonnegative
numbers. Think of b as the function from Z
+ to [0, ∞) defined by b(k) = bk
. Then
Z
b dµ =
∞
∑
k=1
bk
,
How we solve it
hello I think what you describe at around 12min might be a bit inaccurate. the Riemann integral is not defined via upper and lower sums. in the Riemann integral the Riemann sum converges to the integral of a function as the partition norm / mesh -> 0. What you describe at min 12 are technically the Darboux sums. however due to equivalence of the two it doesn't really matter. great video.
Yes, I call the Darboux integral also just Riemann integral :)
I typically think of the Riemann integral as being more naturally related to adding a large but finite number of discrete terms represented within a sigma term. Is that just a prejudice of mine? If not, is there any finite analogue to the Lebesgue integral, (I guess being one where we index over terms with respect to the output of a function rather than the arguments fed into it?), and does that analogue have any unique niceness to it? I can see that this matches the notation that you use near the end of the video, but what I mean is, this seems like maybe an idea that has utility even outside the context of calculus?
Yes, that is just a prejudice of yours. At no point in the definition of the Lebesgue integral are you ever adding infinitely many terms over the σ-algebra, so the Lebesgue integral is also a sum of an arbitrarily large but finite number of terms. To be precise, both the Riemann and Lebesgue integral can be defined as suprema of finite sums.
Hi, this μ( { x∣f(x) > y } ) is the way to choose partition A in video 16:29, right? Which one would it be for another partition? I get confused because the next partition would not meet the condition of f(x) > y, it would be mixed with the previous one, partition A meets it because it is the highest partition with the highest c_i but another one would not meet it, I don't know if I made myself clear, Thanks anyway.
It's not quite clear what you mean. We just choose the x with f(x) between c_i and c_{i+1}.
Thank u sir ur explanation is very easy to understand
Brilliant sir! Thank you.
Glad it was helpful!