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!
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 👍🙏
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 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.
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 RU-vid 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.
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 ☺️🙌🏽❤️
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 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.
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.
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 ☺️
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, +♾].
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 :)
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
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.
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.
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.
There is something I don't get it.... in min 17:00 in the Lebesgue Integral defined by a Summatory...¿it should be (c_[i] - c_[i-1])*mhu(A_[i]) ???... in the way stated in the video its looks that you are adding several times every area under c_[i] for every c_[j] with 0
This is quite nice, but it seems to me that you glossed over one aspect: you said that it is hard to extend the Riemann integral to R^2, say, due to the requirement for limit processes in defining the intervals - however, this is only a problem if, in fact, it is straightforward to extend the Lebesgue measure to R^2. If it is no easier to do that than to define the 2D intervals required for Riemann, then Lebesgue gives no advantage, from that point of view at least. Perhaps you could do a lecture showing how the Lebesgue measure is defined on R^n, to cover this?
Thank you for the comment! Of course, here you are absolutely right. I don't go into details in this video, and the 2D example for Riemann was just an illustration how technical the definition of the Riemann integral has to be. In fact, this looks much nicer for the Lebesgue integral and works immediately for R^n. I have same videos in measure theory that go into more details and I am still working on more :)
So the difference between Lebesgue and Riemann is that a Riemann integral is defined using a longitudinal form of integration versus the Lebesgue's transverse form of integration?
Short answer: sort of Long answer: No, it is still a partition in the x-axis, but Lebesgue integration is more general with the types of partitions allowed.
Thank you very much for this excellent playlist of videos. I have a small question if you don't mind: Shouldn't we add a "lim mu -> 0" prior to "sum c_i * mu(A_i)" so that it equals "int f d mu" just like we do with Riemann integral?
That's super cool! Because with this Lebesgue integral is valid on any support, could you tell us about Lebesgue integrals on a fractal set ? (for ex Cantor dust)
Countably many discontinuities should also be Riemann integrable because the function is still continuous almost everywhere at countable discontinuities.
Hey great video, except I spotted a mistake. The Upper Sum U(P) is the sum of the infinums of the functions multiplied by the displacement, not the supernums. And for the Lower Sum L(P) it should be supernums not infinums.
One immediate difficulty that I anticipate with Lebesgue integral is that it looks like you are gonna need to have access to the inverse of the function you are trying to integrate. Is that true?
@@angelmendez-rivera351 sure, but you still need the inverse, right? That is, for each segment in the range, you should be able to find all the corresponding segments in the domain.
@@siarez Are you talking about the definition of a measurable function? This does not require finding the inverse function at all, this only requires that you know how to find pre-images. Also, at no point do you ever actually need to calculate the pre-images in order to evaluate the Lebesgue integral. The pre-images being measurable sets is only for proving a function is measurable, which, in the year 2021, is not really necessary, since almost every function worth studying has already been studied and determined to be measurable or not with respect to a given σ-algebra.
@@angelmendez-rivera351 Thanks for your reply. I'm talking about a situation where I'm given an actual function and I want to compute its integral using Lebesgue instead of Riemann. How can you find pre-images without the inverse?
I think the problem the author mentioned was when you had a subdomain of R^2, such as a disk. I suppose one could get around this by extending the integrand to be zero outside the disk, and partitioning R^2, but then the integrand would no longer be continuous. So it would seem to involve at the least lots of explanation and a great temptation to hand wave. :)
Instead of explaining the problem with partitioning R^d I'll instead explain why the Riemann integral is not fit to tackle all sets... in the easy case of one dimension. So with the Riemann integral we can integrate ANY continous function over some arbitrary set, as long as this set is some union of intervals. We can also add and remove arbitrary points to these sets, without changing the value of that integral. Now imagine some guy walks up to you and asks: "what about integrating over some interval [a,b] where we remove all rational points from it?" and with the notion of the Riemann integral you can just go ¯\_(ツ)_/¯. With the concept of measure spaces we can basically extend the notion of integral to any set that is measurable (which is a whole lot more) in a clean way, without having to actually think about partitioning our base space (especially in a case where this is not possible, like in the case of [a,b]\ℚ).
Dan Hitt I think simple 1/n side cubes are pretty a straightforward basis for a grid in any R^n. But I do see the point that the definition of regions that have to be filled with them is neat. Howerver, to actually calculate their measure, one usually would have to have to actually resort to such a grid at least on Lebesgue measurable full measure sets. Again, that is of course far fro a general situation, but bloody important in applications :)
@@Zaustie Hi, thanks for interest in my unwitty comment :) The thing is, that I really only comment on the partition of R^n and only that. But, since you mentioned it, imagine instead integrating a very specific integral, nothing fancy though, but definitely something that you cannot do with pen and paper, like some likelihood function in statistic. There are no problems with continuity, nor any problems with the countable subsets of measure zero. What you have, is just a plane ugly, unromantic function to integrate. Now, in that case, one would most likely do some numerical approximation and used the very well defined finite grid, which is, again, well defined. I don't know, if that would solve the problem as dimensions would grow, however. Then, again, one might say that Lebesgue would beat Rieman, because you would have to do some Monte Carlo to represent the measure by points. Anyway, independent of that it is easy to define a meaningful partition of R^d :) Best wishes!
@@matteolacki4533 You are wrong. No such well-defined grid is even required to calculate the Lebesgue measure of any Borel set, because the Lebesgue measure itself is well-defined.
Best case scenario: In the first year (maybe just for the one-dimensional Lebesgue measure). Most likely: In the first or second Stochastic course. Worst case scenario: Never.
It's called an introduction for a reason. I get that the video is not exactly in-depth, but this kind of criticism is not constructive at all and is ultimately kind of stupid.
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.
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.
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.
@@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?
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 :)
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.
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
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
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!!!
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!
