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!
@cliffordino4 жыл бұрын
Thank you for clearly explaining the *motivation* behind the Lebesque integral.
@active2854 жыл бұрын
The mathematician's name was Henri Léon Lebesgue! Common, but awful mistake ;).
@cliffordino4 жыл бұрын
@@active285 Thank you for correcting me!
@monicadelpilar234 жыл бұрын
Und jetzt es ist auch hier: @mathupdate
@diptree2 жыл бұрын
You cut the abstract clutter and clearly explained the concepts. Brilliant !
@johncreet12544 жыл бұрын
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.
@brightsideofmaths4 жыл бұрын
Thank you very much! Please use my videos to refresh your Math memories :)
@sundareshvenugopal65753 жыл бұрын
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.
@dananajj4 жыл бұрын
I like how you used a yellow background; easier on my eyes when watching this in the dark.
@brightsideofmaths4 жыл бұрын
Thank you. That is my style :)
@mahendrapathak26583 жыл бұрын
MCQ on Riemann integral pathaksir2.blogspot.com/2020/11/mcq-on-riemann-integral.html?m=1.
@ozzyfromspace4 жыл бұрын
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 ☺️ 🙏🏽💯🙌🏽
@brightsideofmaths4 жыл бұрын
Thank you :)
@alkankondo894 жыл бұрын
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!!
@angelmendez-rivera3513 жыл бұрын
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.
@fatemekashkouie36622 жыл бұрын
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 👍🙏
@slothochdonut3099 Жыл бұрын
Me too, real analysis was one of my nightmares
@veruskapaninforni64394 жыл бұрын
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.
@mahendrapathak26583 жыл бұрын
MCQ on Riemann integral pathaksir2.blogspot.com/2020/11/mcq-on-riemann-integral.html?m=1.
@veruskapaninforni64393 жыл бұрын
Yes, but be careful to Dirichlet function, where only L integral can be applied.
@angelmendez-rivera3513 жыл бұрын
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.
@rajarshighoshal62564 жыл бұрын
ahh, your channel is a gem....
@angelmendez-rivera3513 жыл бұрын
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.
@Harmonica20005 жыл бұрын
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!
@r.d.75754 жыл бұрын
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 !!!
@gyeonghunkang72153 жыл бұрын
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!!!
@Jay-ms1dv10 ай бұрын
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!
@brightsideofmaths10 ай бұрын
Thank you very much. You can check the map here to see what is required :) tbsom.de/startpage
@Jay-ms1dv10 ай бұрын
@@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!
@gunhasirac3 жыл бұрын
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.
@xavierplatiau46354 жыл бұрын
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 KZbin 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.
@beo_wee4 жыл бұрын
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!
@kaiwalpanchal58724 жыл бұрын
what is the name of that book?
@beo_wee4 жыл бұрын
@@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.
@baoshanzhang69129 ай бұрын
SUPER CLEAR AND INSIGHTFUL. LOVE!
@brightsideofmaths9 ай бұрын
Thanks a lot :)
@spin77653 жыл бұрын
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.
@choxhicken284 жыл бұрын
I wish every teacher would be as good as you at explaining things. Thank you very much!
@AllisinMusic5 жыл бұрын
Wow, i graduated in applied physics and I've never heard of Lebesgue. I'm constantly baffled @ how much there is to learn. Thanks!
@peterbonnema89134 жыл бұрын
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
@thedoublehelix56614 жыл бұрын
@@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
@mastershooter644 жыл бұрын
@@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
@jesiryt85834 жыл бұрын
master shooter64 It’s easy. You just say: assuming the necessary assumptions, let H be a Hilbert space
@thedoublehelix56614 жыл бұрын
@@jesiryt8583 You watch Andrew Dotson and Flammable maths too!
@ozzyfromspace4 жыл бұрын
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 ☺️🙌🏽❤️
@zakiullah81895 жыл бұрын
Thank you sir. Very very good empressive explanation
@dennishui11024 жыл бұрын
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.
@Pklrs4 жыл бұрын
Why do you think it is a mess?
@Kartik-yi5ki4 жыл бұрын
@@Pklrs perhaps because how we define the thickness of the rectangles
@w1darr4 жыл бұрын
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?
@SwissBarracuda4 жыл бұрын
@@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.
@Pklrs4 жыл бұрын
Kartik Nair i guess the improper integral is like a limit with two variables going to infinity.
@achimbuchweisel27364 жыл бұрын
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!
@MultiWilliam155 жыл бұрын
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.
@angelmendez-rivera3513 жыл бұрын
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.
@AdventurersTavern4 жыл бұрын
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.
@nintorws4 жыл бұрын
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!
@ozzyfromspace4 жыл бұрын
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 ☺️
@naman40672 жыл бұрын
👍
@MrBaBaBlackSheep13 жыл бұрын
This is a lucid yet insightful introduction to the idea behind Lebesgue integral. Amazing! Keep up the great work!
@brightsideofmaths3 жыл бұрын
Thank you!
@mastershooter647 ай бұрын
19:05 "okay to sum it up" haha I see what you did there
@Tommaso_Paoli5 жыл бұрын
Wow! Thanks for the video, it was really helpful for understanding the basic idea behind the Lebesgue integral.
@syc806610 ай бұрын
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.
@brightsideofmaths10 ай бұрын
Wow. I am happy to help on your journey :)
@parabatai25213 жыл бұрын
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!!
@o.biertrinker96495 жыл бұрын
Thank you! People like you are the only ones restoring this world
@rahulshetty93353 жыл бұрын
I was surfing privately and found your channel, but yourcontent made me login to subscribe. Gr8 content man
@brightsideofmaths3 жыл бұрын
Thanks :)
@funtofun3215 жыл бұрын
A function having infinitely many discontinuities is also Riemann integrable(5:05) provided the measure of the set of discontinuities is zero.
@brightsideofmaths5 жыл бұрын
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).
@61rmd14 жыл бұрын
Very instructive, especially for the consequences, which are sometimes hidden by the hardness of theorems' demonstrations
@abdouabdelrazek6037 Жыл бұрын
I would love to thank you for your efforts to explain this important mathematics topic. All my respect.
@brightsideofmaths Жыл бұрын
Thank you very much :)
@vector83104 жыл бұрын
Finally! A well - motivated explanation of the Lesbesgue integral.
@jaimelima24204 жыл бұрын
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.
@angelmendez-rivera3513 жыл бұрын
Correct. Probability theory is a special case of measure theory.
@anuragsaraf24246 ай бұрын
Thank you so much for this explanation.
@brightsideofmaths6 ай бұрын
You're very welcome! :)
@levimungai184610 күн бұрын
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
@masonpiatt27984 жыл бұрын
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!
@brightsideofmaths4 жыл бұрын
Thank you very much! Yeah, my German accent will never vanish :D
@2funky4u884 жыл бұрын
@@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?
@hr3nk2 жыл бұрын
Thank you for this clear explanation! I was struggling a bit through wikipedia definitions, but you gave very clear motivation behind this concept!
@brightsideofmaths2 жыл бұрын
Great to hear!
@peterpankert38102 жыл бұрын
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.
@brightsideofmaths2 жыл бұрын
Thanks a lot :)
@KarLEASty4 жыл бұрын
Thanks. This also gives a motivation for Measure Theory, which is very nice!
@islsaka15473 жыл бұрын
Thank you for the nicest introduction for lebesgue integration. A lesson series about measure theoretical probability would be great btw :)
@perkelele4 жыл бұрын
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
@angelmendez-rivera3513 жыл бұрын
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, +♾].
@johndougherty72164 жыл бұрын
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.
@mohammadmahmood82555 жыл бұрын
Great explanation sir...Now I have to subscribe your channel to learn more nice mathematics topics
@hamzabouzid48954 жыл бұрын
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
@elliotxie56255 жыл бұрын
way better than my teacher's explain!
@ompatel80914 жыл бұрын
I just found your channel, but this is awesome! Subscribed!
@s.meritnihitha13835 жыл бұрын
great explanation ...its very easy to understood more than studying definition. make more video in real and algebra tooo thank you
@kirstenwilliams60563 жыл бұрын
Amazing video - I wish I could have seen this during my undergraduate studies!!
@fattahsakuldee9534 жыл бұрын
the most comprehensible explanation I've ever heard
@Ab-ii4oc2 жыл бұрын
I wish you were my maths teacher.....such nice intuitive explanation
@brightsideofmaths2 жыл бұрын
Wow, thanks
@RSLT2 ай бұрын
Thanks ❤
@brightsideofmaths2 ай бұрын
You're welcome 😊
@martinepstein98264 жыл бұрын
Extremely clear explanation. Thank you!
@kabsantoor32514 жыл бұрын
The one thing I learned from this video is that in Lebesque 's' is silent
@DrJGang4 жыл бұрын
Now learn it's spelled with a g rather than a q
@navidemami2433 жыл бұрын
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
@brightsideofmaths3 жыл бұрын
Gerne! Ich habe eine ganze Video-Reihe über Maßtheorie: kzbin.info/www/bejne/anWrgWuZd8ialdE (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.
@Florian.Dalwigk2 жыл бұрын
All time classic. Just rewatched it :)
@brightsideofmaths2 жыл бұрын
Thanks :)
@MathLab4u4 жыл бұрын
I appreciate you
@alxjones2 жыл бұрын
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.
@brightsideofmaths2 жыл бұрын
Thank you very much. I always try not confuse students :)
@nerdygreg41984 жыл бұрын
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.
@brightsideofmaths4 жыл бұрын
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 :)
@andravelea12724 жыл бұрын
Nice presentation! To complete it, it would be nice to add an example on how to calculate the Lebesgue integral. Thank you.
@brightsideofmaths4 жыл бұрын
Thank you. You can see examples and more in my measure theory series: kzbin.info/aero/PLBh2i93oe2qvMVqAzsX1Kuv6-4fjazZ8j
@thomasyoung398 Жыл бұрын
Excellent work man, thank you
@brightsideofmaths Жыл бұрын
Glad you liked it!
@musicalBurr5 жыл бұрын
Nice explanation! Thanks so much!!!
@Mulkek3 жыл бұрын
Thank you so much for this awesome video
@brightsideofmaths3 жыл бұрын
You are welcome :) Thanks for watching!
@lijisisi4 жыл бұрын
really awesome work, definitely a hidden gem, thanks for sharing!
@mortimertz6660 Жыл бұрын
Thanks for all you videos. I hope you can someday make video about Haar Measure as well!
@brightsideofmaths Жыл бұрын
On my list :)
@wojciechwisniewski61804 жыл бұрын
Your english is so german-ish :D "Let's schooz partitions" :D
@brightsideofmaths4 жыл бұрын
I am so sorry about it :D I try my best.
@edgardmacena-ac43224 жыл бұрын
@@brightsideofmaths No reason to be ashamed of your accent! It's lovely
@sb_dunk4 жыл бұрын
@@brightsideofmaths Your English is clear and your accent has character! Don't change.
@douglasstrother65844 жыл бұрын
Just like Papa Flammy with his "serious series"!
@douglasstrother65844 жыл бұрын
You're guaranteed that "Riemann" will be pronounced correctly!
@DylanD-v9g Жыл бұрын
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 Жыл бұрын
How about my videos about it? tbsom.de/s/mt
@ashoktutorial72734 жыл бұрын
A lot of thanks for good representation and explanation.
@galgrunfeld99544 жыл бұрын
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
@brightsideofmaths4 жыл бұрын
I use Xournal :)
@davidmachorro60588 ай бұрын
Is this playlist finished? I would like to watch a full playlist from this channel that is already finished
@brightsideofmaths8 ай бұрын
See here: tbsom.de/s/mt
@lamorfati13023 жыл бұрын
Very clear ! Thank you so much from France !
@yahya-yl3rd2 жыл бұрын
Brilliant sir! Thank you.
@brightsideofmaths2 жыл бұрын
Glad it was helpful!
@MHRAJAI5 жыл бұрын
Thanks for the explanation
@selvav41045 жыл бұрын
Thank u sir ur explanation is very easy to understand
@venkatbabu1864 жыл бұрын
Most integral in partitions can be created as function of a function recursive. Ordinary partitions are just like quantised elementary partitions.
@angelmendez-rivera3513 жыл бұрын
Not really. That is just a very specific type of high-dimensional integral in which you pre-assume Fubini's theorem.
@tunafish12743 жыл бұрын
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.
@brightsideofmaths3 жыл бұрын
Thank you. However, I don't know where this would fit in :) Suggestions?
@tunafish12743 жыл бұрын
@@brightsideofmaths I was thinking just after Part 6 perhaps as a part 2. Thank you for such a wonderful series by the way :)
@DJFriendZone5 жыл бұрын
thank you, very very helpful
@ammarawaseer97793 жыл бұрын
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
@olivermechling797510 ай бұрын
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.
@brightsideofmaths10 ай бұрын
Yes, I call the Darboux integral also just Riemann integral :)
@anshumanagrawal34611 ай бұрын
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.
@brightsideofmaths11 ай бұрын
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.
@frankd11563 жыл бұрын
Wow awesome...you deserve 1M subscribers keep it up
@aghamehmanhajiyev37054 жыл бұрын
very good explanation! Thank you.
@michaelzhao31164 жыл бұрын
My question is how to obtain Lebesgue integral numerically?
@BenOgorek4 жыл бұрын
Michael Zhao I think it exists entirely for analytical purposes. You can integrate over the irrational numbers with Lebesgue integral, for God’a sake. Not sure you could ever program that into a computer. Anytime where you could numerically implement it, I believe the result would be the same as Riemann. I believe, but I do not prove. Please let me know if you find out otherwise.
@angelmendez-rivera3513 жыл бұрын
@@BenOgorek You can do numerical Lebesgue integration, it is not solely analytical. However, numerical Lebesgue integration is more annoying to program.
@davidefrey3434 Жыл бұрын
Nice explanation, but there's just one thing that I think you should clarify. From your drawing, it looks like your c_i is the point on the y-axis, not the width of the horizontal stripe. But then you treat it as if it was the width of the horizontal stripe. So either you clarify the drawing or use c_i - c_{i-1}.
@brightsideofmaths Жыл бұрын
No, c_i is the value at the y-axis and not the width of a horizontal stripe. I have my measure theory theory video course that goes in the detail there: tbsom.de/s/mt
@davidefrey3434 Жыл бұрын
@@brightsideofmaths but in the video you say that c_i is a stripe.. And indeed you do a \sum_i c_i * \mu(A). If c_i is the point shouldn't you have \sum_i (c_i - c_{i-1}) * \mu(A)?
@SmileyHuN3 жыл бұрын
Very nice video! Actually, in the Riemann case we have a measure too, the Jordan measure, as you wrote delta(x_i)
@sitrakamatthieu2 жыл бұрын
So clear...Bravo and thank you.
@brightsideofmaths2 жыл бұрын
Glad you enjoyed it!
@ShivamMishra-yv7zg3 жыл бұрын
Thankyou. really enjoyed it.
@sumitdas92046 ай бұрын
You should speak more clearly. Some of your words are not understandable....either way the videos made are really great and knowledgeable. Thanks
@brightsideofmaths6 ай бұрын
Thanks! The new videos should be better. Here, you can use the subtitles.
@drottercat4 жыл бұрын
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.
@brightsideofmaths4 жыл бұрын
For that question, I have a whole lecture series prepared :) kzbin.info/aero/PLBh2i93oe2qvMVqAzsX1Kuv6-4fjazZ8j
@benheideveld46178 ай бұрын
So give me an example where the Riemann integral cannot be computed, but the Lebesgue integral can. Or an instance that both integrals exist, but do not have the same value.
@brightsideofmaths8 ай бұрын
For your first statement: Indicator function of the rational numbers.
@stopaskingmetousemyrealnam38104 жыл бұрын
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?
@angelmendez-rivera3513 жыл бұрын
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.
@tiernanalbanese73605 жыл бұрын
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.
@brightsideofmaths5 жыл бұрын
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.
@tiernanalbanese73605 жыл бұрын
@@brightsideofmaths Thank you! I hope you continue to make great videos in English :)