Measure Theory 11 | Proof of Lebesgue's Dominated Convergence Theorem

[@DanielKRui]

I'm really grateful you are doing all this -- every time I have a question about what's going on in the video, a couple of seconds later you address it! Truly amazing.

[@JohnDoe927]

Thanks for an intro to measure theory, It's so much better having someone talk through things, than having to slog through a book on my own :)

[@mathijs1987j]

Thanks for this great series! Yes, please show some applications! It's the main reason I am watching this series.

[@RangQuid]

This theorem and its proof are indeed extremely beautiful and elegant.

[@SimpMaker]

I really like how the pedagogical order gives intuitive connections and further cements previous concepts.

[@krisztiannagy5515]

I just took the exam from measure theory yesterday and your videos helped a lot actually. Amazing content! I really like the balance between formal proofs and intuitive explanations. Keep up the good work!

[@PunmasterSTP]

Majorant? More like "My appreciation? It's hard to contain all of it!" Thanks again for making so many educational and intriguing videos.

[@toneromartins5608]

I really appreciate your video.I would like to see its continuation

[@monicadelpilar23]

I just started reading - and liking - Rudin´s Real & Complex Analysis! So, I am very glad to have found your great videos! ... I would love to see something on Hausdorff measures and Fractal!

[@tommyconnolly9427]

Same book for me! This video perfectly complements that book.

[@vilmaroroboff1438]

Excelent videos and explanations!!! I’m starting a course in the next week with uses Measure and Theory and just because of your videos I’m going to be able to learn the theory needed in one week (of course I already done the pre-requisites of measure theory before). Thanks from Brazil! Obrigado!

[@PunmasterSTP]

That's awesome! I'm just curious; how did your course go?

[@alexbrodbelt297]

thank you so much for this video series!! it has been very helpful

[@brightsideofmaths]

Glad it was helpful!

[@alemonta97]

I've just finished watching your Measure Theory videos and I wanted to thank you! They've been extremely helpful.

[@brightsideofmaths]

Thank you very much. However, this is not the last video in the series: kzbin.info/www/bejne/rouZan57nJyWmbc

[@wangdave5574]

Wow thank you for the detailed explanation, this the first time I have seen this proof and its absolutely beautiful.

[@friendoquail]

Really enjoying these videos! Very clear explanations!

[@choungyoungjae8271]

thank you for good tutorials

[@BrainGainzOfficial]

These are great explanations. I've been using them to supplement the measure theory course I'm taking and they've really been helping. I was wondering if you could tell me what program you use to write and record? The quality is great! Thanks!

[@brightsideofmaths]

Thank you! I use Xournal.

[@PunmasterSTP]

I know it's been awhile, but I was just curious; how did the rest of the course go?

[@harryliu4907]

Absolutely beautiful!

[@kkkk-oy9qv]

Thank you, you are the best

[@leopoldocatania2748]

Hi, thanks for the video. In the last step of the proof you write | int (f_n - f) dmu | and apply triangle inequality. But (f_n - f)(x) can be negative (isn't it?), whereas you have defined the Lebesgue's integral for positive function (mu-a.e). Is there a step I am missing?

[@brightsideofmaths]

Thanks for the comment! We discussed this in the last video: kzbin.info/www/bejne/m6aQZ5Ktpbl4Y6M

[@mustafayasir4113]

thanks paul

[@brightsideofmaths]

Who is Paul? :)

[@DDranks]

I was initially confused *why* we need the integrable majorant, but after realising a thing about Fatou's lemmas inequality (explained in my comment here: kzbin.info/www/bejne/p3K8iWyBpNRrbdE&lc=UgzPqSAp__m1-9kATaR4AaABAg ), it became clearer to me. Because it dominates *every* function in the function series 𝑓𝑛, it ensures that the series can't do anything "too" weird as n→∞, like spurt towards infinity for some measurable subsets of the domain, to mess up things. This ensures that the limit of the integral is well-defined and converges to the same value for both 𝑓𝑛 and 𝑓.

[@mathiasbarreto9633]

Excellent video! But I did not understand how liminf got replaced by limsup. Why exactly is the first integral equal to the second one? Shouldn’t it be just less or equal? Thanks!

[@brightsideofmaths]

Thank you very much :) The liminf got a limsup because of the minus sign.

[@luizfernandogoncalves8108]

Very nice! Do you have a video with applications of this thm?

[@brightsideofmaths]

I don't have an explicit video about it but it turns up in other videos again.

[@VictorHugo-xn9jz]

What textbook do you use or would you recommend? It could be a standalone or a chapter of a book, either would be fine. I need to practice with some exercises. Thanks

[@brightsideofmaths]

I recommend Elstrodt's book :)

[@MrWater2]

Do you plan to add some applications :) ? Would be nice

[@nabla2061]

Great video! But I have a quick question. Aren't you doing the proof with the assumptions true "everywhere" and not "almost-everywhere" ? Ty :))

[@brightsideofmaths]

Thanks for the question! Since everything works in the integral, "almost-everywhere" is sufficient here.

[@sinclairabraxas3555]

as always very good stuff

[@brightsideofmaths]

Glad you think so!

[@Heisenberg8307]

Sir, i know i'm 2 years late on this, and it's kind of redundant, but do we have to consider only absolute values of each function in the sequence, i.e. |f_n| to be less than the majorant g in the L1 space?? as part of hypothesis in the LDC theorem...because as per definition of L1 space, it doesn't specify anything about f_n for every natural number n.

[@brightsideofmaths]

Questions are always welcome! Even 2 years after the video got published. However, I don't get your question completely. The absolute value is needed.

[@Heisenberg8307]

@@brightsideofmaths we all know f_n is less than or equal to the majorant g where g is positive lebesgue integrable function in L1 space. My question is, is that |f_n| necessary instead of just f_n less than or equal to the majorant g ?? Because we are going to show f_1, f_2, f_3, and so on are going to be in L1 space anyway.

[@brightsideofmaths]

@@Heisenberg8307 The absolute value is needed to guarantee that f_n also has a majorant on the negative part.

[@Heisenberg8307]

@@brightsideofmaths ok Sir, now i get it, thank you.

[@haggaisimon7748]

It's not clear to me how the switch from lininf to limsup was done?

[@Leslie.Green_CEng_MIEE]

This is explained from 8:46 _Neglecting the indexing for simplicity:_ Consider the value ( |A| - |B| ). A is positive anyway so we get ( A - |B| ) We want the smallest value of ( A - |B| ) which is the *lim inf* of the sequence Since we are separating the terms, we pick the smallest A and the largest |B| to get the smallest possible result. Hence we pick the largest value of |B| which is the *lim sup* of the sequence containing just |B|. *BUT* … consider *LHS < RHS* In order to preserve the inequality we could make the *RHS larger* (or the same). *It is not legal to make the RHS smaller* , as it may no longer be big enough, and that is what has been done. The same happened to the LHS earlier. It is allowed to be made smaller (or the same), but by neglecting the subtracted modulus term it got bigger (or the same).

[@rajeshsalvi8249]

how lim inf hn=2g.I mean is it true lim inf (an + bn)=lim inf an +lim inf bn?

[@brightsideofmaths]

Maybe look again at the definition of h_n. Then you see what happens.

[@rafaelb.333]

We know that lim hn exists and it's 2g, because |fn - f| goes to 0 when n goes to infinity and g doesn't depend on n. And when some sequence converges, we have that lim inf hn = lim sup hn = lim hn, I guess that's why lim inf hn is 2g. Sorry If I'm wrong

[@sanathabbas3659]

I think the key is you have to consider n->infinity in the first place. In that case fn-f is just 0, because f is fn's upper boundary. And the statement you made: 'lim inf (an + bn)=lim inf an +lim inf bn' which I think is not suit for the cases. Because for the second term -absolute(fn-f), there is a minus '-' which implies the inf of it should be smaller if the value in the absolute bracket is larger. Hence I think the key is you need to substitute the value of n first. What confused me is why he didn't substitute n->infinity for the right hand side. I reckon the reason is there is an integral after the lim inf?

[@fept4043]

8:04 what if f_n was equal to f apart from a spike above f that gets increasingly thinner as n gets larger. The theorem would still hold but the spike would never disappear and its peak would change place since the convergence is only pointwise. Shouldn't this mean that the lim inf of h_n is not 2g, but 2g - (height of the spike)?

[@brightsideofmaths]

f_n is bounded from above by g :)

[@fept4043]

@@brightsideofmaths What if the spike was always as high as g? If f is identically 0 and g is identically 1 and the spike of f_n always peaks at 1? Then surely the lim inf of h_n would be 2g - (height of spike) = 2 - 1 = 1. I'm trying to prove this result myself but I keep thinking of these (supposed) counterexamples!

[@brightsideofmaths]

@@fept4043 If the spike is below g, it does not really matter. I don't really see a problem there.

[@fept4043]

@@brightsideofmaths But the spike is there for every n, so h_n is always 1, which is less than 2g = 2, so surely lim inf = 2 - 1 = 1 which is not 2g = 2.

[@brightsideofmaths]

@@fept4043 Who says something about h_n = 2 g?