Measure Theory 10 | Lebesgue's Dominated Convergence Theorem

[@willmurphy8650]

I work as a data scientist /ML engineer and am relearning measure theory for research. During graduate school, I took a class on pure measure theory and a class on measure theoretic probability. I did well but never really understood the proofs and the usefulness of MCT, Fatou’s Lemma, and DCT. Your videos have really changed that. Thank you so much for this content!

[@brightsideofmaths]

You are welcome! And thanks for your support :)

[@qrubmeeaz]

This series is brilliant! More please!

[@Spandan_Ghoshal]

I always regretted that I didn't know much about Measure Theory and always thought it is difficult and technical but now my idea has completely changed. Thanks for putting soo much effort! I really love your videos and finally, I am learning Measure theory now....thanks again! :) :)

[@brightsideofmaths]

Great to hear! I am always happy to help!

[@Spandan_Ghoshal]

@@brightsideofmaths 🥰🥰🥰❤️❤️❤️❤️

[@padraighill4558]

Man, this video series is amazing! I love that your measure space is always abstract. Maybe I am asking too much, but it would have been the cherry on top if your maps were in general complex valued.

[@brightsideofmaths]

Great suggestion! :)

[@PunmasterSTP]

Dominated convergence theorem? More like "Definitely cool information on 'em!" Thanks so much for putting these wonderful videos together that cover so many different topics.

[@What-bw2wk]

Amazing. I wish this bro was my teacher. He explains soooo good. Amazing work!

[@brightsideofmaths]

Thanks a lot :D

[@NathanCrock]

Very helpful videos! Thank you. A video on transformed measures, change of variables and applications might be interesting!

[@brightsideofmaths]

Thank you! This is very good idea. I can do this immediately after the proof of Lebesgue's theorem.

[@houdanaitelbarj8068]

Thank you so much for all your videos, thy are incredible!!!

[@MrOvipare]

Very well explained! I'm just a bit unsure of what is very different from the monotone convergence theorem... The end result is the same, justifying that we can insert the limit inside the integral. Could we say that this theorem gives "another way" to justifying that operation, with only the need of an integrable majorant? Why can't we simply only use the monotone convergence theorem?

[@brightsideofmaths]

Yes, it is another way for this. Often, your sequence of functions is just not monotonic so you cannot use the monotone convergence theorem.

[@crossvalidation1040]

Love your videos, thanks!

[@DDranks]

Does the "power" here mentioned in 2:30 mean the value of the function risen to some power pointwise? Why would the integrability of the function change if you rise the values to some finite power?

[@DDranks]

Oh, wait a sec. Is the point that there could be some functions that converge to zero at infinite limit, but _so_ slowly, that making a "finitary" adjustment like raising to a power, is enough to change the interval from finite to infinite? That sounds very counterintuitive to me at first, but infinities and convergerence are kinda counterintuitive anyway. If somebody could answer me, I'd be very grateful. One more thing, that makes me to doubt such functions could exist: if a function converges towards zero at infinite limit, I'd imagine that after some finite point (let's call it p), it will be continuously under one. Since the integral is overall finite, the area under the curve up to point p is finite, since a finite number to finite power is finite. From point p onwards, the curve is under one, and raising a number under one to a power bigger than one, diminishes its value. So if anything, the area after point p should get smaller. So, I don't really understand how raising a function to power could change its integral from finite to infinite.

[@Christian-xo9ud]

​@@DDranks Suppose f(x) := x^(-1/2), then f is obviously in the space L^1([0,1],\lambda) (you can check easily calculating the antiderivative). But if you square the integrand, the Integral fails to converge due to a huge singularity at 0.

[@RealMcDudu]

This is very similar to the monotone convergence theorem, only here instead of monotone series of functions, you can have alternating functions, which are still bounded/dominated by some function g, and that their limit is a single function. Correct?

[@quantitativeease]

7:00 I am not sure if your picture covers all the generalities. Couldn't the sequence f_n on x_0 wobble above and below the limit in a Cauchy-esque way? Also, I'm pretty sure the f_n should be able to intersect and overlap, so long as the limit exists. This series is the best thing on KZbin since The Great War. I would like to support you but I am disabled and currently underemployed, and I only have a high school diploma so this math is recreational to me, sadly. Lots of people take their education for granted and have no idea what it is like if one lives in the United States and has a passion for learning but one's parents are frightened or scornful of that fervor.

[@brightsideofmaths]

You also support this channel by doing comments and watching it :) Thank you very much for it! For the picture: Of course, there, a lot is possible. The key feature is, however, that the function g lies above all the other functions. That is the key ingredient and that is what I wanted to show with the picture.

[@rick4135]

Great content!!!! Does this fn ->f has from an increasing sequence??? The way you draw the functions getting near to f pointwise which are bounded by g make me think fn has to be an increasing sequence. Thanks

[@chunchuanlv3211]

It's not necessary.

[@kashmirientertainmentchann2067]

Amazing sense of maths

[@kkkk-oy9qv]

Thank you, you are the best

[@salehgholamzadeh3368]

Thanks for your nice videos. I have a quick question. What new stuff does "dominated convergence" bring to "monotone convergence". does it just relax the condition of f1

[@brightsideofmaths]

Thanks! It's a different condition for the sequence of functions :)

[@bdnetplayer]

How can the function f^- be greater or equal to zero if it is below the x-axis as in the example?

[@brightsideofmaths]

- f^- is below the x-axis :) This means that we multiply the part below the x-axis by (-1) to get a non-negative function.

[@helviohild7384]

@@brightsideofmaths you wrote f+,f- >0 and must be f+,-f->0. I really apreciate your work

[@eddycabello3627]

@@helviohild7384 f- is positive by definition, f-(x)= max{ -f(x), 0}

[@angelmendez-rivera351]

@@helviohild7384 No. -(f-)(x) < 0, so (f-)(x) > 0. What he wrote is completely correct. Look at how he defined f-. He defined (f-)(x) := max{-f(x), 0}, so if f(x) < 0, -f(x) > 0, so (f-)(x) = -f(x), while if f(x) > 0, -f(x) < 0, so (f-)(x) = 0. In both cases, f- is nonnegative, which is precisely what he wrote on the screen. He made no mistakes. The part that is below the graph (and therefore nonpositive) is -(f-)(x), *not* (f-)(x) itself.

[@TaxpayerMoney]

Thank you

[@quanganhta5900]

May I ask, does the theorem works with sigma notation?

[@brightsideofmaths]

What do you mean?

[@quanganhta5900]

@@brightsideofmaths I mean suppose that {xn} satisfy all the conditions of dominated convergence theorem does lim ∑{xn} = ∑lim{xn}?

[@anthonyhu4950]

Thank you for saving me in probability class LOL!

[@guohaodou5635]

MCT does look like a special case of DCT, no?

[@SuperSerbia123]

Why is the assumption that f_{n}:X->\mathbb{R} needs to be MEASURABLE for all n\in \mathbb{N} necessary?

[@brightsideofmaths]

Otherwise, the integral wouldn't make sense.

[@JoseCarlos-vb3bj]

Good

[@carmencalungui7199]

example please, i badly need an example of ldct

[@duckymomo7935]

Lebesgue Dominated vs monotone convergence?

[@MrOvipare]

I'm a bit unsure too... the result is the same : you can insert the limit inside the integral. However I believe the usefulness of the Lebesgue dominated convergence comes with the premise that you just have to find a function g that is an "integrable majorant" to justify inserting the limit inside the integral.

[@JoseCarlos-vb3bj]

I need your help. I ask your email

[@mauryasachin7950]

Sir please change in language hindi

[@mohammedabdiwahid6053]

Excellent series. Thanks