Measure Theory 7 | Monotone Convergence Theorem (and more)

[@darrenpeck156]

I really feel this channel is a worthy successor of the late Hilbert Gross's pioneering work with MIT on Calculus Revisted in the late 60s and early seventies. The video library being established here is vital for online learners across the world. I am curious to see how this channel evolves! With the ambitious choice of topics and gifted pedagogy, I feel that this channel could become the most important online resource for undergraduate and masters-level mathematics students. Thank you

[@brightsideofmaths]

Thank you very much! I really put a lot of work into the channel and have a lot of plans for future videos :)

[@nicolasmenet2471]

This channel is way too underrated. Great job!

[@brightsideofmaths]

Thank you. You can promote my channel and tell your friends about it :)

[@nicolasmenet2471]

@@brightsideofmaths I will haha

[@bronsonjosue7423]

I know im randomly asking but does anybody know of a tool to log back into an instagram account? I stupidly lost the password. I would appreciate any help you can give me

[@felipebishop9145]

@Bronson Josue instablaster :)

[@kentkoleslau7390]

@@bronsonjosue7423 creep

[@aquamanGR]

Fantastic series. Thanks so much for doing this, it's one of the best presentations of MT that I have seen.

[@brightsideofmaths]

Thank you very much :)

[@olivergojkovic1345]

Its very good that you not only explain the definitions, but also do some of the proofs!Thanks for this channel

[@nicolaslopez3361]

I love you translate this amazing series in english. I am a native spanish speaker (Latin Amerca-Colombia), otherwise I would miss these terrific videos. Keep up with the good work

[@TheRealSlimPiggy]

Amazing work, I really appreciate how much I am learning for free about high level maths. If I were more rich I'd certainly be a member. Thank you!

[@brightsideofmaths]

Thank you very much! If I were more rich, I wouldn't need any memberships to offer free Maths content :)

[@PunmasterSTP]

Monotone convergence? More like "Magnificent coverage"...of many difficult topics. Thanks so much for making and sharing all of these amazing videos!

[@kentkoleslau7390]

The proof of part (b) should follow very simply from the simple function approximations to f and g. Consider partitioning the set X (domain) into the sets (A_1, ..., A_n) and let h_f = \sum_{i=1}^n c_i I(A_i) and h_g = \sum_{i=1}^n c_i I(A_i), where I() represents the indicator function. Since f(x)

[@eduardocambiaso9579]

You are awesome. I will support you as soon as I can. Hope you a wonderful life and that you are able to keep making these videos.

[@brightsideofmaths]

Thank you so much!

[@Y-us8hg]

thank you sir Love from India

[@SamSarwat90]

I just wonder, the subset of X where f and g differ, how do we know that it is in the sigma algebra?

[@brightsideofmaths]

That is a really question and I think that you should be able to answer it. You need to use the fact that we consider measurable maps.

[@vihdzp]

It's the preimage of {0}^C under f - g.

[@carlosraventosprieto2065]

You are a great teacher. Thank you so much!

[@brightsideofmaths]

Thanks a lot :) And thanks for the support!

[@pramodkhade2554]

Outstanding explaination

[@johnstewart5651]

Amazingly clear and useful. Thank you.

[@brightsideofmaths]

Glad it was helpful!

[@mirak76]

Well explained ,thank you very much

[@viktorhansen3331]

Thank you so much for these videos :-) I just became a Pi supporter!!!

[@brightsideofmaths]

Thank you very much :)

[@devkar233]

In which textbook I can find proof of (a) and (b)?

[@harisfawad1738]

At 10:05, why do you sum over all values of h in the image of X, rather than over all values of h in the image of X-tilde?

[@angelmendez-rivera351]

Because μ(complement(X~)) = 0

[@wronski11]

You are the best teacher ever. Could you please tell me which book your course is based on.

[@林泓均-q4j]

Hi, thanks for your awesome video. But I have question on 2:57, where is the definition of \mu-a.e.. I found that from Wikipedia, it said that It is ''not'' required that the set \{x\in X: eg P(x)\} has measure 0; it may not belong to \Sigma . I'm not sure what is the difference. Is it because f,g are both measurable?

[@brightsideofmaths]

Can you link the wikipedia article? Then I can explain what the exact meaning is.

[@MagicBoterham]

Should the image at 9:04 not have x̃ᶜ and X swapped?

[@brightsideofmaths]

Why do you think so?

[@duckymomo7935]

Ooh exciting to see new video!

[@HUEHUEUHEPony]

at 16:00 why was converting h to ~h necessary, wouldn't you also be able to argue it without this? If not, why not? Also, why when you convert it to ~h, it belongs to S but not on S^+? We're trying to prove that f

[@brightsideofmaths]

The keyword is "almost everywhere" here :)

[@EulerGauss13]

We can't do the same argument because we don't necessarily have f

[@brightsideofmaths]

@@EulerGauss13 Thanks. Sorry, yeah, I missed a plus sign in one formula.

[@ysun0]

Thx for the videos! Q: at 13:02, I suspect the countable sum of 0 is 0, but do we need to prove this first? (is it even a countable sum?)

[@brightsideofmaths]

Yes, the countable sum of zeros is zero. Here, we have just a normal finite sum, which means you sum up some zeros. Of course, then the result is still zero. The proof is very easy in this case :)

[@ysun0]

@@brightsideofmathsMakes sense, the sum doesn't stop being countable just because we write it a different way :) thank you and thanks for the videos! They are very clear.

[@siddharthjoshi9543]

Sorry if this was implicit. You proved the result for simple functions, but how did you go from simple function to any general borel-measurable function, using dominated convergence probably?

[@quantitativeease]

3:36 I have been familiar with this concept a long time. However, when I saw this illustration, I was somehow reminded of the Dirac delta function. I do not understand that function I must admit, but what I do know is that it is value is zero everywhere but zero, and its integral is one. That makes me think, what if a function F has integral a, and G is some function which is F plus an affine translation of the Dirac function? Wouldn't the integral of G be a+1? Is the point not considered because it is measure 0 on the domain and therefore the Dirac function is not Lebesgue-integrable?

[@DDranks]

I think that with Dirac function you are supposed to use the Dirac measure that was mentioned briefly in an earlier video. (A measure that is always zero expect for the exact point the "spike" is at, where it's one.) However, I don't know how would the measures "combined" when you sum the Dirac function and some other functions.

[@angelmendez-rivera351]

The Dirac delta you speak of is not a function, so this idea is of adding it to other functions is just not defined in a vector space of functions. Instead, you need to define a space of linear functionals, some of which are canonically represented by multiplying by a function and integrating with respect to a measure. Then the Dirac delta can be defined as a linear functional which integrates functions with respect to the Dirac measure. Of course, this is all under the assumption that a set of measure spaces has been defined a priori.

[@DDranks]

At point 7:08, I get that it's just another representation, but defining h(x) as the sum of all t × χ{a set defined in terms of h(x)} feels... very circular.

[@brightsideofmaths]

Yeah, it it not a definition. It is similar to saying: I have a function f and now I look at the value f(1). You still don't know what the value f(1) is neither you are defining it. You just want to calculate with f(1). Nothing more :)

[@angelmendez-rivera351]

There are alternative ways to define it using weighted sums, but this is more tedious.

[@xuhu8148]

Thanks for the great video! When we apply \mu to the set of x where f(x) is not equal to g(x), how do we know this set belongs to the sigma algebra so that we can apply \mu to it? My intuition is we can find a series of subsets with measures which are supersets of that set, and the measure converges to 0 ==> almost everywhere equality.

[@josearmandovivero408]

Remember that f and g are measurable functions, so is f-g, so the set can be written as (f-g)^(-1)(R\{0}) and this set belongs to the sigma algebra.

[@kaminimangal8702]

pls provide all the proofs 5:52.

[@MrWater2]

Sorry just to confirm, in the last part of proof b, you first use the definition of lebesgue integral, then that for previus result if you EXclude zero measure sets the result of the integral it's the same. Then you just use that the set X_tilde is such that f

[@akanksha8311]

Thankyou so much , it was a great help

[@hansenpen9833]

Thank you so much for such great explanation! Could you please explain how we do integral on bounded functions instead of simple functions?

[@angelmendez-rivera351]

We perform integration on measurable functions, which is not equivalent to be bounded.

[@brightlin777]

Wait, is that how you can find the integral for the Dirichlet function? With property (a) ? (assuming we use the Borel sets)

[@brightsideofmaths]

Yes, exactly.

[@Juanbrestrepob]

can i know whats the software you use to do the whiteboard in your videos i am starting virtual education courses and i have had trouble finding such good software

[@brightsideofmaths]

This is the beautiful but old program Xournal.

[@PunmasterSTP]

I'm just curious; how have your courses been going?

[@Independent_Man3]

Can you post PDFs of what you write in the videos in the links below the video? Maybe public Google Docs of something. That will be extremely helpful.

[@arjunsambhi5081]

Surely if the measure of the set X Tilda complement is 0, then there is at most countable points x at which the value of h Tilda differs from h. So why is it ok to assume h Tilda takes on the value 'a' at these points? (I understand that a simple function takes on finitely many values). What would happen if h tilda differed from h at countably many points, with countably many different values, say a_1,a_2,... instead of just a? (Also, these videos are great, very helpful!)

[@marios1861]

you don't really care about the value a since the measure is still 0 thus the integral does not change.

[@josearmandovivero408]

That last thing you say cannot happen since h tilda would not be simple. Simple functions only can take finitely many values, it doesn't matter if the domain is countable or not.

[@vibingwithndt8908]

you rock bro

[@brightsideofmaths]

Thanks :)

[@husniyatojiboyeva1154]

Thank you so much for the great explanation! What you are doing is amazing. But I have a question. In property (a) you mentioned M(myu) is almost everywhere. What does this mean? I did not understand.

[@brightsideofmaths]

Thanks for your support! µ-almost everywhere means that the property holds with the exception of a set with measure 0.

[@juniorsimo7054]

Thank you very much for your videos, they've been nothing but fantastic thus far. Are there any particular textbook(s) you would recommend for anyone looking for additional practice on measure theory ?

[@brightsideofmaths]

Oh, the best textbooks I know are in German but Schilling has one in English that is very fitting. You can check it out.

[@fleurblanche3727]

Thank you I 'm so grateful to you, I like your method of explanation 🌹

[@brightsideofmaths]

Thank you! 😃

[@harryhirsch2024]

But why does the implication in b not go the other way, why don't we have f

[@theo_pap_]

Consider this counterexample. Let f, g be defined in X = [0,2] with f(x) = 1 everywhere, g(x) = 0 in [0,1] and g(x) = 3 in (1,2]. Let μ be the normal length. Then Int f = 2 < 3 = Int g, but in the set [0,1], which has nonzero measure, f > g.

[@HUEHUEUHEPony]

@@theo_pap_ what if you just restrict to the same domain per comparison.

[@angelmendez-rivera351]

@@HUEHUEUHEPony Both functions have the same domain. The domain is X = [0, 2].

[@ravishsingh3571]

Thanks sir

[@akanksha8311]

Hi can you make a video about the realtion between series and integrals with the help of measure theory? My Prof said something about counting measure being used but i didnt get it also someone has asked samething on mathstackexchange but its not explained fully Thanks for this channel and your videos

[@angelmendez-rivera351]

When we integrate in most applications, we typivally work with the measure space (R, B(R), λ), where R is the set of real numbers, B(R) is the Borel algebra of R, and λ is the Lebesgue measure. This makes the Lebesgue integral a direct generalization of Darboux integration. When you work with series, you typically work with the measure space (N, P(N), c), where N is the set of natural numbers, P(N) is the power set of the natural numbers, which is the discrete σ-algebra of N, and c is the counting measure, defined by c(S) = card(S). Integrating a measurable function f : N -> R with respect to the measure c is exactly equivalent to evaluating a series of elements of R. In fact, series are just a special case of integration.

[@kkkk-oy9qv]

Thank you, you are the best

[@pramilakumari4565]

Thank u so much sir

[@macaco3001]

How is this result any different from simply applying the definition of int(f)?

[@brightsideofmaths]

How mean the difference between int(f) and lim int(f_n)?

[@angelmendez-rivera351]

I(lim f[n]) = I(f) is trivially true, since lim f[n] = f. However, what is not trivial is that lim I(f[n]) = I(lim f[n]). This is what the theorem is about.

[@I_Adarsh_kumar]

Love this channel But from where we shall practice questions

[@marcusjohnpeterson2302]

Try "Real Analysis" by Halsey L. Royden . It's a good textbook for this and more.

[@anton96121]

Isn't the mu-a.e. condition for (c) indifferent? Doesn't the function f has zero measure everywhere for the lebesgue integral to be zero?

[@brightsideofmaths]

The condition with mu-a.e is a weaker assumption than having it everywhere.

[@anton96121]

@@brightsideofmaths but what I mean is that for the integral to equal zero the measure space X needs to have zero measure everywhere. So the condition mu-a.e. is implicit due to the the measure space having zero measure everywhere? Or have I missed something? And thank you so much for the answer and the videos!

[@brightsideofmaths]

@@anton96121 Okay. I see the confusion: The integral is not just zero for all functions. Just for this particular function (with the mu-a.e. assumption), the integral is zero. You are welcome :)

[@anton96121]

@@brightsideofmaths Arh yes I get it now. Thank you!

[@HoloBoss]

Muss fn gleichmäßig gegen f gehen oder reicht punktweise?

[@brightsideofmaths]

Punktweise reicht.

[@EulerGauss13]

I don't understand how you write M{x\in X|f(x) eq g(x)}=0. It assumes that the set inside M is in sigma algebra?

[@brightsideofmaths]

Can you give me a timestamp?

[@EulerGauss13]

@@brightsideofmaths It's at 2:42, regarding the definition of almost everywhere. I thought the measure is only defined on sigma algebra, and I don't know how the sigma algebra is defined on X. Maybe I'm confused on something basic and I'm trying to figure out!

[@brightsideofmaths]

@@EulerGauss13 Thanks. The Sigma-Algebra consists of subsets of X. And here we measure such a subset.

[@angelmendez-rivera351]

@@EulerGauss13 The set is measurable, since f and g are measurable functions.

[@sehernaaz6263]

Sir kindly send proof of a and c properties

[@brightsideofmaths]

Send where?