Measure Theory 7 | Monotone Convergence Theorem (and more)

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This is part 7 of 22 videos.

#MeasureTheory
#Analysis
#Integral
#Calculus
#Measures
#Mathematics
#Probability

I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[darrenpeck]

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

[nicolasmenet]

This channel is way too underrated. Great job!

[aquamanGR]

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

[olivergojkovic]

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

[nicolaslopez]

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!

[PunmasterSTP]

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

[kentkoleslau]

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) <= g(x) a.e., then c_i <= d_i for all i. Therefore, I(h_f) <= I(h_g). Then we note that both of these are functions of n. Collecting all such values for n = 1, 2, ..., and taking the supremum of both sets, we have that I(f) = sup(I(h_f)) <= sup(I(h_g)) = I(g).

[eduardocambiaso]

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.

[SamSarwat]

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

[MrWater]

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 <= g (so the inequality) and finally that INcluding measure zero sets does not change the result of the integral right?

And the last one, in the second equality (of integral f dmu) you use S instead of S+ that's a typo or that's mean the c_i (or t) could be negative? (I understand that that S is an S+ but maybe I miss something important)

[wronski]

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

[devkar]

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

[juniorsimo]

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 ?

[johnstewart]

Amazingly clear and useful. Thank you.

[husniyatojiboyeva]

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.

[xuhu]

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.

[viktorhansen]

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

[林泓均-qj]

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 <math>\{x\in X: \neg P(x)\} </math> has measure 0; it may not belong to <math>\Sigma </math>. I'm not sure what is the difference. Is it because f, g are both measurable?

[Independent_Man]

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.