Integration of measurable functions - Lec06 - Frederic Schuller

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  • Опубліковано 11 бер 2016
  • This is from a series of lectures - "Lectures on Quantum Theory" delivered by Dr.Frederic P Schuller

КОМЕНТАРІ • 57

  • @Afxonidis
    @Afxonidis 3 роки тому +1

    That was maybe the best explanation I have ever heard of the quotient space. Brilliant work!

  • @mathjitsuteacher
    @mathjitsuteacher 4 роки тому +8

    It is important to notice that the pointwise supremum of nonnegative measurable functions is also a nonnegative measurable function and this allows you to integrate it.

  • @timelsen2236
    @timelsen2236 2 роки тому +1

    Greatest mathphysics on line, and loved the student interaction.

  • @jean-pierrecoffe6666
    @jean-pierrecoffe6666 5 років тому +12

    Such a great lecture. Absolute clarity in all of the explanations

  • @shuaiwang5606
    @shuaiwang5606 3 роки тому +9

    I tried to self study and different videos, your explanation is sooo clear and easy to understand, wish I could attend your class!!!!!

  • @user-hp1xk3vv7y
    @user-hp1xk3vv7y Рік тому

    Wonderful teachig style with sufficient content covered less than 2 hours. Thank you sir.

  • @lordeuler2912
    @lordeuler2912 8 років тому +62

    Hölders and Cauchys inequality should both be multiplications and not additions.

    • @antoniolewis1016
      @antoniolewis1016 7 років тому +2

      yes

    • @benpovar3914
      @benpovar3914 6 років тому +4

      The prof. meant Minkowski inequality

    • @alexgoldhaber1786
      @alexgoldhaber1786 5 років тому +4

      The important thing now is that I learned how to pronounce ö as in Hölder!

    • @connorfrankston5548
      @connorfrankston5548 4 роки тому +4

      @@benpovar3914 no, he must have made a mistake. The Cauchy-Schwarz inequality is definitely with multiplication, and the Hölder inequality too.

    • @adambruce1688
      @adambruce1688 3 роки тому +2

      @@benpovar3914 Minkowski is slightly different. ||f+g||

  • @jacoboribilik3253
    @jacoboribilik3253 27 днів тому

    Great lecture. Much appreciated.

  • @circuithead94
    @circuithead94 3 роки тому +3

    The clarity of Schuller's explanation is amazing.

  • @Ciupakabrinas
    @Ciupakabrinas 6 років тому

    Thanks for the lecture! You make Lebesgue integration seem easy.:)

  • @stefanogioberti
    @stefanogioberti 4 роки тому +3

    Such a clear lecturer. Makes the difficult easy. Many thanks.

  • @lisaking3996
    @lisaking3996 5 років тому +5

    I like his wavy norm sign

  • @paulserna2104
    @paulserna2104 4 роки тому

    It is clear and tidy.

  • @user-bj5xr
    @user-bj5xr 3 місяці тому

    At 1:09:43, (ii) lim \int |fn-f| = 0 , not only finite. (corrected)
    At 1:11:11 it can be R bar, with infinity since f,g are integrable, and therefore \int f is finite by definition.

  • @AmatisoveLove
    @AmatisoveLove 7 років тому +9

    This really is a great video, I'll definitely watch more of your lectures! Helped me a lot, thanks.

  • @abdellatifelgrou7772
    @abdellatifelgrou7772 7 років тому +5

    Just a remark teacher chaucy inequality with product not a sum : ||

  • @farshadnoravesh
    @farshadnoravesh 4 роки тому

    a great lecture

  • @jwp4016
    @jwp4016 7 років тому +13

    35:06 the measure of the pre image of z? Shouldn't it be the measure of the pre image of [z, infinity) ?
    Anyway, he is easily the best math teacher I've ever seen in my math learning journey.

    • @jwp4016
      @jwp4016 7 років тому +5

      I missed 1:13:36

    • @joaodfbravo
      @joaodfbravo 5 років тому

      @@jwp4016 Why is it not just preim(z)?

    • @sardanapale2302
      @sardanapale2302 3 роки тому

      Yes ... in the sketch he did preim(z) has zero measure

    • @sardanapale2302
      @sardanapale2302 3 роки тому +1

      @@jwp4016 Its still false, on preim[0,z) f will be < z... you need preim[z,oo) so that on that set f >= z and you will get a lower bound of the integral of f...

    • @sardanapale2302
      @sardanapale2302 3 роки тому +1

      lol he finally corrects that afterwards :P

  • @allwanamar1
    @allwanamar1 7 років тому +11

    this is an excellent lecture sir.

  • @sisayketema3415
    @sisayketema3415 4 роки тому +2

    Dear Dr. i glads to follow such interesting lecture ...really really it was nice lecture. before this lecture i am not familiar with this concept after following your lecture things was clear for me thank you very much.keep it up

  • @wenzhang365
    @wenzhang365 6 місяців тому

    Amazing lectures. Is there a textbook(s) for this course?

  • @user-bj5xr
    @user-bj5xr Рік тому

    Very clear. The set of your mistakes has measure zero.

  • @fredxu9826
    @fredxu9826 Рік тому

    QM lectures turn out to be great for anyone interested to start functional analysis

  • @user-pi7do8dv4y
    @user-pi7do8dv4y 6 років тому +4

    1:20:50 Progress to the direction of Hilbert and Banach spaces... and why this matters when one does solid state physics!

  • @ryanchiang9587
    @ryanchiang9587 6 років тому

    i remember these terminologies such as borel sets, sigma-algebras...etc.

  • @minexe
    @minexe 2 роки тому

    thanks

  • @millerfour2071
    @millerfour2071 3 роки тому +1

    25:03, 42:35, 1:12:28 (35:58), 1:25:17, 1:32:00, 1:35:10, 1:38:38, 1:44:03, 1:48:48

  • @michielsnoeken5596
    @michielsnoeken5596 3 роки тому +2

    Why is the singleton {s_i} measurable, since a singleton in the standard topology is not open and therefore also not an element of the sigma-algebra?

    • @HilbertXVI
      @HilbertXVI 2 роки тому +5

      But the borel algebra contains a lot more than just open sets. You could write {s_i} as a countable intersection of the sets (s_i - 1/n, s_i + 1/n), each of which is open, and hence measurable. So the intersection {s_i} is also measurable.

    • @MrLikon7
      @MrLikon7 2 роки тому +5

      there is an easier way to see this: the complement of a singleton is open in the standard topology. but the sigma-algebra is closed under complement-taking, so the singleton is in it. in fact, the borel sigma-algebra contains all the open AND closed sets, furthermore all the half-open intervals and so on

  • @juliogodel
    @juliogodel 6 років тому +2

    Excellent lecture, Thanks.

  • @jiahao2709
    @jiahao2709 3 роки тому

    why these lectures, youtube don't have automatical translation?

  • @Anthony-db7ou
    @Anthony-db7ou 2 роки тому +3

    I was the one who could be easily mislead by this. I haven’t taken measure theory formally, but need to learn more for my upcoming program in finance! You’re series is definitely unambiguously presenting the math, and I won’t say that about anything else I’ve come across. Any book recommendations?

    • @benwincelberg9684
      @benwincelberg9684 2 роки тому +1

      Real Analysis by Royden

    • @lugia8888
      @lugia8888 Рік тому +1

      Needs more examples

    • @mastershooter64
      @mastershooter64 7 місяців тому +1

      "A User-Friendly Introduction to Lebesgue Measure and Integration" by Gail S. Nelson, it is a really nice and friendly introduction to measure theory. Has lots of intuitive explanations and includes illustrations, all the while being completely rigorous.

  • @ChaimaChaima-gb4ng
    @ChaimaChaima-gb4ng 5 років тому

    Good🤓

  • @peterpalumbo3644
    @peterpalumbo3644 5 років тому +2

    Are there any problem sheets for people viewing these lectures on You-Tube, are the Experimental lectures on You-Tube?

  • @jonathanstudentkit
    @jonathanstudentkit 4 роки тому

    height times width (not with :D)

  • @gooomaaal
    @gooomaaal 6 років тому +3

    good joke 1:11:48

  • @user-uq7fb8eh5v
    @user-uq7fb8eh5v 3 роки тому

    Please doctor, l need to solve this equations. :
    1: show that the step function f (X) =[x] is measurable function on [1,3]
    And
    2: if f is measurable function over a measurable set E then f is measurable over any measurable subset A of E

  • @maciej12345678
    @maciej12345678 Рік тому

    very fine but where are proof of theorems

  • @ciruzzobello3108
    @ciruzzobello3108 4 роки тому +1

    1:05 point aus... Oops, wrong language :-)

  • @atzenkack
    @atzenkack 7 років тому +2

    cocaine? :D

  • @hassaannaeem4374
    @hassaannaeem4374 Рік тому +1

    the gender jabs never get old