Measure Theory 23 | Proof of Carathéodory's Extension Theorem

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  • Опубликовано: 15 ноя 2024

Комментарии • 12

  • @benjaminlloyd-hughes5621
    @benjaminlloyd-hughes5621 2 месяца назад +1

    Thank you so much! A wonderfully clear exposition which rounds off the series perfectly.

    • @brightsideofmaths
      @brightsideofmaths  2 месяца назад

      I am happy to help :) And thanks for your support!

  • @Grateful92
    @Grateful92 4 месяца назад +3

    I love your videos, thanks for existing here on RUclips

  • @Algebraictivist
    @Algebraictivist 4 месяца назад +3

    Well, got more things to study about here.

  • @waqarsoomro2298
    @waqarsoomro2298 3 месяца назад +1

    Thank You, Man! out there for this masterpiece.
    Finally, I completed this Playlist.
    Thank You again.

    • @brightsideofmaths
      @brightsideofmaths  3 месяца назад +1

      You're very welcome! :) And thanks for your support!

  • @vagelisxanthopoulos9171
    @vagelisxanthopoulos9171 23 дня назад

    TL;DR: At 6.34 how do we know that Aj\A belongs in the semiring? It should in order to apply additivity.
    I have been studying your videos on measure theory. They are very good. I have one question here. At 6.34, you apply additivity for the pre-measure μ and do: μ[(Αj∩A)U(Aj\A)] = μ(Αj∩A) + μ(Aj\A).
    However, in the additivity property of the pre-measure --> there was the requirement that both the separate sets and their union --> should belong in the semiring --> because a semiring is not closed under union. Let's check that.
    Aj belongs in the semiring and thus --> (Αj∩A)U(Aj\A)=Aj belongs also. Next Aj∩A belongs in the semiring because Aj and A belong in the semiring and a semiring is closed under intersection.
    My question is how do we know that Aj\A= Aj∩A' belongs in the semiring? Semiring is not closed under complement. And Aj, A are totally irrelevant with each other, since they were picked arbitrarily. Am I missing something?

    • @brightsideofmaths
      @brightsideofmaths  22 дня назад

      Good question! At this point, I was a little bit sloppy to make the whole proof less complicated. Actually, we have to extend to semiring to a ring such that we can do that calculation. The given pre-measure extends uniquely to a pre-measure on the ring.

  • @britoalv2790
    @britoalv2790 2 месяца назад

    Very good series, do you know a source of the proof of the uniqueness, part of the proof with semi-rings ??

  • @bilaltaskin0
    @bilaltaskin0 Месяц назад

    Life saver...