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Math Help - Integrade with disjoint sets

  1. #1
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    Integrade with disjoint sets

    Suppose that f is an integrable non-negative function mapping from a set X to all reals.

    Let  E_n be disjoint sets in X, why is  \int _{ \cup E_n } f = \sum \int _{E_n} f ?

    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tttcomrader View Post
    Suppose that f is an integrable non-negative function mapping from a set X to all reals.

    Let  E_n be disjoint sets in X, why is  \int _{ \cup E_n } f = \sum \int _{E_n} f ?

    Thanks.
    I don't think I have ever seen you post your work. What have you done up to this point?
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  3. #3
    Moo
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    Oh yes he did at least several times

    http://www.mathhelpforum.com/math-he...w-measure.html
    http://www.mathhelpforum.com/math-he...rsections.html
    http://www.mathhelpforum.com/math-he...t-algebra.html
    http://www.mathhelpforum.com/math-he...open-sets.html
    http://www.mathhelpforum.com/math-he...-infinity.html
    http://www.mathhelpforum.com/math-he...ous-below.html
    http://www.mathhelpforum.com/math-he...ng-unions.html
    http://www.mathhelpforum.com/math-he...ric-space.html
    http://www.mathhelpforum.com/math-he...sure-sums.html

    Do you want more ?
    I think it's rather scarce to see someone detailing this way, with good latex, what he's done so far, eh ?
    So I guess you'd better check before saying such a thing.


    Hint :
    Spoiler:
    prove that 1_{\cup E_n}=\sum 1_{E_n} (when are they equal to 0, when are they equal to 1)
    Last edited by Moo; November 10th 2009 at 12:58 PM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Moo View Post
    Oh yes he did at least several times

    http://www.mathhelpforum.com/math-he...w-measure.html
    http://www.mathhelpforum.com/math-he...rsections.html
    http://www.mathhelpforum.com/math-he...t-algebra.html
    http://www.mathhelpforum.com/math-he...open-sets.html
    http://www.mathhelpforum.com/math-he...-infinity.html
    http://www.mathhelpforum.com/math-he...ous-below.html
    http://www.mathhelpforum.com/math-he...ng-unions.html
    http://www.mathhelpforum.com/math-he...ric-space.html
    http://www.mathhelpforum.com/math-he...sure-sums.html

    Do you want more ?
    I think it's rather scarce to see someone detailing this way, with good latex, what he's done so far, eh ?
    So I guess you'd better check before saying such a thing.


    Hint :
    Spoiler:
    prove that 1_{\cup E_n}=\sum 1_{E_n} (when are they equal to 0, when are they equal to 1)
    tttcommander,transgalatic same thing.
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  5. #5
    Moo
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    Quote Originally Posted by Drexel28 View Post
    tttcommander,transgalatic same thing.
    Not at all !
    They certainly don't have the same syntax. One speaks a correct English, the other one is obviously a non native speaker !
    The use of the LaTeX is not the same. And the level of the questions asked in the forums is completely different from one to another.
    And I showed you 9 proofs that tttcomrader \neq transgalactic.
    Please check before saying that.

    Coming from someone who's only been a member for 10 days and who apparently wasn't in this forum before that, it's quite inappropriate
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    lol, sorry about the lack of the work that I showed in this thread, this question is actually part of a proof that I read from the current chapter. I didn't type the whole proof out since it isn't really my original work and I only have question on that particular part.

    Anyway, thank you and I appreicate the help, and again I apologize for the confusion.

    So my work following the hint:

    Prove that  1_{ \cup E_n } = \sum 1_{ E_n }

    Proof.

    If  1_{ \cup E_n } (x) = 1 , that means  x \in \cup E_n , implies that  x \in E_n for some n, and that is the only one since all En are disjoint.

    Then  \sum 1_{ E_n } (x) =  1, and the arguement is the same the other way around.

    If  1_{ \cup E_n } (x) = 0 , then x is not in any of the E_n, so of course  \sum 1_{ E_n } (x) =  0

    So now, with that established, I need to think about:

     \int _{ \cup E_n } f = sup \{ \int _{ \cup E_n } s : 0 \leq s \leq f , s \ simple \ \}

     \sum \int _{ E_n } f = \sum sup \{ \int _{ E_n } s : 0 \leq s \leq f , s \ simple \ \}

    Let  s = \sum a_k 1_k (x)

    And  \int _ { \cup E_n} s = \sum _k a_k m( \cup E_n)

     \sum \int _ {E_n} s= \sum _n \sum _k a_k m( E_n )

    Now, can I say that  \sum _n m (E_n) = m( \cup E_n ) to conclude the proof?
    Last edited by tttcomrader; November 11th 2009 at 07:17 AM.
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  7. #7
    Moo
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    No,

    Once you've proved (well) the preliminary thing, just note that :

    \int_{\cup E_n} f =\int f \cdot 1_{\cup E_n}=\int \sum_{n} f \cdot 1_{E_n}=\sum \int f \cdot 1_{E_n}=\sum \int_{E_n} f

    and of course, justify the inversion of sum and integral (in that case it's simple since f is positive)
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