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Thread: Measure question

  1. #1
    May 2012

    Measure question

    Assume $A_j,j\geq 1,j\in\Bbb N$ are measurable sets. Let $m \in N$, and let $E_m$ be the set defined as follows : $x \in E_m \Longleftrightarrow x$ is a member of at least $m$ of the sets $A_k$.

    I wanna know how to prove that

    1. $E_m$ is measurable.
    2. $m\lambda(E_m)\le\sum^{\infty}_{k=1}\lambda(A_k)$.

    Thank you for your help.
    Last edited by Wowhapjs; May 7th 2012 at 09:40 PM.
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  2. #2
    Mar 2012

    Re: Measure question

    Well, the property of x \in E_m as defined above simply means that if you have a look at at least m intersecting sets,
    then x will be in this intersection. And if there is another intersection, then x could also be in there.

    In other words:  E_m is the set of the union of all intersections including m sets.
    Since the  A_j, j \in \mathbb{N} are measurable and there is a defined measure \lambda,
    there must be a \sigma - algebra  \ \Sigma which means that if  A_j is in \Sigma,
    so is any countable intersection and countable union.

     m \in \mathbb{N} , so E_m lies in the sigma-algebra and is therefore measurable.

    (Or otherwise stated: Any countable intersection and union of countable sets is countable.)


    Measures are subadditive:

     \lambda(\bigcup_{i = 1}^N A_i) \ \leq \ \sum_{i = 1}^N \lambda(A_i)

    For m = 1 the equation is obvious (use subadditivity).
    Since the measure can only become smaller if we raise m, the "at least" does not make any difference to the proof.
    The thing is that one has to prove the "m" on the left side.

    Let  m \geq  1 .

    Consider  A_i , A_j measurable.

    Then  \lambda(A_i \cap A_j) \leq \lambda(A_i)

    The same is true for  A_j .
    Is that clear to you? One might have to prove this relation.

    So it follows that  2\ \lambda(A_i \cap A_j) \leq \lambda(A_i) + \lambda(A_j)

    Here we set m = 2. By induction with respect to m, you should get to the required relation.
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