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Thread: Measure question

  1. #1
    May 2012

    Measure question

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

    I wanna know how to prove that

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

    Thank you for your help.
    Last edited by Wowhapjs; May 7th 2012 at 08:40 PM.
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  2. #2
    Mar 2012

    Re: Measure question

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

    $\displaystyle m \in \mathbb{N} $ , so $\displaystyle E_m$ 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:

    $\displaystyle \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 $\displaystyle m \geq 1$ .

    Consider $\displaystyle A_i , A_j $ measurable.

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

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

    So it follows that $\displaystyle 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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