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Math Help - Lower bound of 1-hausdorff measure

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    Lower bound of 1-hausdorff measure

    Let K to be a Cantor middle-half set. Consider KXK.
    figure 2 shows the 1-stage of this cantor set construction.


    diam(B)=sup{|x-y|:x,ybelongs to B}

    Want to show that
    Ans from the article:
    The lower bound can be derived by considering the probability measure μ on K^2 such that μ(K^2 ∩ Q) = 4^(−n) for each of the 4^n squares Q arising at stage n of the construction. This measure satisfies μ(B) ≤ 9diam(B) for any closed set B, since B can be covered by at most nine dyadic squares with side-length smaller than diam(B). Therefore, for any cover {Bi } of K^2 by disks, \sum_{i}diam(Bi )>= \sum_{i}μ(Bi )/9>=μ(K^2)/9 =1/9


    Question
    I do not understand why μ(B) ≤ 9diam(B). I do not understand why probability measure is related to the diameter.
    Please help me!
    Last edited by ldpsong; April 6th 2010 at 09:25 AM.
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