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Math Help - Hausdorff Measure

  1. #1
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    Hausdorff Measure

    The question I am trying to answer is as follows

    Let F={(a,b): 0< a, b<7} \subset . Show that Hs(F)=0 if s>2, 0<H2(F)< \infty and Hs(F) = \infty if s<2. (You may need to use the fact that if U is a subset of , then the area of U is at most equation to |U|2 x \pi/4

    I have managed to show Hs(F)=0 if s>2 but showing 0<H2(F)< \infty if s<2 I do not know how to do or even get started.

    Any help greatfully appreciated
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  2. #2
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    Re: Hausdorff Measure

    The following is a simple consequence of the definition of Hausdorff measure: if H^s(E) > 0 and  t < s, then H^t(E) = \infty, and if H^s(E) < \infty and  t > s then H^t(E) = 0 . This reduces your present situation to showing that 0 < H^2(F) < \infty , which is a straightforward computation.

    Edit:

    Fix a  \delta > 0 and let  \{E_j\}_{j=1}^{\infty} be any covering of F so that diam(E_j) < \delta for each j. Now  diam(E_j)^2 = diam(E_j)^{2-s} diam(E_j)^s < \delta^{2-s} diam(E_j)^s .

    So by definition of the Hausdorff measure,  H^2_{\delta}(F) \le \sum_{j=1}^{\infty} diam(E_j)^2 \le \delta^{2-s} \sum_{j=1}^{\infty} diam(E_j)^s .

    That is,  \delta^{s -2} H^2_{\delta}(F) \le \sum_{j=1}^{\infty} diam(E_j)^s .

    Inffing over all such coverings, we're led to

     \delta^{s - 2}H^2_{\delta}(F) \le H_{\delta}^s (F) .

    By definition,  H^s(F) = \lim_{\delta \to 0^{+}} H_{\delta}^s(F) . Take the limit as delta tends to 0 and conclude that since s < 2, the left hand side tends to infinity if the 2 dimensional Hausdorff measure of F is positive.
    Last edited by JakeBarnes; April 5th 2012 at 03:43 AM. Reason: Error
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