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Math Help - Lebesgue measure and Hausdorff dimension

  1. #1
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    Lebesgue measure and Hausdorff dimension

    Let E\subset\mathbb{R}^{n} be a set that is not measurable with respect to the n-dimensional Lebesgue measure. Show that E has Hausdorff dimension n.
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  2. #2
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    Re: Lebesgue measure and Hausdorff dimension

    Assuming that the Hausdorff dimension is the number s= \inf_{t} \{ \mathcal{H}_t(E)=0\} = \sup_t \{ \mathcal{H}_t(E)=\infty \}, then we have \mathcal{H}_{s'}(\mathbb{R}^n)=0 for all s'>n (To see this note that the s'-Hausdorff measure of an n-cube is zero for s'>n and use subadditivity of the outer measure), and by monotonicity \mathcal{H}_{s'}(E)=0, so s\leq n. If s<n then \mathcal{H}_n(E)=\lambda_n(E)=0, but Lebesgue measure is complete so E is measurable, a contradiction.
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