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Thread: Lebesgue measure and Hausdorff dimension

  1. #1
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    Lebesgue measure and Hausdorff dimension

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