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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- Dec 2nd 2011, 10:05 AMkierkegaardLebesgue 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.

- Dec 3rd 2011, 12:04 PMJose27Re: 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.