Hausdorff measure of Lebesgue non-measurable set In $\mathbb{R}^{N}$, if $E \subset \mathbb{R}^{N}$ is not Lebesgue measurable and $\alpha < N$, then $m^{*}_{\alpha}(E)=\infty$.
I have started with the definition of a non-measurable set: there exists a constant $c$ s.t. $\forall \mathcal{O}$ an open set with $E \subset \mathcal{O}$ which satisfies
$$m^{*}(\mathcal{O} \setminus E)>c$$
I know that $m^{*}_{\alpha}(\mathcal{O})=\infty$ and I am trying to show that $m^{*}_{\alpha}(E)\geq m^{*}_{\alpha}(\mathcal{O})$ but I don't know if this approach is correct and I don't know how to continue if so is.
I would appreciate if someone could help me. Thanks.
 A: I would try to show the contrapositive.  Suppose that $E$ is a set with finite $\alpha$-dimensional Hausdorff measure.  You can actually show that $E$ must have Lebesgue outer measure zero, and in particular it is Lebesgue measurable.
To get started, note that by definition, there must exist $M < \infty$ such that for all sufficiently small $\delta$, there is a covering of $E$ by a sequence of sets $C_i$ such that $d_i := \operatorname{diam}(C_i) \le \delta$ and $\sum_i d_i^\alpha < M$.  Now for each $i$, let $B_i$ be a closed ball of radius $d_i$ centered at some point of $C_i$.  Then $C_i \subset B_i$, so the balls $B_i$ also cover $E$, and we have $m(B_i) = \omega_N d_i^N$ where $\omega_N$ is the volume of the unit $N$-ball.  Now the Lebesgue outer measure $m^*(E)$ can be bounded by
$$m^*(E) \le \sum_i m(B_i) = \omega_N \sum_i d_i^N \le \omega_N \delta^{N-\alpha} \sum_i d_i^\alpha \le \omega_N \delta^{N-\alpha} M.$$
Since $\alpha < N$ by assumption, letting $\delta \to 0$ we have $m^*(E)=0$.
