How prove that $m_\alpha ^*(\mathcal C)\geq 1$ where $\mathcal C$ is the Cantor set, and $m_\alpha $ the Hausdorff measure. Let $\mathcal C$ the Cantor set and $m_\alpha ^*$ the $\alpha$-Hausdorff measure with $\alpha =\frac{\ln(2)}{\ln(3)}$,  i.e. $$m_\alpha ^*(E)=\lim_{\delta\to 0}\mathcal H_\alpha ^\delta(E)$$
where $$\mathcal H_\alpha ^\delta(E)=\inf \left\{\sum_{k}\operatorname{diam}(F_i)^\alpha \mid E\subset \bigcup_{i=1}^\infty F_i, \operatorname{diam}(F_i)<\delta\right\}$$
and $\operatorname{diam}(A)=\{\|x-y\|: x,y\in A\}$ the diameter.
I had no problem to show that $m_\alpha ^*(\mathcal C)\leq 1$, but I have to show that $m_\alpha ^*(\mathcal C)=1$, then, I would like to show that $m_\alpha ^*(\mathcal C)\geq 1$. I really have no idea on how to proceed.
 A: For my own sake, I will use the more common notation $\mathcal{H}^{\alpha}$ for the $\alpha$-Hausdorff measure.

To show that $\mathcal{H}^{\alpha}(\mathcal{C}) \geq 1$ it is enough to show that $\sum_{j}\operatorname{diam}(I_j)^\alpha \geq 1$, whenever open intervals $I_j$ cover $\mathcal{C}$. As $\mathcal{C}$ is compact, it can be covered by finitely many $I_j$'s, so without loss of generality we can assume we only have $I_1, \dots, I_n$, as this can only decrease the sum. 
Moreover, we can take each $I$ ($I=I_j$ for some $j=1,\dots,n$) to be the smallest interval containing two intervals $J$ and $J'$, which appear in the construction of the Cantor set (they don't need to appear at the same stage of the construction). Again, to do so we may shrink $I$, but this only decreases the sum.
If $J$ and $J'$ are the largest such intervals, by the construction of the Cantor set we know that $I$ must be made up of $J$, followed by an interval $K$ in the complement of $\mathcal{C}$, followed by $J'$. By construction, we also have $\operatorname{diam}{K} \geq \operatorname{diam}(J), \operatorname{diam}(J')$ and therefore $\operatorname{diam}(K) \geq \frac12 (\operatorname{diam}(J) + \operatorname{diam}(J'))$.
Then, we have
\begin{align}
\operatorname{diam}(I)^\alpha & = (\operatorname{diam}(J) + \operatorname{diam}(K) + \operatorname{diam}(J'))^\alpha \geq \\
& \geq \left(\frac32 (\operatorname{diam}(J) + \operatorname{diam}(J'))\right)^\alpha= \\
& = 2 \left(\frac12 (\operatorname{diam}(J) + \operatorname{diam}(J'))\right)^\alpha \geq \\
& \geq \operatorname{diam}(J)^\alpha + \operatorname{diam}(J')^\alpha,
\end{align}
using the fact that $3^\alpha=2$ and that $t^{\alpha}$ is a concave function.
This allows us to replace each $I_j$ with $J_j$ and $J'_j$, without increasing the sum of $\alpha$-th power of the diameters.
We proceed in this way, until, after finitely many steps,  we can cover $\mathcal{C}$ with intervals of equal lengths $3^{-i}$. This must include all the intervals in the $i$-th stage of the construction of $\mathcal{C}$. Since for such intervals we have $\sum \operatorname{diam}(I)^\alpha \geq 2^i 3^{-\alpha i}= 2^i 3^{\log_3 2^{-i}}= 1$, the same inequality must hold for the initial intervals.

This proof can be found almost verbatim in "The geometry of fractal sets", by Falconer.
