I need to show that if $\alpha=\log2/\log3$ then $m_{\alpha}^{*}(\mathcal{C})=1$.
I know how to prove that it is $\leq 1$, and I know that I need to prove that it is $ > 0$, namely $\geq c$ for some constant $c$. I have read the proof in the book "The geometry of fractal sets" but I don't understand it. I don't understand the step after the inequality (1.22), in the first lower bound.
Moreover, why the fact that the measure is greater than $0$ implies that the dimension is greater than $\log2/\log3$ implies that the measure is $1$?
We have $|J| \le |K|$ and $|J'| \le |K|$, therefore
$$|J| + |K| + |J'| \ge |J| + \frac{1}{2}(|J| + |J'|) + |J'| = \frac{3}{2} (|J| + |J'|).$$
There is a typo in the immediate equality thereafter in the book. It should be
$$\left(\frac{3}{2} (|J| + |J'|)\right)^s = 2 \left(\frac{|J| + |J'|}{2}\right)^s \ge 2 \left(\frac{|J|^s + |J'|^s}{2}\right).$$