I have to show that $0<\mathcal H_{\alpha }^\delta(\mathcal C)<\infty $ where $\mathcal C$ is the cantor set, $\alpha =\frac{\ln(2)}{\ln(3)}$ and $$\mathcal H_\alpha ^\delta(E)=\left\{\sum_{k=1}^\infty |U_i|^\alpha \mid E\subset \bigcup_{i=1}^\infty U_i, |U_i|<\delta\right\}$$ and $|A|=\sup\{|x-y|\mid x,y\in A\}$. So the fact $\mathcal H_\alpha ^\delta(\mathcal C)\leq 1$ is fine. I'm trying now to show that $\mathcal H_\alpha ^\delta(\mathcal C)\geq \frac{1}{2}$. My proof goes like:
Let $\{U_i\}$ a cover of $\mathcal C$. Since $\mathcal C$ is compact, we can suppose that $\{U_i\}$ is a finit collection of sub-interval of $[0,1]$. For all $i$, let $k$ s.t. $\frac{1}{3^{k+1}}\leq |U_i|<\frac{1}{3^k}$. Let recall that $\mathcal C=\bigcap_{i=0}^\infty C_i$ where $C_i$ is a set of $2^i$ intervals of length $3^{-i}$. Then $U_i$ intersect at most one interval of $C_k$. For $j\geq k$, $U_i$ intersect $2^{j-k}=2^j3^{-k\alpha }\leq 2^j3^\alpha |U_i|^\alpha $ interval of $C_j$. For $j$ big enough, since $\{U_i\}$ is finite, $3^{-(j+1)}\leq |U_i|$ for all $i$, and thus $$2^j\leq \sum_{i}2^j3^\alpha |U_i|^\alpha $$ and thus $\sum_{i}|U_i|^\alpha \geq 3^{-\alpha }=1/2$.
I just don't understand how we get $$2^j\leq \sum_{i}2^j3^\alpha |U_i|^\alpha.$$ I'm sure it's obvious, but I really don't see how we get this formula.