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Jan 14, 2013 at 12:38 comment added R W You are right - the dimension formula I use also holds just under the assumption $\delta_n\to 0$.
Jan 14, 2013 at 6:28 comment added Tapio Rajala With your fast convergence of $\delta_n$ the measure $m$ has also the property that $m(B(x,r)) \le cr^s$ for all $x \in \mathbb{R}^2$ (with $s = \log 2/ \log 3$). Therefore by Frostman's lemma $\mathcal{H}^s(C_\delta)>0$. For $\dim_\mathcal{H}(C_\delta) = s$ it is enough to assume $\delta_n \to 0$ and $\delta_n \in [0,2/3)$. No fast convergence is needed for this.
Jan 13, 2013 at 23:08 vote accept Nikita Sidorov
Jan 13, 2013 at 17:38 comment added Nikita Sidorov Thanks. Yes, introducing measures looks like a good approach here.
Jan 13, 2013 at 15:33 history answered R W CC BY-SA 3.0