given the (ternary) Cantor set $\mathcal{C}$ it is well known that its Hausdorff dimension is given by $\dim_\mathcal{H}(\mathcal{C})=ln(2)/ln(3)$, which I am going to denote by $\alpha$. I am especially interested in the computation of the following lower bound: $$ \mathcal{H}^{\alpha}(\mathcal{C}) \geq 1/2. $$ I can provide a proof applying basic properties and the definition, however I am especially interested in providing a proof using the mass distribution principle (similiar to the energy method presented here). Let us (briefly) recall the Mass distribution principle first.
Mass Distribution Principle: Let $E$ be a set in a metric space $(X,d)$ that supports a finite Borel measure $\mu$ such that for all $x \in E$ and $r>0$ one has $$ \mu(B(x,r)) \leq C r^{s} $$ Then one has $$ \mathcal{H}^{s}(E) \geq \mu(E)/C > 0, $$ and hence one has $\dim_{\mathcal{H}} \geq s$.
Setup-Cantor Set:
I want to provide a "probabilistic" setting for this task where one uses the representation of the Cantor set in terms of $3$-adic expansion, for this I believe the following points are valid:
- Consider the Cantor set $\mathcal{C} = \{ \sum^{\infty}_{n=1} \frac{2 a_n}{3^n} : a_n \in \{0,2\}\}$, where each number is presented in terms of its $3$-adic expansion.
- Using the measure descripted here and denote this by $\mu$, that is constructed as follows: Take $\Omega = \{0,1\}^{\mathbb N}$, the sigma Algebra$\mathcal{A}$ that is generated by the following cylinder sets $$ \{\omega \in \Omega : \forall S \subseteq \mathbb N \text{ finite} : a_n \in \{0,2\}\}. $$ And finally the measure $\mu = \prod_{n \in \mathbb N} \mu_n$ where for each $n$ one has $\mu_n(\omega_n = 1) = 1/2$ and then extend this to all Borel sets $B \subset \mathcal{A}$ by Kolmogorov extension theorem. This is the "natural" measure that we want to use for the Mass distribution principle.
- We need to estimate $\mu(B(x,r))$ for arbitary $r>0$ and $x \in \mathcal{C}$, however one can show that a calculation on $3$-adic grids on generation $n$ sufficies where $I_n(x) = [\frac{k-1}{3^n},\frac{k}{3^n})$ contains $x$. Thus we only need to estimate on balls of radius $B(x,3^{-n})$ for some $n$ (In case the ball has radius bigger than 1/3 one can reduce it to the 1/3 case by using similiarity of the left and right side of the cantor set).
We are now ready to give the proof. Take arbitary $x \in \mathcal{C}$, by the last point it sufficies to proof that $$ \mu(B(x,3^{-n})) \leq C 3^{-\alpha n} = C 2^{-n}. $$ But this is easy now, since $B(x,3^{-n})$ contains all the points $y \in \mathcal{C}$ such that there $3$-adic expansion agrees on the first $n$ digits and then starts to differ. Using $(2)$ there are exactly $\frac{1}{2^{-n}}$ choices and hence the above inequality holds. Applying the mass distribution principle yields $$ \dim_{\mathcal{H}}(\mathcal{C}) \geq \alpha. $$
Question: Is this correct? Is there a cleaner way to write this in terms of $(3)$, i.e. a direct argument why one can only take balls of radius $r=3^{-n}$. Thanks in advance.