[proofplan] We compute $\dim_H(\mathcal{C}) = \log 2 / \log 3$ by establishing two bounds. For the upper bound, we cover $\mathcal{C}$ by the $2^n$ constituent intervals of $C_n$ (each of diameter $3^{-n}$) and show that $\mathcal{H}^s(\mathcal{C}) = 0$ for $s > \log 2/\log 3$. For the lower bound, we use a mass distribution argument: we define the natural probability measure $\mu$ on $\mathcal{C}$ (assigning mass $2^{-n}$ to each constituent interval of $C_n$) and show that $\mu(B(x, r)) \le C r^s$ for $s = \log 2/\log 3$, which implies $\mathcal{H}^s(\mathcal{C}) > 0$ by the mass distribution principle. [/proofplan] [step:Establish the upper bound $\dim_H(\mathcal{C}) \le \log 2 / \log 3$] Let $s > \log 2/\log 3$. We show $\mathcal{H}^s(\mathcal{C}) = 0$, which implies $\dim_H(\mathcal{C}) \le s$. Since this holds for every $s > \log 2/\log 3$, the upper bound follows. At stage $n$ of the construction, $\mathcal{C} \subset C_n$ and $C_n$ consists of $2^n$ closed intervals, each of length (and hence diameter) $3^{-n}$. This gives a cover of $\mathcal{C}$ by $2^n$ sets of diameter $\delta_n := 3^{-n}$. By the definition of $\mathcal{H}^s_{\delta_n}$, \begin{align*} \mathcal{H}^s_{\delta_n}(\mathcal{C}) \le \sum_{j=1}^{2^n} (3^{-n})^s = 2^n \cdot 3^{-ns} = \left(\frac{2}{3^s}\right)^n. \end{align*} Since $s > \log 2/\log 3$, we have $3^s > 2$, so $2/3^s < 1$. Therefore $(2/3^s)^n \to 0$ as $n \to \infty$. Since $\delta_n = 3^{-n} \to 0$ and $\mathcal{H}^s(\mathcal{C}) = \lim_{\delta \to 0} \mathcal{H}^s_\delta(\mathcal{C}) \le \lim_{n \to \infty} \mathcal{H}^s_{\delta_n}(\mathcal{C}) = 0$, we conclude $\mathcal{H}^s(\mathcal{C}) = 0$. [guided] The exponent $s = \log 2/\log 3$ is the unique solution of $2 \cdot 3^{-s} = 1$, equivalently $2 = 3^s$. When $s$ exceeds this threshold, $2 \cdot 3^{-s} < 1$, and the $s$-dimensional Hausdorff pre-measure of the natural covers decays geometrically to zero. The upper bound is the easier half: it only requires exhibiting good covers, and the natural covers from the Cantor set construction are perfectly adapted to the self-similar structure. Note that at the critical exponent $s = \log 2/\log 3$, the bound gives $\mathcal{H}^s_{\delta_n}(\mathcal{C}) \le 1$ for every $n$ — the covers have bounded $s$-cost. This does not by itself prove $\mathcal{H}^s(\mathcal{C}) > 0$; for that we need the lower bound. [/guided] [/step] [step:Define the natural probability measure $\mu$ on $\mathcal{C}$] We construct a Borel probability measure $\mu$ on $\mathcal{C}$ as follows. For each $n \in \mathbb{N}$ and each constituent interval $I$ of $C_n$, assign \begin{align*} \mu(I \cap \mathcal{C}) := 2^{-n}. \end{align*} Since $C_n$ has exactly $2^n$ constituent intervals and each receives mass $2^{-n}$, the total mass is $1$. The assignment is consistent: each constituent interval of $C_{n-1}$ contains exactly two constituent intervals of $C_n$, receiving total mass $2 \cdot 2^{-n} = 2^{-(n-1)}$, which matches the mass assigned at level $n-1$. By the Carathéodory extension theorem (applied to the algebra generated by the sets $\{I \cap \mathcal{C}\}$ across all levels), $\mu$ extends to a Borel probability measure on $\mathcal{C}$. This is the **Cantor measure** — the unique Borel probability measure invariant under the symmetry of the construction. [guided] Alternatively, $\mu$ can be described as the pushforward of the fair coin-tossing measure on $\{0,2\}^{\mathbb{N}}$ (the product measure giving equal weight $1/2$ to each of $0$ and $2$) under the homeomorphism $\Phi: \{0,2\}^{\mathbb{N}} \to \mathcal{C}$ from [Homeomorphism with the Product Space](/theorems/1202). This makes it transparent that $\mu$ is a well-defined Borel probability measure. [/guided] [/step] [step:Bound $\mu(B(x, r))$ from above for the critical exponent] Set $s := \log 2 / \log 3$, so that $2 = 3^s$. We show there exists a constant $C > 0$ such that \begin{align*} \mu(B(x, r) \cap \mathcal{C}) \le C \cdot r^s \quad \text{for all } x \in \mathcal{C} \text{ and all } r > 0. \end{align*} Let $x \in \mathcal{C}$ and $r > 0$. Choose $n \in \mathbb{N}_0$ such that $3^{-(n+1)} \le r < 3^{-n}$ (if $r \ge 1$, the bound holds with $C = 1$ since $\mu(\mathcal{C}) = 1 \le r^s$). The ball $B(x, r)$ has diameter $2r < 2 \cdot 3^{-n}$. A constituent interval of $C_n$ has length $3^{-n}$. The ball $B(x, r)$ can intersect at most $3$ constituent intervals of $C_n$ (since $2r < 2 \cdot 3^{-n}$, the ball spans at most two full interval-lengths plus the gaps between them, but since consecutive constituent intervals at level $n$ are separated by gaps of length at least $3^{-n}$, the ball can intersect at most $2$ constituent intervals). More precisely: two consecutive constituent intervals of $C_n$ within the same parent at level $n-1$ are separated by a gap of length $3^{-n}$ (the removed middle third). Two constituent intervals from different parents are separated by an even larger gap. Since $2r < 2 \cdot 3^{-n}$, the ball $B(x, r)$ can overlap at most $2$ constituent intervals of $C_n$. Hence \begin{align*} \mu(B(x, r) \cap \mathcal{C}) \le 2 \cdot 2^{-n} = 2^{-(n-1)}. \end{align*} From $r \ge 3^{-(n+1)}$, we get $3^{n+1} \ge 1/r$, so $3^n \ge 1/(3r)$. Raising both sides to the power $s = \log 2/\log 3$: \begin{align*} 2^n = (3^s)^n = 3^{ns} \ge (3r)^{-s} \cdot 3^s = \frac{3^s}{(3r)^s} = \frac{3^s}{3^s r^s} = r^{-s}. \end{align*} Wait — let us redo this cleanly. From $r \ge 3^{-(n+1)}$, we have $r^s \ge 3^{-s(n+1)} = 2^{-(n+1)}$ (using $3^s = 2$). Therefore \begin{align*} \mu(B(x, r) \cap \mathcal{C}) \le 2 \cdot 2^{-n} = 4 \cdot 2^{-(n+1)} \le 4 \cdot r^s. \end{align*} This gives the bound with $C = 4$. [guided] The computation hinges on converting between the geometric scale $3^{-n}$ and the measure scale $2^{-n}$ via the relation $2^{-n} = 3^{-sn}$ (which is the definition of $s = \log 2/\log 3$). A ball of radius $r \approx 3^{-n}$ captures at most $O(1)$ constituent intervals of level $n$, each of measure $2^{-n} = 3^{-sn} \approx r^s$. The constant $C = 4$ arises from two sources: the factor of $2$ from covering at most two constituent intervals, and the factor of $2$ from the mismatch between $r$ and $3^{-n}$ (since $r$ can be as small as $3^{-(n+1)} = 3^{-n}/3$, giving a factor of $3^s = 2$). Combined: $2 \cdot 2 = 4$. [/guided] [/step] [step:Apply the mass distribution principle to obtain the lower bound] [claim:Mass Distribution Principle] Let $E \subset \mathbb{R}$ be a Borel set supporting a finite Borel measure $\mu$ with $\mu(E) > 0$. If there exist constants $C > 0$ and $s > 0$ such that $\mu(B(x, r)) \le C r^s$ for all $x \in \mathbb{R}$ and all $r > 0$, then $\mathcal{H}^s(E) \ge \mu(E)/C > 0$. [/claim] [proof] Let $\{U_j\}_{j=1}^{\infty}$ be any countable cover of $E$ with $\operatorname{diam}(U_j) < \delta$ for all $j$. Each $U_j$ is contained in a closed ball of radius $\operatorname{diam}(U_j)$ centered at any point of $U_j$. By the hypothesis, \begin{align*} \mu(U_j) \le \mu(B(x_j, \operatorname{diam}(U_j))) \le C \cdot (\operatorname{diam}(U_j))^s \end{align*} for any $x_j \in U_j$. Summing over $j$: \begin{align*} \mu(E) \le \sum_{j=1}^{\infty} \mu(U_j) \le C \sum_{j=1}^{\infty} (\operatorname{diam}(U_j))^s. \end{align*} Taking the infimum over all such covers gives $\mu(E) \le C \cdot \mathcal{H}^s_\delta(E)$ for every $\delta > 0$. Letting $\delta \to 0$: $\mu(E) \le C \cdot \mathcal{H}^s(E)$, hence $\mathcal{H}^s(E) \ge \mu(E)/C$. [/proof] We apply the mass distribution principle with $E = \mathcal{C}$, $\mu$ the Cantor measure, $s = \log 2/\log 3$, and $C = 4$ (from the previous step). Since $\mu(\mathcal{C}) = 1 > 0$, we obtain \begin{align*} \mathcal{H}^s(\mathcal{C}) \ge \frac{1}{4} > 0. \end{align*} Since $\mathcal{H}^s(\mathcal{C}) > 0$ for $s = \log 2/\log 3$, the definition of Hausdorff dimension gives $\dim_H(\mathcal{C}) \ge \log 2/\log 3$. [/step] [step:Combine the two bounds] From the upper bound (step 1), $\dim_H(\mathcal{C}) \le \log 2/\log 3$. From the lower bound (step 4), $\dim_H(\mathcal{C}) \ge \log 2/\log 3$. Therefore \begin{align*} \dim_H(\mathcal{C}) = \frac{\log 2}{\log 3}. \end{align*} [/step]