[proofplan] We compute the total Lebesgue measure of the complement $[0,1] \setminus \mathcal{C}$ by summing the lengths of all removed intervals across all stages. The geometric series sums to $1$, giving $\mathcal{L}^1([0,1] \setminus \mathcal{C}) = 1$, and hence $\mathcal{L}^1(\mathcal{C}) = 0$. [/proofplan] [step:Compute the total length removed at each stage] At stage $n$ of the construction (passing from $C_{n-1}$ to $C_n$), we remove the open middle third of each of the $2^{n-1}$ constituent intervals of $C_{n-1}$. Each constituent interval of $C_{n-1}$ has length $3^{-(n-1)}$, so each removed middle third has length \begin{align*} \frac{1}{3} \cdot 3^{-(n-1)} = 3^{-n}. \end{align*} The total length removed at stage $n$ is therefore $2^{n-1} \cdot 3^{-n}$. [/step] [step:Sum the geometric series to obtain $\mathcal{L}^1([0,1] \setminus \mathcal{C}) = 1$] The removed intervals across all stages are pairwise disjoint (each is contained in a constituent interval of the previous stage and is removed permanently). By countable additivity of Lebesgue measure, \begin{align*} \mathcal{L}^1([0,1] \setminus \mathcal{C}) = \sum_{n=1}^{\infty} 2^{n-1} \cdot 3^{-n} = \frac{1}{3} \sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^{n-1} = \frac{1}{3} \cdot \frac{1}{1 - 2/3} = \frac{1}{3} \cdot 3 = 1. \end{align*} [guided] We use the geometric series formula $\sum_{m=0}^{\infty} r^m = 1/(1-r)$ for $|r| < 1$, applied with $r = 2/3$. The substitution $m = n - 1$ gives \begin{align*} \sum_{n=1}^{\infty} 2^{n-1} \cdot 3^{-n} = \sum_{n=1}^{\infty} \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{n-1} = \frac{1}{3} \sum_{m=0}^{\infty} \left(\frac{2}{3}\right)^m = \frac{1}{3} \cdot \frac{1}{1 - 2/3} = 1. \end{align*} This confirms that the removed intervals account for the entire length of $[0,1]$. [/guided] [/step] [step:Conclude $\mathcal{L}^1(\mathcal{C}) = 0$] Since $\mathcal{C} \subset [0,1]$ and $[0,1] \setminus \mathcal{C}$ is the disjoint union of the removed open intervals, \begin{align*} \mathcal{L}^1(\mathcal{C}) = \mathcal{L}^1([0,1]) - \mathcal{L}^1([0,1] \setminus \mathcal{C}) = 1 - 1 = 0. \end{align*} This uses the fact that $\mathcal{C}$ is Lebesgue measurable (as a closed set, hence a Borel set) and the finite additivity of $\mathcal{L}^1$ on the partition $[0,1] = \mathcal{C} \sqcup ([0,1] \setminus \mathcal{C})$. [/step]