Is the Cantor set uncountable but still a null set under Lebesgue measure?

The apparent paradox of an uncountable set with zero Lebesgue measure arises from conflating **cardinality** with **measure**. These are fundamentally different notions of "size". **Cardinality:** The Cantor set $\mathcal{C}$ consists of all numbers in $[0,1]$ whose ternary expansion uses only digits $0$ and $2$. The map $$\varphi: \mathcal{C} \to [0,1], \quad \varphi(0.a_1 a_2 a_3 \ldots_{(3)}) = 0.(a_1/2)(a_2/2)(a_3/2)\ldots_{(2)}$$ where $a_i \in \{0,2\}$, is a bijection (interpret the ternary digits $0 \to 0$, $2 \to 1$ as binary). Hence $|\mathcal{C}| = |[0,1]| = \mathfrak{c}$. **Measure:** At stage $n$, the set $C_n$ consists of $2^n$ intervals each of length $3^{-n}$, so $\lambda(C_n) = (2/3)^n$. Since $\mathcal{C} = \bigcap_n C_n$ and $C_{n+1} \subseteq C_n$, $$\lambda(\mathcal{C}) = \lim_{n\to\infty} \lambda(C_n) = \lim_{n\to\infty} \left(\frac{2}{3}\right)^n = 0.$$ **Intuition:** The Cantor set is a **thin** but **numerous** set. It has no interior — every point is a boundary point — yet it has the same cardinality as the continuum. It is a perfect set (closed with no isolated points) that is nowhere dense and totally disconnected. Think of it as a "dust" of points: infinitely many points scattered so sparsely that they occupy no total length. The construction removes "almost all" length while preserving the cardinality of the continuum. **Higher dimensions:** The Cantor dust $\mathcal{C} \times \mathcal{C} \subset [0,1]^2$ is also uncountable with $\lambda_2(\mathcal{C} \times \mathcal{C}) = 0$. More generally, one can construct fat Cantor sets with positive measure by removing proportionally smaller middle sections at each stage.