The Cantor Set
Table of Contents

The Cantor Set

We have already proven that if $E$ is a countable set then $m^*(E) = \infty$. In particular, every Lebesgue measurable countable set has Lebesgue measure $0$.

However, is it possible to construct an uncountable Lebesgue measurable set that has Lebesgue measure $0$? The answer is yes, and we construct it now.


\begin{align} \quad C_0 &= [0, 1] \\ \quad C_1 &= \left [ 0, \frac{1}{3} \right ] \cup \left [ \frac{2}{3}, 1 \right ] \\ \quad C_2 &= \left [ 0, \frac{1}{9} \right ] \cup \left [ \frac{2}{9}, \frac{3}{9} \right ] \cup \left [ \frac{6}{9}, \frac{7}{9} \right ] \cup \left [ \frac{8}{9}, 1 \right ] \\ \quad & \vdots \end{align}

That is, for each $n \in \mathbb{N}$, $C_n$ is obtained by deleting the middle third of each closed interval in the union of $C_{n-1}$. We are now ready to define the Cantor set.

Definition: The Cantor Set is defined as $\displaystyle{C = \bigcap_{n=0}^{\infty} C_n}$.

We will now look at two important results regarding the Cantor set.

Theorem 1: The Cantor set has Lebesgue measure $0$.
  • Proof: $\{ C_n \}_{n=0}^{\infty}$ is a collection of Lebesgue measurable sets such that for all $n \in \mathbb{N} \cup \{ 0 \}$ we have that $C_n \supseteq C_{n+1}$ and furthermore, $m(C_0) = m([0,1]) = 1$. So by The Continuity Properties of the Lebesgue Measure we have that:
\begin{align} \quad m(C) = m \left ( \bigcap_{n=0}^{\infty} C_n \right ) = \lim_{n \to \infty} m(C_n) \end{align}
  • For each $n \in \mathbb{N}$ we have that:
\begin{align} \quad m(C_n) = \left ( \frac{2}{3} \right )^n \end{align}
  • Therefore:
\begin{align} \quad m(C) = \lim_{n \to \infty} \left ( \frac{2}{3} \right)^n = 0 \end{align}
  • So the Cantor set has Lebesgue measure $0$. $\blacksquare$
Theorem 2: The Cantor set is uncountable.
  • Proof: Suppose instead that the Cantor set is in fact countable. Then $C = \{ c_n \}_{n=1}^{\infty}$. The point $c_1$ is contained in one of the intervals in the union $C_1$. Let $F_1$ denote the interval that does not contain $c_1$. Furthermore, at least one of the two intervals in $C_2$ whose union is $F_1$ does not contain $c_2$. Let $F_2$ denote the interval that does not contain $c_2$. We continue on in this process and contain a collection of closed and bounded intervals $(F_n)_{n=1}^{\infty}$ that are nested. By the nested intervals theorem this set is nonempty but by construction this set does not contain any element of $C$. Therefore the assumption that $C$ is countable must have been false. So the Cantor set is uncountable. $\blacksquare$
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