Exercise 1.2.9 (Middle thirds Cantor set)

Let $I_0:=[0,1]$ be the unit interval, let $I_1:=[0,1∕3]\cup [2∕3,1]$ be $I_0$ with the interior of the middle third of the interval removed. Let $I_2:=[0,1∕9]\cup [2∕9,3∕9]\cup [6∕9,7∕9]\cup [8∕9,1]$ be $I_1$ with the interior of the middle third of each of the two intervals of $I_1$ removed, and so forth. We obtain

(i)

(Lebesgue measure)
Notice that in each $n\in \mathbb{N}$ we take the union of at most ${(\#\{0,2\})}^n=2^n$ combinations $a_1,\dots ,a_n\in \{0,2\}$. Each of these intervals has a Lebesgue outer measure of $1∕3^n$ and they are obviously disjoint. Thus, the measure of $I_n$ cannot exceed $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:m^{\text{∗}}(I_n)\le 2^n\cdot \frac{1}{3^n}$$

But $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^n{I}_i\subset I_n$; taking supremum in $n$ thus yields

$$m^{\text{∗}}\left(\bigcap _{i=1}^n{I}_i\right)\le 0$$

since $\underset{n\in \mathbb{N}}{\sup <!--\; FUNCTION\; APPLICATION\; -->}{(2∕3)}^n=0$. By Lemma 1.2.13, $C$ is Lebesgue measurable.

(ii)

(Compactness)
We use the Heine-Borel criteria. Obviously, $C\subseteq [0,1]$ is a bounded set. Furthermore, $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}$ $I_n$ is closed; a countable intersection of closed sets $C$ must hence be also closed. Thus, $C$ is both closed and bounded; therefore, it is compact.

(iii)

(Cardinality)

Lemma 1. The Cantor set $C$ is equal to

$$\left\{\sum _{n=1}^{\infty }\frac{a_n}{3^n}:a_n\in \{0,2\}\right\}$$

Proof. Actually for the purpose of demonstrating that $C$ is uncountable, it is enough to demonstrate that the above set is a subset of $C$. Pick an arbitrary sequence ${(a_n)}_{n\in \mathbb{N}}$ in ${\{0,2\}}^{\mathbb{N}}$. □

Lemma 2. The set $C=\left\{{\sum <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }\frac{a_n}{3^n}:a_n\in \{0,2\}\right\}$ is uncountable.

Proof. One way to intuitively understand the cardinality of the Cantor set is to think of it as essentially the set of *ternary* (base 3) representations of all $x\in [0,1]\subset ℝ$ which can be expressed using *only* the digits 0 and 2. That is, it is the set of all "trecimal expansions” of $x\in [0,1]\subset ℝ$ that consist of only zeros and twos: e.g. $0.0{\overline{0}}_{\text{ base 3}}=0\in C,0.2{\overline{2}}_{\text{ base 3}}=1\in C,0.2020202020202{\dots }_{\text{(base 3)}}\in C$, etc. The set of all such $x$ is uncountable: There are $2^{\left|\mathbb{N}\right|}$ distinct sequences consisting of only zeros and twos: for each $n\in \mathbb{N}$, there are two possible choices for any given $a_n:0$ or $2$. Since there are uncountably many distinct sequences $\{a_n\}\in {\{0,2\}}^{\mathbb{N}}$, there are uncountably many series $\sum <!--\; FUNCTION\; APPLICATION\; -->\frac{a_n}{3^n}$, each of which represents a *single point* in the Cantor set. Hence, there are uncountably many points in the Cantor Set. □