[Measure Theory] Cantor Set

Cantor set

Cantor set $C$ is the standard example of Lebesgue null sets that are uncountable. Construct

\[\begin{aligned} C_0 & = [0, 1] \\ C_1 & = C_0 - (\frac{1}{3}, \frac{2}{3}) \\ C_2 & = C_1 - (\frac{1}{9}, \frac{2}{9}) - (\frac{7}{9}, \frac{8}{9}) \\ \vdots & \end{aligned}\]

by continuing removing middle thirds. Then, we call $C = \bigcap_n C_n$ Cantor set.

$\mathbf{Fig\ 1.}$ Cantor set

Here are some basic and useful properties for Cantor set:

1. Cantor set is closed
   $\mathbf{Proof.}$

   It is because Cantor set is intersection of closed set (complement of $(a, b)$). Note that intersections of closed subsets are also closed. This can be derived using De Morgan's Law and the axiom of a topology that unions of open subsets are also open. $$\tag*{$\blacksquare$}$$
2. Cantor set is uncountable
   $\mathbf{Proof.}$

   Let any element $a \in C$ be given. Then, we can represent any element $a$ with binary number. Precisely, if $a$ is contained in the first third part of $C_1$, then choose $a_1 = 0$. Otherwise, choose $a_1 = 2$. Without loss of generality, suppose $a_1 = 0$. Then, if $a$ is contained in the first third part of the first third part of $C_1$, choose $a_2 = 0$. Otherwise, if it is contained in the second third part of the first third part of $C_1$, choose $a_2 = 2$. Continuing this procedure, we obtain $a_1, a_2, \cdots \in \{0, 2\}$. Let $f(x) = \sum_{j=1}^\infty b_j 2^{-j}$ where $b_j = a_j /2$. Then the series defining $f(x)$ is the base-2 expansion of a number in $[0, 1]$. Hence, $f$ maps $C$ onto $[0, 1]$, thus $$ \begin{aligned} 2^{| \mathbb{N} |} \leq | F | \end{aligned} $$ which implies $C$ is uncountable. $$\tag*{$\blacksquare$}$$
3. Cantor set doesn't contain any interval that is subset of $[0, 1]$.
   $\mathbf{Proof.}$

   Note that the length of $C_n$ is $$ \begin{aligned} \sum_{i=1}^n \frac{2^{n-1}}{3^n} = \frac{2}{3}^n \end{aligned} $$ Thus, the length of $C$ is $$ \begin{aligned} \lim_{n \to \infty} \frac{2}{3}^n = 0. \end{aligned} $$ In other words, Lebesgue measure of Cantor set $C$ is zero. $$\tag*{$\blacksquare$}$$
4. and every point is a limit point
   $\mathbf{Proof.}$

   Let $x \in C$, and let $S$ be an open interval containing $x$. Let $I_n$ be the closed interval in $C_n$ (of which there are $2^n$) that contains $x$. Choose $n$ large enough so that $I_n \subset S$. Note that it always exist such $n$ since Cantor set doesn't contain any open interval. Let $x_n$ be an endpoint of $I_n$ such that $x_n \neq x$. Now, by construction, $x_n \in C$. Hence, $$(S - \{x \}) \cap C \neq \emptyset$$, i.e. $x$ is a limit point: an arbitrary open interval $S$ containing $x$ contains at least one element of $C$. $$\tag*{$\blacksquare$}$$

Cantor Function

As we obtained from the proof of uncountability of $C$, it will also be useful to observe that each $x \in [0, 1]$ in Cantor set has a base-3 expansion $x = \sum_{j=1}^\infty a_j 3^{-j}$ with $a_j = 0, 2$. Likewise, we can assign a 2-base representation to the removed interval while constructing Cantor set.

In $n$-th step of removal, note that total $\sum_{j=1}^n 2^{j - 1} = 2^n - 1$ number of open intervals are removed. Then, we can assign a number $a_n = 1, 2, \dots 2^n - 1$ from the front. For instance, $\frac{1}{2}$ on the interval $( \frac{1}{3}, \frac{2}{3})$, to be $\frac{1}{4}$ on $(\frac{1}{9}, \frac{2}{9})$, to be $\frac{3}{4}$ on $(\frac{7}{9}, \frac{8}{9})$, and so forth. Let $f_0: C^c \to [0, 1]$ be such a function.

Then, we extend $f_0: C \to [0, 1]$ to a function $f: [0, 1] \to [0, 1]$ by

\[\begin{aligned} f (x) = \begin{cases} \text{inf } \left\{ f_0 (y) \mid y \geq x, y \neq C \right\} & \text{ if } x < 1\\ 1 & \text{ if } x = 1\\ \end{cases} \end{aligned}\]

And this definition implies that $f = f_0$ on the complement of the Cantor set.

Since $f$ is increasing, so it has only jump discontinuities. And if it has a jump discontinuity, there will be a rational of the form $k / 2^n$ with $k \leq 2^n$ that is not in the range of $f$. But by the construction, each of the values \(\{ k/2^n \mid n \geq 0, k \leq 2^n \}\) is taken by $f_0$ for some point in the complement of $C$, and so is taken by $f$. The only way this can happen is if $f$ is continuous. And this function $f$ is called Cantor-Lebesgue function or sometimes simply the Cantor function

$\mathbf{Fig\ 2.}$ Cantor function $f$

Generalized Cantor set

The construction of the Cantor set by starting with $[0, 1]$ and successively removing open middle thirds of intervals can be obviously generalized as follows.

For example, instead of removing the middle third at the first stage, remove that middle fourth $(\frac{3}{8}, \frac{5}{8})$. On each of the two intervals that remain, remove the middle sixtennths. On each of the four intervals that remain, remove the middle interval of length $\frac{1}{64}$, and so on. Then the total length of intervals that we remove is

\[\begin{aligned} \sum_{n=1}^\infty 2^{n - 1} \times \frac{1}{4^n} = \frac{1}{2}. \end{aligned}\]

But, again the remaining set contains no intervals, is closed and uncountable, every point is a limit point, and has measure $1/2$. Of course, other choices than $\frac{1}{4}$, $\frac{1}{16}$, etc. are possible.

Let $I$ is a bounded interval, not $[0, 1]$, and $\alpha \in (0, 1)$. Then, let us call the open interval with the middle length equal to $\alpha$ times the length of $I$ the open middle $\alpha$-th of $I$. If \(\{ \alpha_j \}_{j=1}^\infty\) is any sequence of numbers in $(0, 1)$, let us define a decreasing sequence \(\{ F_j \}\) of closed set as

\[\begin{aligned} & F_0 = [0, 1] \\ & \quad \vdots \\ & F_j = F_{j-1} - (\text{open middle } \alpha_j \text{th from each of the intervals that make up } F_{j - 1}) \end{aligned}\]

The resulting limiting set $F = \bigcap_{j=1}^\infty F_{j}$ is called a generalized Cantor set.

Non-Borel but Lebesgue measurable set

With Cantor function $f$, we can construct an example that shows not all Lebesgue measurable set is Borel measurable.

Let $f: [0, 1] \to [0, 1]$ be the Cantor function, and let $g(x) = f(x) + x$.

First of all, it is obvious that $g$ is a bijection from $[0, 1]$ to $[0, 2]$ and $h = g^{-1}$ is continuous from $[0, 2]$ to $[0, 1]$.

$\mathbf{Proof.}$

Since $f$ is monotone, $g(x) = f(x) + x$ is strictly monotone from $[0, 1]$ to $[0, 2]$ as

\[\begin{aligned} x < y \Rightarrow g(y) - g(x) = f(y) - f(x) + (y - x) > 0 \end{aligned}\]

which shows $g$ is bijective.

Note that for metric space $E$ and $E^\prime$, if $f : E \to E^\prime$ is a continuous function, $E$ is compact, and $f$ is bijective then $f^{-1}: E^\prime \to E$ is continuous:

Let $F$ be a closed set in $E$. Since $E$ is compact, $F$ is compact. hence $f(F)$ is compact and hence closed. Since $f$ is continuous, therefore $g$ is continuous, and since $g$ is bijection, we show that $g^{-1}$ is continuous.

\[\tag*{$\blacksquare$}\]

Next, we claim that $m(g(C)) = 1$.

$\mathbf{Proof.}$

Let $K = [0, 1] - C$. From the construction of Cantor set, $K$ be written as the countable union of disjoint intervals, i.e. $K = \bigcup_{n=1}^\infty I_n$. Why? It is because, each of the open intervals removed in the construction of the middle-thirds Cantor set must contain a rational number, and these intervals are pairwise disjoint, so each must contain a different rational number. There are only countably many rational numbers, so there can be only countably many of these intervals.

Also, by construction of Cantor function $f$, it is clear that $g(I_n) = c_n + I_n$ for some constant $c_n$. Thus $m(g(I_n)) = m(I_n)$. It follows that

\[\begin{aligned} 2 = m (g ([0, 1]) = [0, 2]) = m (g(K) \cup g(C)) = m(g(K)) + m(g(C)) = \sum_{n=1}^\infty m(I_n) + m(g(C)) \end{aligned}\]

From $[0, 1] = K \cup C$ and $m(C) = 0$:

\[\begin{aligned} 1 = m ([0, 1]) = m(C) + m(K) = \sum_{n=1}^\infty m(I_n) \end{aligned}\]

Thus, we see that $m(g(C)) = 1$.

\[\tag*{$\blacksquare$}\]

Since $m(g(C)) > 0$, from the previous post we know that there exists a non-measurable set $A$ contained in $g(C)$.

Let $B = h(A) = g^{-1} (A) \subset g^{-1}(g(C)) = C$. Since $C$ has measure $0$, so does $B$ and since Lebesgue measure $m$ is complete, $B$ is Lebesgue measurable.

But $B$ is not Borel measurable. Suppose $B$ is a Borel set. Since $g^{-1}$ is continuous, it is measurable. (We will discuss the concept of measurability of function in the next post.) Thus $(g^{-1})^{-1} (B) = A$ should be measurable, a contradiction.

Reference

[1] Richard F. Bass, Real Analysis for Graduate Students, Version 4.3
[2] Wikipedia, Cantor set
[3] Wikipedia, Cantor function
[4] Show that every point of cantor set is a limit point