# Category Archives: Measure Theory

## The Cantor Set

This is a post regarding the basic properties of the Cantor set.

We start with the definition: The Cantor set is obtained by first constructing a sequence $(C_n)$ of closed sets and then taking the intersection of the sets in this sequence. The sequence is constructed as follows:

1. Start with $[0,1]$ and remove the middle third, i.e. the interval $(\frac{1}{3},\frac{2}{3})$ to obtain the set $C_1=[0,\frac{1}{3}]\cup[\frac{2}{3},1]$. So $C_1$ contains $2$ disjoint closed intervals each of length $\frac{1}{3}$.

2. Next remove the middle third of each of these two intervals leaving $C_2=[0,\frac{1}{9}]\cup[\frac{2}{9},\frac{1}{3}]\cup[\frac{2}{3},\frac{7}{9}]\cup[\frac{8}{9},1]$ consisting of $2^2$ disjoint closed intervals each of length $\frac{1}{3^2}$.

3. Assuming $C_n$ has been constructed and consists of $2^n$ disjoint closed intervals each of length $\frac{1}{3^n}$, remove the middle thirds of all these intervals to obtain $C_{n+1}$ consisting of $2^{n+1}$ disjoint closed intervals each of length $\frac{1}{3^{n+1}}$.

By induction we have our sequence $(C_n)$.

Now, we define the Cantor set $C$ as $C=\cap_{n=1}^\infty C_n$.

We now describe the properties of the Cantor set:

1. In the usual topology on $\mathbb{R}$ the Cantor set is closed (being an intersection of closed sets). Moreover since it is bounded so by the Heine Borel theorem it is compact. What’s extraordinary is that every compact metric space is a continuous image of the Cantor set! The proof may be found here.

2. The Cantor set is uncountable. We prove this by considering the ternary expansion of all numbers in $[0,1]$. All such numbers may have the form $0.a_1a_2\cdots$ with $a_i\in\{0,1,2\}$ (including $1=0.22\cdots$). We claim that $x\in C$ iff $x$ has a ternary expansion of the form $0.a_1a_2\cdots$ with $a_i\in\{0,2\}$.

To prove this consider the construction of $C$ through the ternary lens: In ternary the construction of $C_1$ involves removing all numbers in $(0.011\cdots,0.2)$ from $[0.00\cdots,0.22\cdots]$. This shows that in $C_1$ every number has a ternary expansion of the form $0.a_1a_2\cdots$ with $a_1\in\{0,2\}$ and $a_i\in\{0,1,2\}\forall i>1$. Conversely every number of this form is in $C_1$. Likewise the construction of $C_2$ involves removing all numbers in $(0.0011\cdots,0.02)$ and in $(0.2011\cdots,0.22)$ from $C_1$. This shows that in $C_2$ every number has a ternary expansion of the form $0.a_1a_2\cdots$ with $a_1,a_2\in\{0,2\}$ and $a_i\in\{0,1,2\}\forall i>2$. Again conversely every number of this form is in $C_2$. Continuing in this way we can conclude that $x\in C_n$ iff $x=0.a_1\cdots a_na_{n+1}a_{n+2}\cdots$ with $a_1\cdots a_n\in\{0,2\}$ and $a_i\in\{0,1,2\}\forall i>n$. Hence by definition of $C$ our claim is established.

Now assume that $C$ is countable and has been enumerated in the list described below:

$0.d_1^{(1)}d_2^{(1)}d_3^{(1)}\cdots$
$0.d_1^{(2)}d_2^{(2)}d_3^{(2)}\cdots$
$0.d_1^{(3)}d_2^{(3)}d_3^{(3)}\cdots$
$\cdots$

Here each $d_i^{(j)}\in \{0,2\}$.

Now let $x=0.a_1a_2a_3\cdots$ where $a_i=0$ if $d_i^{(i)}=2$ and $a_i=2$ otherwise. Then $x\notin C$ despite being of the requisite ternary form. Hence $C$ is uncountable.

3. The Cantor set has Lebesgue measure zero. We define a set $A$ to be of Lebesgue measure zero if $\forall \epsilon>0\exists$ a sequence of intervals $(I_n)$ such that $A\subset\cup_{n=1}^\infty I_n$ and $\sum_{n=1}^\infty l(I_n)<\epsilon$. Here if the end points of $I_n$ are $a_n$ and $b_n$ (with $a_n\le b_n)$ then $l(I_n)=b_n-a_n$. (The proof that this definition is consistent with the Lebesgue measure may be found in this book.) Now given $\epsilon>0$ choose $n$ so that $(\frac{2}{3})^n<\epsilon$. Since $C\subset C_n$ and the total length of $C_n$ is $(\frac{2}{3})^n$, so it is clear that $C$ is a set of Lebesgue measure zero. Together with point 2, we see that this establishes the existence of uncountable sets of zero measure!