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I have to prove two things.

First is that the Cantor set has a lebesgue measure of 0. If we regard the supersets $C_n$, where $C_0 = [0,1]$, $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$ and so on. Each containig interals of length $3^{-n}$ and by construction there are $2^n$ such intervals. The lebesgue measure of each such interval is $\lambda ( [x, x + 3^{-n}]) = 3^{-n}$ therefore the measure of $C_n$ is $\frac{2^n}{3^n} = e^{(\ln(2)-\ln(3)) n }$ which goes to zero with $n \rightarrow \infty$. But does this prove it?

The other thing I have to prove is that the Cantor set is uncountable. I found that I should contruct a surjectiv function to $[0,1]$. But im totaly puzzeld how to do this.

Thanks for help

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up vote 5 down vote accepted

Hint: - There is a theorem that if $C_n$ is a descending sequence of measurable sets, $C = \bigcap C_n$ then $\lim_{n \rightarrow \infty}m(C_n)= m(C)$. Here, you know $C_n$ is a descending sequence, $m(C_n) $= (2/3)^n, and you want to know m(C)

  • to prove $C$ is uncountable:

Express a number $x$ between $[0,1]$ in base $3$:

$x =0.x_1x_2x_3...$

In the 1st step, we remove the middle third from $[0,1]$

We express 0=.0, $\frac{1}{3} = .1, \frac{2}{3} = .2$, 1= .222222...

We have 3 intervals: [.0 , .1] , (.1, .2), [.2 , .22222...] and we remove the middle interval. The removed interval (.1 , .2) consists of all numbers with $x_1$ = 1, except the endpoint $.1$ of [.0 , .1 ], however, we can express .1 as .02222.....

So we can use the rule: whenever we have a number of form 0.x1(x is a sequence consisting of 0,1,2) as the end point of an interval, we express as 0.x0222....So in this step, we remove all numbers with $x_1$ = 1. The remaining intervals are [.0, .0222...] and [.2, .222...]

Similarly, we can prove that in the n-th step, we keep only those numbers with $x_n$ = 0 or 2:

So the Cantor set contains of all numbers of the form $.x_1x_2..$ with $x_i =0 $ or 2.

There exists a bijection between $E$ and $[0,1]$: If you consider the new set E, with each member of E is a member of the Cantor set with every digit is divided by 2. E consists of all sequence with $x_i$ is 0 or 1. $|E| = |C|$

There exists an injective map from $[0,1]$ to this new set E. So you can prove it's uncountable.

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Thanks this helps. Do you know a name of the theorem you mentioned at the beginning? Don't think we proved it our lectures. – Haatschii May 16 '12 at 20:07
It's theorem 1.19e in Rudin's Real and Complex Analysis. It's an important theorem, I saw it used in couple of places. It's not hard to prove. the way to prove it is: let $B_1 = C_1, B_2 = C_2\setminus C_1, B_3 = C_3\setminus (C_1 \cup C_2 )$,...hence $B_1,B_2,B_3,...$ are pairwise disjoint, and you can use the property of measure of union of disjoint sets. – Long Mai May 16 '12 at 20:34

You may notice that the Cantor set is homeomorphic (which is even more than required) to $\{0,1\}^{\mathbb{N}}$, which is uncountable. A homeomorphism is established e.g. with $f: \{0,1\}^{\mathbb{N}}\to C$ so that \begin{align*} f((x_{i})_{i=1}^{\infty})=\sum_{i=1}^{\infty}\frac{2x_{i}}{3^{i}}. \end{align*} Since it has a converging geometric serie as a majorant; it definitely converges and even uniformly. Since the components are continuous then $f$ is continuous. You may conclude injectivity by noticing that $|f(x)-f(y)|>0$ if $x\neq y$. Surjectivity is also quite straightforward from the definition since you may produce any element of the cantor set with such a serie by choosing $x_{i}$'s wisely (if $y\in C$, choose $x_{i}=0$ if at $i$:th step $y$ is located in the left part of the $i-1$:th step interval being split to thirds, and $x_{i}=1$ if on the right). Since every continuous bijection from a compact set to a Hausdorff space is homeomorphism, we conclude that $f$ is a homeomorphism.

Since every $C_{n}$ has Lebesgue measure of $(\frac{2}{3})^{n}$ (at each step the measure of the previous step is multiplied with $\frac{2}{3}$) and $\frac{2}{3}<1$, it follows by convergence of measure that \begin{align*} m_{1}(C)=m_{1}(\bigcap_{n=1}^{\infty}C_{n})=\lim_{n\to\infty}m_{1}(C_{n})=\lim_{n\to\infty}(\frac{2}{3})^{n}=0. \end{align*}

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