# Finding measure of Cantor set

Let us consider the following function designed to measure the length of subsets of $\mathbb R$:

$l^∗ : 2^{\mathbb R} → [0,+∞]$ satisﬁes

1. $l^∗(∅) = 0$,

2. $l^∗(S) \leq l^∗(T)$, if $S ⊆ T$,

3. $l^∗(\bigcup \limits_{i=1}^∞ S_i) \leq \sum \limits_{i=1}^∞ l^∗(S_i)$ for any $S_i ⊆\mathbb R$,

4. $l^∗([a,b]) = b − a$ for any $a \leq b$.

The Cantor set is deﬁned in the following way: Consider the interval [0,1] and remove its middle third to get $$A_1 =[0, 1/3] ∪ [2/3 ,1]$$

Then from each of the intervals remove their middle third to get $$A_2 =[0, 1 /9]∪[2/9 , 1/3]∪[2 /3 , 7 /9]∪[8/ 9 ,1]$$ If we continue in that way indeﬁnitely, on each subsequent step removing a middle third from each of the closed intervals remaining in the union, the set of remaining points is called the Cantor set: $$C = \bigcap \limits_{n=1}^{\infty} A_n$$ Find $l^∗(C)$, where $C$ is the Cantor set.

If $n=1$ then we would have $C=A_1=S_1 \cup S_2$ (say) which is $l^∗(\bigcup \limits_{i=1}^2 S_i) \leq \sum \limits_{i=1}^2 l^∗(S_i)=l^*(S_1)+l^*(S_2)=(1/3-0)+(1-2/3)=2/3$

So am I correct in saying if n=3 say, it would be the intersection of $A_1,A_2,A_3$ which would be equal to $A_3$ wouldn't it? Since that is the only set contained in the other sets.

Notice that $A_n$ is compound of $2^n$ intervals of length $\frac{1}{3^n}$ so by item $(3)$ we have that $l^*(A_n) \le \left(\frac{2}{3}\right)^n$ now since $C \subset A_n$ for any $n$ positive integer, by item $(2)$ follows that $l^*(C) \le l^*(A_n) \le \left(\frac{2}{3}\right)^n$ which means that $l^*(C) \le \left(\frac{2}{3}\right)^n$ for any $n$ so the measure of the Cantor set is...
• Yes that's right, so $l^*(C)=0$ – JulianP Jan 16 '16 at 18:37
• wait that is kinda confusing me. So the final answer would be $l^*(C) \geq 0$, as $n$ tends to ininity? – snowman Jan 16 '16 at 18:54