the cantor set is constructed by removing the middle part of the previous cantor sets. So $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$
$C_2 = [0, \frac{1}{9}] \cup [\frac{2}{9},\frac{1}{3}] \cup[\frac{2}{3},\frac{7}{9} ] \cup [\frac{8}{9},1]$
....
and $C = \cap C_n$
So far i've tried it like this:
Define $D_n = [0,1] \backslash C_n$ and $D = \cup D_n$.
Notice that $[0,1]\backslash C = D$
So what i've tried is constructing a increasing sequence $A_1 = D_1 \dots A_{j+1} = D_{j+1}\backslash D_{j}$, obviously $A_i$ are disjunct so i can use the continuity of measures for below. but now we see:
$\mu(D) = \lim_{j\rightarrow \infty}\mu(A_j)$
so we get :
$\lim_{j\rightarrow \infty}\mu(A_j) = \lim_{j\rightarrow \infty}\mu(D_{j}\backslash D_{j-1})$.
and here i'm a little bit stuck, anyone any thoughts on how i get to the magic number 1?
Kees
Since $C \subseteq C_n$ for all $n$, $\mu(C) \le \mu(C_n)$ for all $n$. Now since $\mu(C_n) = \left(\frac{2}{3}\right)^n$ for all $n$, we have $\mu(C) \le \left(\frac{2}{3}\right)^n \to 0$ as $n\to \infty$. Hence, $\mu(C) = 0$.