I'm learning about Measure Theory and need some help with this problem:
Let $0 < \alpha < 1$. We construct a set $C_\alpha$ (Cantor type) as follows: In the first step we remove from the interval $[0, 1]$ a "middle" open interval of length $(1 - \alpha)3^{-1}$. In the nth step we remove $2^{n-1}$ open intervals of length $(1 - \alpha)3^{-n}$. Find the Lebesgue measure of $C_\alpha$.
My work and thoughts:
If we remove the set $C_\alpha$ from the close interval $[0, 1]$ we are left with the union of pairwise disjoint intervals. In more traditionally formulaic notation, we can write:
$$[0, 1] \setminus C_\alpha = E_1 \cup E_2 \cup E_3 \cup \ldots$$
Therefore, taking the Lebesgue measure on both side of the preivous equality we get:
$$\mu \left([0, 1] \setminus C_\alpha \right) = \mu \left( \bigcup_{n=1}^{+\infty} E_n \right) = \sum_{n = 1}^{+\infty} \mu(E_n).$$
I need to find a way to express $\mu(E_n)$ and calculate the above series. If I can do so then the result is immediate since:
$$\mu(C_\alpha) = \mu([0, 1]) - \mu([0, 1]\setminus C_\alpha)$$
where $\mu([0, 1]) = 1$.
As to the way you construct $C_\alpha$, if you denote by $E_n$ (hopefully I'm not misinterpreting it) the removed part during the $n$th step, then clearly $$\mu(E_n)=\frac12(1-\alpha)\left(\frac23\right)^n$$ Hence $$\mu(C_\alpha)=\mu([0,1]\setminus\cup E_n)=1-\sum\mu(E_n)=1-(1-\alpha)=\alpha$$