Fat cantor set has positive lebesgue measure

Question

This is probably a duplicate of some other question, but it's not immediately obvious which. The fat cantor set is constructed by removing smaller fractions of the center in each stage of the cantor set construction, and it is claimed that this set has positive measure. To be precise, let $C_0=[0,1]$ and $C_n$ be the result of removing the central $1/2^{n+2}$-th of each interval in $C_{n-1}$, and let $C=\bigcap_{n=0}^\infty C_n$. Then $\mu(C_n)=\frac12(1+2^{-n})$, and as a countable intersection of measurable sets $C$ is measurable. Since each $C_n$ is a countable (finite, even) interval cover of $C$, we have $\mu(C)\le\frac12$. But how does one prove $\mu(C)\ge\frac12$? The only proof I have seen that bounds a measure away from zero is the proof that $\mu([a,b])=b-a$, and it's not a simple proof. What argument gives this lower bound on $\mu(C)$? Obviously I can't use monotonicity since $C$ has no interior.

Answer 1

Here you find the Monotone convergence theorem from Lebesgue. In wikipedia is explained as integral of functions, but it can be stated also like that:

Let $A_1, \cdots, A_n, \cdots$ be a decreasing sequence of measurable sets, such that $m(A_1) < \infty$. Then (a) $A = \cap_{n=1}^{\infty} A_n$ is measurable, and (b) $m(A) = \lim_{n\to \infty}m( A_n) $.

The proof of (a) is quite simple from the basic definition, and (b) from monotony and $\sigma$-subaditivity.

A direct proof of $\mu ( C ) \geq \frac{1}{2} $ can be given, then, by $\sigma$-subaditivity of $\mu$ : As $C_1 \subseteq C \cup \bigcup_{n=1}^{\infty} (C_n \setminus C_{n+1})$, then $\frac{1 + 2^{-1}}{2} \leq \mu(C) + \sum_{n=1}^{\infty} \frac{2^{-n} - 2^{-n-1}}{2} = \mu (C) + \frac{ 2^{-1}}{2} $ if i'm not mistaken.