Fat cantor set has positive lebesgue measure

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.

• The measure of the intersection of a decreasing sequence of measurable sets, the first of which has finite measure, is the infimum of their measures. You can prove this as a corollary of the fact that the measure of an increasing union of measurable sets is the supremum of their measures, which is in turn not too hard to verify. – Andrés E. Caicedo Jun 16 '14 at 1:59
• @AndresCaicedo Does that theorem follow from the axioms of measure (monotonicity, shift invariance, countable additivity, etc.) or does it require consideration of interval coverings specific to lebesgue measure? – Mario Carneiro Jun 16 '14 at 2:12
• It is valid for all measures. – Andrés E. Caicedo Jun 16 '14 at 2:12
• – Will Jagy Jun 16 '14 at 3:02

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.