Proof that the Cantor set has measure zero.

Question

Consider a closed and bounded set $ F $ in the open interval $ (-n,n) $ of $ \mathbb{R} $. Then in the usual topology with Lebesgue measure $ \mu(F) = 2n - \mu((-n,n) \setminus F) $. The Cantor set satisfies this property so it can be our $ F $. That is, $ C $ is in $ [0,1] $ and can also be said to be in $ [-1,1] $. $ C $ is bounded because it has a supremum and infimum, in this case $ 1 $ and $ -1 $.

Write $ (-1,1) \setminus C = (-1,0) \cup [0,1] \setminus C $.

So $ \mu((-1,1) \setminus C) = 1 + \mu([0,1] \setminus C) $, and this apparently implies $ C $ has measure zero.

This proof was given to me and I am not sure where the errors are if there are any. Would anyone be able to explain or correct this proof?

Answer 1

That is not a proof, you use the fact that the Cantor set has a measure zero in order to prove that it has measure zero.

Your argument is essentially "Because $[0,1]\setminus C$ has measure $1$, the Cantor set has measure $0$", but you don't know that $[0,1]\setminus C$ has measure zero.

The correct proof would be to show that the Cantor set satisfies the definition of a measure zero set; or that it is a subset of a measure zero set; or that its complement in $[0,1]$ has measure $1$.

Answer 2

Look at the unit interval after the $n$th chopping. Let $C_n$ be the length of the set resulting from $n$ removal of the middle third. Then $$|C_n| = \left({2\over 3}\right)^n.$$ Now arrive at your conclusion. (I am using $|\cdot |$ for Lebesgue measure).