# Proof that the Cantor set has measure zero.

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?

• That's nothing like a proof. It can't be, since it uses no properties of the Cantor set. It "proves" the measure of the interval $[0,1/2]$ is zero. Feb 27, 2013 at 2:17
• If you want a proof that $C$ has measure zero, see math.stackexchange.com/questions/145803/…? Feb 27, 2013 at 2:21
• Thank you, I will take a look. I was hoping to clarify why something like this, ostensibly from lecture notes, looked so strange.
– 114
Feb 27, 2013 at 2:27

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).
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 $1$.
• @Stopwatch: Not really, because your text really doesn't contain a proof. I suppose the simplest way would be to show that in the inductive definition of the Cantor set we remove open intervals whose aggregated lengths sum to $1$, therefore $[0,1]\setminus C$ must have measure $1$. Feb 27, 2013 at 2:25