4
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Feb 27, 2013 at 2:17
  • $\begingroup$ If you want a proof that $C$ has measure zero, see math.stackexchange.com/questions/145803/…? $\endgroup$ Feb 27, 2013 at 2:21
  • $\begingroup$ Thank you, I will take a look. I was hoping to clarify why something like this, ostensibly from lecture notes, looked so strange. $\endgroup$
    – 114
    Feb 27, 2013 at 2:27

2 Answers 2

14
$\begingroup$

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).

$\endgroup$
2
$\begingroup$

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 $1$.

$\endgroup$
2
  • 1
    $\begingroup$ So it is circular then, thanks. Great, it was confusing me for the longest time. Would you happen to have a suggestion for a proof that stays as close to the language above as possible, minus the glaring errors? $\endgroup$
    – 114
    Feb 27, 2013 at 2:24
  • 1
    $\begingroup$ @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 $1$. $\endgroup$
    – Asaf Karagila
    Feb 27, 2013 at 2:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.