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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?

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  • $\begingroup$ 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. $\endgroup$
    – Gerry Myerson
    Feb 27, 2013 at 2:17
  • $\begingroup$ If you want a proof that $C$ has measure zero, see math.stackexchange.com/questions/145803/…? $\endgroup$
    – Gerry Myerson
    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

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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).

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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 $1$.

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    $\begingroup$ 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
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    $\begingroup$ @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

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