# Question on the proof that the Cantor set has measure zero

I was reading the proof that the Cantor set has measure zero here. It all made sense until the very last step. Basically, the proof goes like this:

1. First show that the total lengths of all the removed intervals converge to $1$.

2. Show that as the geometric series of the sum converges to 1, at step N of the removal process, we can always cover the set with the intervals that have yet to be removed.

3. The difference between the total lengths of all the intervals that are meant to be removed vs the intervals that have already been removed can be made arbitrarily small (by the convergence of the geometric series).
4. We can then cover whatever is in the cantor set by the intervals that have yet to be removed, which can be made arbitrarily small.

Question on 4: How do we know that we can actually cover the cantor set with the intervals that have yet to be removed? I know that the lengths sum up to $1$, but the Cantor set is not empty. Even after you remove all the intervals, there's still something left! Isn't there a contradiction?

• I personally wouldn't even bother with the geometric series part - just say that the intervals left after step $n$ have total measure $(2/3)^n$, and the Cantor set is contained in that union of intervals so its measure is at most $(2/3)^n$ for each $n$. – Daniel Schepler Apr 26 '18 at 0:12
• Seems that everything after 1. can be dispensed with. If $[0,1] \setminus C$ has measure $1$, then of course $C$ has measure $0$. – grndl Apr 26 '18 at 0:24

Fortunately, as aduh pointed out in the comments, the whole business of covering the Cantor set is unnecessary here. From Step 1 we can conclude that the complement of the Cantor set has measure $1$. Since $[0,1]$ has measure $1$, and the union of the Cantor set and its complement is $[0,1]$, the Cantor set must have measure $0$.