# Help with the middle-quarter Cantor set--show that it has measure zero

I am stuck with this problem-- it involves the middle fourths cantor set. I am having trouble constructing this set. It is formed by removing the middle quarter from $[0,1]$, then removing the middle quarter from each of the remaining two intervals and so on.

A quick question: is this what they call the fat cantor set?

If we have to prove that the set has measure zero, then we add up the chunks that we remove and show that they add up to $1$, but here is what I have:

$\frac{1}{4}$ + $2 \frac{3}{32}$ + $...$

Sadly, I could not make them into a geometric series that added to 1, which made me realize I am going about the process the wrong way.

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What is the general term of the length you remove? If you want to sum a geometric series, you need this. – Ross Millikan Feb 4 '13 at 5:01
That is where I am facing the issue: I cannot find a general term for the length I am removing. It has been very hard for me to computationally visualize the construction of the first few terms. – user43901 Feb 4 '13 at 5:18
At the first step you remove the central quarter, leaving two segments of length $\frac 38$. You get this, as shown by the second term in your sum. Now go one step farther. As you say, you remove two segments of length $\frac 3{32}$. That leaves you four segments of what length? At each step you will have $2^n$ segments of some length and remove $2^n$ segments of length $2^{n+2}$ of that length. – Ross Millikan Feb 4 '13 at 5:32

Say you develop this cantor set in stages, $C_n$, with $C_0 = [0,1]$. It's not hard to see that $\lambda C_n = \frac{3}{4} \lambda C_{n-1}$, and the measure therefore goes to zero. So, the answer is no, this is a "skinny" cantor set.
If you want a fat one, you need to vary the relative measure you're carving out. Think about products of the form $\prod_{n = 1}^{\infty} (1 - \alpha_n)$; what conditions on $\{\alpha_n\}$ do you need to ensure this product converges to a positive value?