“Fat” Cantor Set

So the standard Cantor set has an outer measure equal to $$0$$, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of $$1$$. I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of $$0$$ as well...

Are there other constraints that need to be made in order to accomplish this?

• The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.) – user856 Jan 27 '13 at 4:06
• If the bits you remove at each stage have total length less than $1$, then what's left has positive measure. – Gerry Myerson Jan 27 '13 at 4:06
• You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness. – user53153 Jan 27 '13 at 4:09
• You can, however, have outer measure arbitrarily close to one. – Brian M. Scott Jan 27 '13 at 4:10
• Here are some details of what's contained on the Wikipedia page. – Martin Jan 27 '13 at 4:13

Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on. The amount you delete is $\displaystyle\frac13+\frac16+\frac{1}{12}+\cdots= \frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.