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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?

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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 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 at 4:09
You can, however, have outer measure arbitrarily close to one. – Brian M. Scott Jan 27 at 4:10
Here are some details of what's contained on the Wikipedia page. – Martin Jan 27 at 4:13
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Say you delete the middle third.

Then delete middle sixths from the two remaining intervals.

Then delete middle twelfths from 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.

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