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

Say you delete the middle third.

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.


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