"Fat" Cantor Set

Question

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.) — , Jan 27 '13 at 04:06
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 04:06
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness. — , Jan 27 '13 at 04:09
You can, however, have outer measure arbitrarily close to one. — Brian M. Scott, Jan 27 '13 at 04:10
Here are some details of what's contained on the Wikipedia page. — Martin, Jan 27 '13 at 04:13
So, is there no difference in value of the "outer measure" and "measure" when dealing with cantor sets? I've seen some sets with different measure and outer measure but it seem everyone here is using them interchangeably in reference to the cantor set. — Frank, Jan 27 '13 at 04:36

Michael Hardy · Answer 1 · 2014-01-09T03:56:11.210

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.

score 2 · Answer 2 · answered Mar 14 '22 at 11:10

The measure of a Cantor set cannot be 1, but certainly can be arbitrary close to it. Let the total measure of the pieces removed at step $n$ be equal $m_n$, so we remove an open intervalof length $m_1$ in the first step, then remove two open intervals of total length $m_2$ from the two pieces remaining after step 1. Then remove four open intervals of total length $m_3$ from the remaining four pieces left after step 2. Continuing this way, we remove open intervals with total measure $M=m_1+m_2+...$. The measure of the compact set $C$ left is $1-M$. If $M=1$, the measure of $C$ is $0$, if $M<1$ then the measure of $C$ is positive---it is fat.

"Fat" Cantor Set

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