Are there Cantor sets of non-zero measure?

Question

A cantor set is generated by removing a centered open interval from $[0, 1]$ and repeating the process infinitely on the two leftover segments.

What if the first interval we remove is of length $r$, then we remove an interval of length $r^2$ from the two leftovers, $r^3$ from the four leftovers after that, and so on? Obviously we have $r<1$. Then the total amount removed is something like $\sum (2r)^n$ (give or take a few off-by-one errors), which can be made less than $1$ if $r$ is small enough, implying that what's left over (the Cantor set) is of non-zero measure.

(The only thing you have to worry about as far as I can see is if the process might involve trying to remove a segment of length $r^n$ from a segment of length $<r^n$. After a lot of calculation I think I've shown this is impossible if $r<\frac 1 3$)

Answer 1

Yes. These are called "Fat Cantor sets". And for any $x<1$ you can find a fat Cantor set in $[0,1]$ of measure $x$ (but you can't find one of measure $1$).

The construction is the same as the Cantor set, only you remove smaller open intervals. (See Wikipedia.)

Answer 2

Every Cantor sets defined by $C^{1+\epsilon}$, $\epsilon>0$, expanding maps have zero Lebesgue measure. This is a consequence of bounded distortion. This is not true if the Cantor set is defined for a $C^1$ expanding map.