Cantor-like subset on a set of positive measure

Given a set of positive Lebesgue measure, say $E$, with $m(E) = 1$, can you always construct a Cantor-like subset?

My motivation is that, if this were so, it would allow us to easily take a subset of arbitrary measure by adjusting the intervals removed (like the Smith-Volterra-Cantor set). Additionally, the subsets would be closed and have an empty interior.

• By Cantor-like, you mean homeomorphic to the standard Cantor set? – gary Apr 13 '15 at 21:21
• Probably? I have not actually taken a topology class. I would guess that the SVC set is homeomorphic to the Cantor set and both have obvious homeomorphic sets on arbitrary intervals $(a,b)$ or $[a,b]$ or even finite unions of such intervals. I just want to extend it to arbitrary sets of positive measure. So... probably. – Logan Apr 13 '15 at 21:25
• How do you intend to restrict the set and the measure? For instance, if $E$ is a one point set with the atomic measure then the answer to your question is evidently "no", but my guess is that you don't intend to consider such examples. – Lee Mosher Apr 13 '15 at 21:37
• @LeeMosher: I am dealing with Lebesgue measure. I should have mentioned that. – Logan Apr 13 '15 at 22:54

Presumably this is referring to Lebesgue measure on $\mathbb R$.
By inner regularity, any measurable set $E$ of positive measure contains a compact set $K$ of positive measure. You can then inductively construct a sequence of sets $E_n$, each of which will be the union of $2^n$ disjoint closed intervals of length $\le 1/3^n$, and the intersection of $K$ with each of these intervals will have positive measure. $E_0$ is a closed interval of length $1$ whose intersection with $K$ has positive measure; given $E_n$, to find $E_{n+1}$ replace each of the intervals in $E_n$ by two disjoint subintervals, each of length $\le 1/3^n$, whose intersection with $K$ has positive measure.
Then the intersection of the $E_n$ is a compact subset of $K$ homeomorphic to the Cantor set.