Is it possible to construct a Cantor set in a measurable set such that the measure of the constructed Cantor is close to that of the original set?

Question

According to the Cantor–Bendixson theorem, it is possible to find a perfect set in a compact set of $R^n$ and construct a Cantor set in the perfect set. However, if the compact set has positive Lebesgue measure, can we still construct a Cantor set in it while keeping the measure of the constructed Cantor set close to the measure of the original compact set? My guess is Yes. I try to imitate the standard construction of the Fat Cantor set on the perfect set to make sure that its measure is close to that of the compact set. But I have difficulty proving after the construction, the resultant set is still a Cantor set. Is my guess right? Otherwise, how could I modify my process? Thanks.

Answer 1

If we define "Cantor set" to mean a perfect compact set with no interior, then the answer is yes. Suppose for simplicity we're in $\mathbb R,$ $K\subset [0,1]$ is compact, and $m(K)>0.$

Let $\epsilon>0.$ Then there is an open dense subset $U$ of $(0,1)$ such that $m(U)<\epsilon.$ We then have $K\setminus U$ compact and nowhere dense. Furthermore,

$$m(K\setminus U) \ge m(K)-m(U) > m(K)-\epsilon.$$

Now $K\setminus U = P\cup C,$ where $P$ is perfect and $C$ is countable (Cantor-Bendixson). Thus $m(K\setminus U) = m(P).$

So given $\epsilon >0,$ we have found a perfect nowhere dense subset $P$ of $K,$ such that $m(P)\ge m(K)-\epsilon.$