If $K\subset\mathbb{R}$ is compact and has positive Lebesgue measure, is the interior of $K$ non-empty?

Question

Let $\mathbb R$ be the field of real numbers, $\mu$ the Lebesgue measure on it. Let $K$ be a compact subset of $\mathbb R$.

Is the following assertion true?

If $\mu(K) \gt 0$, then the interior of $K$ is non-empty.

This seems to be false, but I was unable to construct a counter-example.

Answer 1

The Smith-Volterra-Cantor set fits the bill. It is a closed subset of $[0, 1]$ (so it compact), with empty interior and measure $\frac{1}{2}$. More generally, you can modify the construction of the Cantor set to obtain so-called 'fat Cantor sets' which have the desired properties.

Answer 2

Let $q_1, q_2, \dots $ be the rationals. Define $U= \cup_{n=1}^\infty(q_n - 1/2^n,q_n + 1/2^n).$ Then $U$ is open, $U$ is dense in $\mathbb {R},$ and $m(U)\le 2.$ It follows that $m([0,3]\setminus U) \ge 3 - m(U)\ge 3-2 =1.$ But $[0,3]\setminus U$ is compact, and contains no rational, hence has empty interior. So $[0,3]\setminus U$ does the job.

Answer 3

Inasmuch as every set with positive Lebesgue measure contains a compact set with positive Lebesgue measure, it's enough to find any set with positive Lebesgue measure and empty interior, say, the set of all irrational numbers.