Let $L^n(A)$ denote the $n$-dimensional Lebesgue measure of a set $A \subset \mathbb R^n$. Suppose that $L^n(A)>0$. Is it true that the closure of $A$ contains an open set?
Asked
Active
Viewed 448 times
3
-
3No, any fat Cantor set provides a counterexample. I'm sure this is a duplicate, but I will wait for someone with more experience in the measure-theory tag to judge which one is best. – KReiser Apr 04 '22 at 18:00
2 Answers
4
This is not true. Consider the fat Cantor set (see https://en.wikipedia.org/wiki/Smith–Volterra–Cantor_set). It is closed and has positive Lebesgue measure but contains no intervals. Thus, it cannot contain an open set since every open set contains an interval.
almosteverywhere
- 2,193
1
Another example: let $\{q_n\}_{n\ge 1} \subset \mathbb{R}$ be the set of rational numbers. For each $q_n$, take a neighborhood of $q_n$ of measure $1/2^n$ and denote it by $V_n$. Then $V= \cup V_n$ is an open set of measure at most $1$, and therefore, $\mathbb{R} - V$ is a closed set of infinite measure, which has empty interior since it is closed and dones't contain any rational number.
gaoqiang
- 370
-
Also worth noting is that the closure of a measure zero set can have positive measure (the rationals). This suggests that the combined notion of a set whose closure has measure zero might be especially small, and in fact this is the case. See this 1 May 2000 sci.math post and my 14 July 2014 answer to Jordan measure zero discontinuities a necessary condition for integrability. – Dave L. Renfro Dec 02 '24 at 12:21