Doubt about Exercise 5, Chapter 1, Shakarchi [RA]

Question

So, I need to find an example of $E$ an open and bounded set, such that if $\mathcal{O}_n:=\{x \in E: d(x,E) < \frac {1}{n}\}$

Then $\lim_{n \to \infty} m(\mathcal{O}_n) \neq m(E)$. Where $m$ is the Lebesgue measure.

My problem is the following, if all the $\mathcal{O}_n$ are measurable sets (Lebesgue measurable), you can apply Corolary 3.3 from the book, in other words, if $\mathcal{O}_n$ are all measurable, then the equality holds.

Because we are taking the limit of $m(\mathcal{O}_n)$, we need infinite many $\mathcal{O}_n$ to be not measurable, the only example of a non-measurable set is a Vitali set. So, how can a Vitali be the $\mathcal{O}_n$ of a set?

Any help would be appreciated.

Answer 1

In $[0,1]$ let $C$ be a Cantor like set of positive measure and $E=C^{c}$. Then $m(O_n) \to m(\overline {E})\neq m(E)$ because $\overline {E}$ is the complement of the interior of $C$ and interior of $C$ is empty.

Answer 2

The idea is to make an assumption of Corollary 3.3 not satisfied. Let $\{x_n\}$ be an enumaration of $[0,1]\cap\mathbb{Q}$. Consider $$ E=\cup_{n=1}^{\infty} (x_n-\frac 1{4^n}, x_n+\frac1{4^n}).$$ This is an open set with Lebesgue measure at most $\sum 2/4^n = 2/3$.

Also, $E$ is dense in $[0,1]$. Then $m(O_n)\geq 1$ for each $n$.