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

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.

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.

• So.. is my statement above wrong, or are the $O_n$ not measurable sets? Why does $m(O_n) \to m(\overline{E})$ ? Jul 11, 2019 at 1:01
• @BajoFondo Hint: $\bigcup_n O_n=\overline E$. Jul 11, 2019 at 1:20
• @DavidC.Ullrich Isn't it for $\cap_n$? Jul 11, 2019 at 1:28
• @BajoFondo That was a typo; I meant $\cap_n O_n=\overline E$. Jul 11, 2019 at 1:38
• @i707107 Oops... thx. Jul 11, 2019 at 1:39

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 $$n$$.

• $E$ is not bounded here. Jul 11, 2019 at 0:56
• Good point!. To make $E$ bounded, take the rational numbers from $[0,1]$ and enumerate them as $\{x_n\}$. Jul 11, 2019 at 1:10