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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.

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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.

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    $\begingroup$ So.. is my statement above wrong, or are the $O_n$ not measurable sets? Why does $m(O_n) \to m(\overline{E})$ ? $\endgroup$ – Bajo Fondo Jul 11 '19 at 1:01
  • $\begingroup$ @BajoFondo Hint: $\bigcup_n O_n=\overline E$. $\endgroup$ – David C. Ullrich Jul 11 '19 at 1:20
  • $\begingroup$ @DavidC.Ullrich Isn't it for $\cap_n$? $\endgroup$ – Sungjin Kim Jul 11 '19 at 1:28
  • $\begingroup$ @BajoFondo That was a typo; I meant $\cap_n O_n=\overline E$. $\endgroup$ – David C. Ullrich Jul 11 '19 at 1:38
  • $\begingroup$ @i707107 Oops... thx. $\endgroup$ – David C. Ullrich Jul 11 '19 at 1:39
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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 $n$.

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    $\begingroup$ $E$ is not bounded here. $\endgroup$ – Bajo Fondo Jul 11 '19 at 0:56
  • $\begingroup$ Good point!. To make $E$ bounded, take the rational numbers from $[0,1]$ and enumerate them as $\{x_n\}$. $\endgroup$ – Sungjin Kim Jul 11 '19 at 1:10

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