5
$\begingroup$

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.

$\endgroup$

2 Answers 2

2
$\begingroup$

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.

$\endgroup$
6
  • 1
    $\begingroup$ 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, 2019 at 1:01
  • $\begingroup$ @BajoFondo Hint: $\bigcup_n O_n=\overline E$. $\endgroup$ Jul 11, 2019 at 1:20
  • $\begingroup$ @DavidC.Ullrich Isn't it for $\cap_n$? $\endgroup$ Jul 11, 2019 at 1:28
  • $\begingroup$ @BajoFondo That was a typo; I meant $\cap_n O_n=\overline E$. $\endgroup$ Jul 11, 2019 at 1:38
  • $\begingroup$ @i707107 Oops... thx. $\endgroup$ Jul 11, 2019 at 1:39
2
$\begingroup$

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 $n$.

$\endgroup$
2
  • 1
    $\begingroup$ $E$ is not bounded here. $\endgroup$
    – Bajo Fondo
    Jul 11, 2019 at 0:56
  • $\begingroup$ Good point!. To make $E$ bounded, take the rational numbers from $[0,1]$ and enumerate them as $\{x_n\}$. $\endgroup$ Jul 11, 2019 at 1:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.