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Following a question posted here: Approximating measures by open sets and compact sets.

I wanted to ask, if I am given a measurable set $E\subseteq \mathbb{R}$ s.t. $m(E)=\infty$, then how can I find an open set $O\supseteq E$ s.t. $m(O- E)<\epsilon$ for a given $\epsilon>0$?

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  • $\begingroup$ What's your definition of measurable? $\endgroup$ – T.J. Gaffney Dec 1 '15 at 17:15
  • $\begingroup$ Use caratheodory. A set $A$ is measurable iff $m*(A)=m*(A\cap E)+m*(A\cap E^c)$ for every subset $E\subseteq \mathbb{R}$. $\endgroup$ – User666x Dec 1 '15 at 17:17
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    $\begingroup$ It's not at all clear what it means to "find" such a set. It's a theorem that it exists... $\endgroup$ – David C. Ullrich Dec 1 '15 at 17:27
  • $\begingroup$ To show there exists such set $\endgroup$ – User666x Dec 2 '15 at 9:05
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Given the formulation of your question, I assume that you already know the claim for sets of finite measure.

Now, for $n\in \Bbb{N}$, there is an open set $O_n \subset (-n,n)$ with $O_n \supset E_n := E\cap(-n,n)$ and $\mu(O_n \setminus E) = \mu(O_n \setminus E_n)<\epsilon/2^n$. Here,I used $O_n \subset (-n,n)$ to get the first equality.

Now, $O:=\bigcup O_n$ is open with $O\supset E$ (why?) and $$ \mu (O\setminus E)\leq \sum_n \mu(O_n \setminus E)\leq \sum_n \epsilon/2^n \leq \epsilon. $$

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  • $\begingroup$ why is it important to assume $O_n\subset (-n,n)$? If we do not assume that, we have μ(On∖E)<μ(On∖En)<ϵ/2n and it is still valid $\endgroup$ – User666x Dec 2 '15 at 12:44
  • $\begingroup$ @User666x: You are right (if we replace "$<$" by "$\leq $"). I somehow thought we would need that property, but in fact we don't. $\endgroup$ – PhoemueX Dec 2 '15 at 13:48

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