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 $\epsilon>0$?

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $$ \mu (O\setminus E)\leq \sum_n \mu(O_n \setminus E)\leq \sum_n \epsilon/2^n \leq \epsilon. $$

  • $\begingroup$ 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

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.