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I have to prove that

If $E \subset \mathbb R$ is measurable and $\mu ^*(E)< \infty$, then for each $\epsilon>0$, $\exists A \subseteq E$ such that $A$ is compact and $\mu^*(E \setminus A)< \epsilon$.

I know that if $E \subseteq \mathbb R$ is measurable, then for each $\epsilon>0$, $\exists C \subseteq E$ such that $C$ is closed and $\mu^*(E \setminus C)< \epsilon$.

So, I thought when $E$ has a finite outer measure, if I am able to prove $C$ is bounded then I am done but I cannot go any further. Can anyone give me a HINT, not an answer?

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    $\begingroup$ If $E$ happens to be bounded, then any closed set $C$ with $C \subseteq E$ is compact, so you are done by the statement you said you know. Can you think of a way to reduce to the bounded case? $\endgroup$ – Ethan Alwaise Mar 18 at 8:35
  • $\begingroup$ @EthanAlwaise Exactly that is where I stuck, I thought if the measure of $E$ is finite then $E$ has to be bounded, but then I found this i.e $\bigcup_{n=1}^\infty (n-\frac{1}{2^{n+1}},n+\frac{1}{2^{n+1}})$ has positive outer measure but is unbounded. $\endgroup$ – thomson Mar 18 at 8:41
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First note that there is no need to use outer measure because all our sets are measurable. $C\cap [-N,N]$ is compact and it is contained in $E$. Also $\mu (E\setminus C) =\lim \mu^(E\setminus C \cap [-N,N])$. Hence $\mu (E\setminus C) <\epsilon$ for $N$ large enough.

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