# A necessary condition for measurability?

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?

• 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? – Ethan Alwaise Mar 18 at 8:35
• @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. – thomson Mar 18 at 8:41

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.