The problem is : Let $E\subset R$ be an arbitrary measurable set. Then, $m^∗(E) = \sup$ { $m^∗(K): K ⊂ E, K :$ closed and bounded(which means it's compact) }
($m^∗$ is Lebesgue outer measure.)
I know how to solve this problem when $m^*(E) < \infty$ (Also, there was such a nice answer when $m^*(E) < \infty$) But, it was hard for me to solve this when the measure of E becomes infinite.(And, there was no idea I could find "$m^*(E) = \infty$")
I tried to get $m^*(E) = N $ and make N sufficiently big, so that it goes to infinity. But, since all $K$ is bounded, I don't know if this idea is right.
Could you give me a nicer solution or good tip to solve in infinity?
Suppose $m^{*}(E)=\infty$. Then, given any $M>0$ there exists $N$ such that $m^{*}(E \cap [-N,N]) >M$. There is a closed and bounded set $K \subseteq E \cap [-N,N]$ such that $m^{*} (K) >m^{*} (E \cap [-N,N])-\epsilon$. Can you finish?