Show that if A is a measurable set then there is a Gδ (a countable intersection of open sets) set G with A ⊂ G and µ ∗ (G − A) = 0.
Where $\mu^{*}$ denotes outer measure.
I understand that since A is countable, its outer measure is 0, but I'm not sure why that would make the outer measure of G-A 0
Let $A_N=A \cap [-N,N]$. For each $k$ there is an open set $U_{N,k}$ such that $A_N \subset U_{N,k}$ and $m(U_{N,k}\setminus A_N) <\frac 1 {2^{N}k}$. Now $m(\bigcup_N U_{N,k}\setminus A) \leq \sum_N \frac 1 {2^{N}k}= \frac 1 k$. Let $U_k=\bigcup_N U_{N,k}$ The $U$ is open, $A \subset U$ and $m(U_k \setminus A) \leq \frac 1 k$. Finally, $G=\bigcap_k U_k$ is a $G_{\delta}$ set containing $A$ such that $m(G\setminus A)=0$.