The first part of the question says to show that the outer measure of $A$is 1. I am pretty sure that my proof is correct for this part. However, the next part of the question says to show that for every $\epsilon > 0$ there is a closed set $F$ with $F \subseteq A$ and The outer measure of $F$ is greater than $1-\epsilon$.
I know that somehow I have to construct an open set $G$ which only contains the rationals between 0 and 1 and that this open set must overlap with $A$ so that $F=[0,1]\setminus G$ and $|F|+\epsilon >1$. I'm stuck because every open set that intersects [0,1] is going to intersect it in an infinite number of points, both irrational and rational. Any guidance or hints for this problem would be greatly appreciated. Thanks.
Arrange the rationals in $[0,1]$ in sequence $(r_n)$ and consider $U=\bigcup_n (r_n-\frac {\epsilon} {2^{n+1}}, r_n+\frac {\epsilon} {2^{n+1}})$. Then the measure of $U$ is at most $\sum_n \frac {\epsilon} {2^{n+1}} \leq \frac {\epsilon} 2 <\epsilon$ Take $F=[0,1] \setminus U$. Since $U$ is open, $F$ is closed and the measure of $F$ is at least $1-\epsilon$. Since $U$ contains all rationals it follows that $F \subseteq A$.