I have the following problem:
Let rational numbers $\mathbb{Q} = \left \{q_1,q_2,q_3,... \right \}$, $G=\bigcup_{n=1}^{\infty} (q_n-\frac{1}{n^2},q_n+\frac{1}{n^2})$. Show that for every closed set F, either $m(G \setminus F) > 0$ or $m(F \setminus G) > 0$ where $m(A)$ is Lebesgue measure of set A.
In the previous part of this problem, I have already shown that G is measurable and $m(G) < \infty$. Also, if $\mathbb{Q} \subseteq F$ where F is a closed set, $F = \mathbb{R}.$ Sets involved are subsets of $\mathbb{R}$.
I tried to attempt this problem as follows:
Case 1: $F = \mathbb{R}$. $m(F \setminus G) = m(\mathbb{R} \setminus G) = m(\mathbb{R}) - m(G)$ = $\infty - m(G)$ > 0 as $m(G) < \infty$
Case 2: $F \neq \mathbb{R}$. As if $\mathbb{Q} \subseteq F$, $F = \mathbb{R}$. We have $F \neq \mathbb{R}$, so $\mathbb{Q} \nsubseteq F$, which is eqivalent to $\exists q_i \in \mathbb{Q}$ such that $q_i \notin F$. But, $q_i \in G$, $G \nsubseteq F$. We have the following two subcases:
Case 2.1: $F \subset G$. $m(G \setminus F) = m (G) - m(F)$ as if $F \subset G$, $m(F) \leq m(G) < \infty$. But, I can only get $m (G) - m(F) \geq 0$ in this case as I understand that $F \subset G$ does not mean $m (F) < m(G), m(G) - m(F) > 0$ from Lebesgue measure of proper subset
Case 2.2: Otherwise, $m(G \setminus F) = m(G) - m(G \cap F)$. As $G \cap F \subseteq G$, $m(G \cap F) \leq m(G)$, $m(G) - m(G \cap F) \geq 0$. I cannot get $m(G) - m(G \cap F) > 0$.
I want to ask if my approach is incorrect? How can I solve this problem?
Note that $m(F \cup G) = m(F \setminus G) + m(F \cap G) + m(G \setminus F)$, as the latter three sets form a disjoint union of $F \cup G$.
So suppose both $m(F \setminus G)$ and $m(G \setminus F)$ are $0$, then $m(F \cup G) = m(F \cap G)$; can you rule this out?