Let $\{r_n\}^\infty_{n=1}$ be all rational numbers in $\mathbb{R}$, and
$$G = \bigcup^\infty_{n=1} \left( r_n-\frac{1}{n^2}, r_n+\frac{1}{n^2} \right)$$
How to prove that for any closed set $F$ in $\mathbb{R}$, $G \Delta F$ must have positive measure?
$G\triangle F=(G\setminus F)\cup(F\setminus G).$
If $G\not\subseteq F$ then $G\setminus F$ is a nonempty open set and has positive measure. So we can assume that $G\subseteq F.$ Since $G$ is dense in $\mathbb R$ and $F$ is closed, this means that $F=\mathbb R.$ So all we have to prove is that the set $G\triangle\mathbb R=\mathbb R\setminus G$ has positive measure. This follows from the fact that $\mathbb R$ has infinite measure while $G$ has finite measure, namely, $$\mu(G)\lt\sum_{n=1}^\infty\frac2{n^2}=\frac{\pi^2}3.$$