There exists a set $A\subset\mathbb{R}$ with $|G-A|=\infty$ for each open set containing $A$. I happened upon an exercise in my book which tasks the reader with proving: "There exists a set $A\subset\mathbb{R}$ with $|G-A|=\infty$ for each open set containing $A$." Here the bars represent outer measure. I know that such an $A$ is not Lebesgue measurable, but I've not had much success in proving this result. This is an interesting and unexpected fact, so I'm pretty curious how one would approach this sort of problem.
 A: Let $X$ be a subset of $(0,1)$ which is not (Lebesgue) measurable. Let $B$ be a measurable set (e.g. a $G_\delta$ set) such that $X\subseteq B$ and $|B|=|X|$.
Claim 1. $|B\setminus X|\gt0$.
Proof. If $|B\setminus X|=0$, then $B\setminus X$ is measurable, and so is $B\setminus(B\setminus X)=X$.
Let $\varepsilon=|B\setminus X|\gt0$.
Claim 2. If $G$ is measurable and $X\subseteq G$, then $|G\setminus X|\ge\varepsilon$.
Proof. Since $X\subseteq B\cap G\subseteq B$ we have $|X|\le|B\cap G|\le|B|=|X|$, so $|B\cap G|=|B|$ and $|B\setminus G|=|B|-|B\cap G|=0$. Since $B\setminus X\subseteq(B\setminus G)\cup(G\setminus X)$, it follows that
$$\varepsilon=|B\setminus X|\le|B\setminus G|+|G\setminus X|=0+|G\setminus X|=|G\setminus X|.$$
Claim 3. If $G$ is measurable and $X+n\subseteq G$, then $|G\setminus(X+n)|\ge\varepsilon$.
Proof. Lebesgue measure and outer measure are translation invariant. Details left to the reader.
Let $A=X+\mathbb Z=\bigcup_{n\in\mathbb Z}(X+n)$.
Claim 4. If $G$ is measurable and $A\subseteq G$ then $|G\setminus A|=\infty$.
Proof.
$$|G\setminus A|=\sum_{n\in\mathbb Z}|(G\setminus A)\cap(n,n+1)|=\sum_{n\in\mathbb Z}|[G\cap(n,n+1)]\setminus(X+n)|\ge\sum_{n\in\mathbb Z}\varepsilon=\infty.$$
