Existence of a set $A\subset\mathbb{R}$ such that $|G\setminus A|=\infty$ for every open set $G\supset A$ I am trying to prove the following statement and I would like to have an hint about how to complete it, thanks:
"Prove that there exists a set $A\subset\mathbb{R}$ such that $|G\setminus A|=\infty$ for every open set $G$ that contains $A$."
What I have tried:
We know that if $A\subset\mathbb{R}$ is Lebesgue measurable iff for each $\varepsilon>0$, there exists an open set $G\supset A$ such that $|G\setminus A|<\varepsilon$ so the set $A$ we are looking for must be non Lebesgue measurable. Now, the only non Lebesgue measurable set I know of is the Vitali set $V$ which is a subset of $[-1,1]$ so although it must have a positive and finite outer measure $|V|$ does not have the property required.
I have thought about $\bigcup_{k=1}^{\infty}(r_k+V)$ (where the $r_k$s are an enumeration of the rationals) but this has infinite outer measure.
I have also thought about the set $\tilde V$ which contain one element from each of the sets $\{V+r:r\in\mathbb{Q}\}$ but one could find an open set $G$ containing it made up of $(-2,2)$ and then of intervals $I_k=(a_k-\frac{\varepsilon}{2^k},a_k+\frac{\varepsilon}{2^k})$ which has a total outer measure of $4$ so it doesn't work.
Thus I am guessing this set needs to contain an uncountable number of elements (to prevent $G$ from having finite measure) and containing ever bigger terms.
Is my intuition correct? I would appreciate any hint about how to find/build this set. Thanks

$|A|$ denotes the outer measure of the set $A$
 A: You are indeed right that your set $A$ need be non-measurable and uncountable (after all any countable set is measurable).
Let $S \subseteq [0,1]$ be a non-measurable set. By the criteria that you stated there must exist $\epsilon_0 >0$ such that $|U \setminus S| > \epsilon_0$ for any open set $U$ containing $S$. I will use the notation $S +n$ for the set $S$ but translated by $n$, i.e. $S+n = \{s + n:s \in S\}$. Note that $|S| = |S+n|$ by translation invariance and $S+n \subseteq [n,n+1]$. Moreover $S+n$ has the property that any open set $U$ containing $S+n$ satisfies $|U \setminus (S+n)| \ge \epsilon_0$.
Now consider
$S \cup (S+ 3)$.
I claim that $|U \setminus (S\cup (S+3))| \ge 2 \epsilon_0$ for any open set $U$ containing $S \cup (S+3)$. Indeed if $U$ is open and contains $S \cup (S+3)$ then since $S \subseteq [0,1]$ and $S+3 \subseteq [3,4]$ we have that $U \cap (-1,2)$ is an open set containing $S$ and $U \cap (2,5)$ is an open set containing $S+3$. Thus
$$|(U\cap(-1,2)) \setminus S| \ge \epsilon_0\quad\text{and}\quad|(U \cap (2,5)) \setminus (S+3)| \ge \epsilon_0$$
And since $S\subseteq (-1,2)$ and $S+3\subseteq (2,5)$ we have that
$$|U \setminus (S \cup (S+3))| \ge |(U\cap(-1,2)) \setminus S|+|(U \cap (2,5)) \setminus (S+3)| \ge 2\epsilon_0$$
Then consider $S \cup (S+3) \cup (S+6)$. A similar argument shows
$$|U\setminus (S \cup (S+3)\cup (S+6))| \ge 3 \epsilon_0$$
for any open set $U$ containing $S \cup (S+3)\cup (S+6)$.
Can you figure out the rest of the construction?
A: Let $A$ be a Bernstein set. That is, if $C$ is any uncountable closed set then $C\cap A\ne\emptyset\ne C\cap A^c.$ (Where $A^c=\Bbb R\setminus A.$)
So if $D$ is closed and  $D\subset A$ or $D\subset A^c$ then $D$ is countable so $|D|=0.$
Now $|A^c|=\infty.$ Because if $E$ is open and $E\supset A^c$ then $\Bbb R\setminus E=D=\overline D\subset A$ so $|D|=0$, hence by the finite sub-additivity of outer Lebesgue measure we have $\infty=|\Bbb R|=|E\cup D|\le |E|+|D|=|E|.$
So if $G$ is open and $G\supset A$ then $A^c\supset A^c\setminus G=\Bbb R\setminus G=\overline {\Bbb R\setminus G},$  so $|A^c\setminus G|=0,$ hence $$\infty=|A^c|\le | A^c\setminus G|+|A^c\cap G|=|A^c\cap G|=|G\setminus A|.$$
