TITLE: Existence of a set $A\subset\mathbb{R}$ such that $|G\setminus A|=\infty$ for every open set $G\supset A$
QUESTION [5 upvotes]: 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 $A$

REPLY [2 votes]: 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 $$\infty=|A^c|\le | A^c\setminus G|+|A^c\cap G|=|A^c\cap G|=|G\setminus A|.$$