Does a set of full measure contain an affine copy of any countable set?

Question

In this paper here (beginning of Section 1.3 at page 4) it is stated that any measureable set $E\subset[0,1]$ with $\lambda(E)=1$ contains an affine copy of any countable set $A\subset \mathbb{R}$. Is the same true in multiple dimensions? I.e. does any measureable set $E\subset[0,1]^n$ with $\lambda^n(E)=1$ contain a copy of any countable set $A\subset \mathbb{R}^n$?

Answer 1

Let $n$ be any natural number. Note first that if an affine copy of a set $A\subset\mathbb R^n$ is contained in $[0,1]^n$ then the set $A$ is bounded. On the other hands, let $A$ be any countable bounded subset of $\mathbb R^n$ and $E$ be any measurable subset of $[0,1]^n$ such that the Lebesgue measure $\lambda(E)$ equals $1$. Let $A'\subset [0,1/2]^n$ be any affine copy of $A$ and $x\in (E\cap [0,1/2]^n)\setminus (([0,1]^n\setminus E)-A')$ be any point. Then $x+A'\subset E$.