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

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 $$A\subset \mathbb{R}^n$$?

• As far as I see, the referenced statement is not at the bottom of page $5$ but at the beginning of Section 1.3 at page 4. Commented Feb 1 at 12:30

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 $$x+A'\subset E$$.