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

LordOfNumbers
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  • 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. – Alex Ravsky Feb 01 '24 at 12:30

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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 $x+A'\subset E$.

Alex Ravsky
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  • Hi there, can you please take a look at this question https://math.stackexchange.com/questions/4851938/revisited-question-lebesgue-measure-of-symmetric-difference-is-0? Thank you! – Hermi Feb 05 '24 at 06:34