Let $\mathbb{R}^d=\Pi_{i=1}^d \mathbb{R},$ does there exist a covering $\{E_n:n\in \mathbb{Z}^d\}$ of subsets $E_n\subseteq \mathbb{R}^d$ that satisfies the following: $1)$ $ E_n$'s are pairwise disjoint, $2) $ for any $x \in E_n, y\in E_m, $ $x+y\in E_{n+m},$ and $3)$ $\sup_{n\in \mathbb{Z}^d}|E_n|<\infty?$ Note that I'm using $|M|$ to denote the Lebesgue measure of any measurable set $M\subseteq \mathbb{R}^d.$

The collections given by $\{n+[0,1)^{d} : n\in \mathbb{Z}^d \}$ easily comes to mind, but it doesn't seem to satisfy condition $2$ with some easy counterexample like $(1+0.5)+(2+0.5)=3+1\neq 3+q_3$ where $0\leq q_3<1$ in the case $d=1$.

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    $\begingroup$ the first equation looks nonsense to me $\endgroup$ – mathworker21 Dec 7 '18 at 9:26
  • $\begingroup$ Does the empty set cover $\mathbb{R}^d?$ $\endgroup$ – Kurome Dec 7 '18 at 9:27
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    $\begingroup$ I don't think so if the sets should be measurable. Say focus on the real line. Some $E_n$ must have positive measure. Then $E_{2n}$ contains $E_n+E_n$, so by Steinhaus theorem (sum version math.stackexchange.com/questions/86209/…) it also contains an interval. But then $E_{4n}, E_{6n}$ and so on must contain longer and longer intervals. I haven't thought the details through though. $\endgroup$ – Michal Adamaszek Dec 7 '18 at 9:40
  • $\begingroup$ Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals. $\endgroup$ – Kurome Dec 7 '18 at 10:00
  • $\begingroup$ No but if $U$ contains an interval of length $a$ and $V$ contains an interval of length $b$ then surely $U+V$ contains an interval of length $a+b$. $\endgroup$ – Michal Adamaszek Dec 7 '18 at 10:03

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