# Partitions of $\mathbb{R}^d$

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 $$d=1$$.

• the first equation looks nonsense to me – mathworker21 Dec 7 at 9:26
• Does the empty set cover $\mathbb{R}^d?$ – Kurome Dec 7 at 9:27
• 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. – Michal Adamaszek Dec 7 at 9:40
• Hmmm, the theorem doesn't seem to say anything about the sizes of the contained intervals. – Kurome Dec 7 at 10:00
• 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 $a+b$. – Michal Adamaszek Dec 7 at 10:03