Let $(U_n)_n$ be an arbitrary sequence of open subsets of the unit disk $D(0,1)\subseteq \mathbb{R}^2$ s.t. $\sum_{n=0}^\infty \lambda(U_n)=\infty$ (where $\lambda$ is the Lebesgue measure). Does there exist a sequence $(q_n)_n$ in $\mathbb{R}^2$ s.t. $D(0,1) \subseteq \bigcup_{n=0}^\infty (q_n+U_n)$?

With the notation $q_n+U_n$, I mean $$q_n+U_n:=\{x\in \mathbb{R}^2|x-q_n\in U_n\}$$

EDIT: Fedor Petrov was quick to find an easy answer to this one and I'm forced to accept it. His method doesn't hold up though if I additionally demand that all the $U_n$ are convex. So, submissions with a take on such a related question are still welcome (although I'll not be able to reward your submission with a "accepted answer" badge)

UPDATE: Acting on popular request, I've reposted the revised question over here.

  • 3
    $\begingroup$ It looks reasonable to ask a separate question for convex sets. $\endgroup$ Nov 10, 2019 at 8:30
  • 1
    $\begingroup$ By Fritz John's theorem, the question involving convex sets is the same as requiring that all of the $U_n$ be open ellipses (or rectangular boxes - not necessarily coordinate aligned) $\endgroup$ Nov 10, 2019 at 14:04
  • $\begingroup$ . . . and then you can assume for each $\epsilon \gt 0$ that they're aligned within $\epsilon$ if that helps. $\endgroup$ Nov 10, 2019 at 15:40
  • $\begingroup$ @FedorPetrov: there is now another question circulating mathoverflow.net/questions/345706/… $\endgroup$
    – Vergilius
    Nov 10, 2019 at 18:17

1 Answer 1


No even in dimension 1 (and multiplying the example for $\mathbb{R}$ by the small segment you get a counterexample in $\mathbb{R}^2$).

Take the set $A_n\subset \mathbb{R}$ defined as $\bigcup_{k\in \mathbb{Z}} (2k\cdot 10^{-n},(2k+1)\cdot 10^{-n})$. I claim that there exists no finite family of translates $\bigcup_{n=1}^\infty (A_n+q_n)$ which covers $[0,1]$. Indeed, we may recursively find a nested family segments $[0,1]\supset \Delta_1\supset \Delta_2\ldots$ such that length of $\Delta_i$ equals $10^{-i}$ and $\Delta_i\cap (A_i+q_i)=\emptyset$. The intersection of $\Delta_i$'s is not covered by our translates.

Now if $U_n=[0,1]\cap A_n$, the measures of $U_n$ are bounded from below.


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