# Covering the disk with a family of infinite total measure

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.

• It looks reasonable to ask a separate question for convex sets. – Fedor Petrov Nov 10 '19 at 8:30
• 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) – Anthony Quas Nov 10 '19 at 14:04
• . . . and then you can assume for each $\epsilon \gt 0$ that they're aligned within $\epsilon$ if that helps. – Noam D. Elkies Nov 10 '19 at 15:40
• @FedorPetrov: there is now another question circulating mathoverflow.net/questions/345706/… – Thibaut Demaerel Nov 10 '19 at 18:17

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.