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](https://mathoverflow.net/questions/345706/covering-the-disk-with-a-family-of-infinite-total-measure-the-convex-sequel).



Covering the disk with a family of infinite total measure
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.


