
> 
> A set $X\subset \mathbb{R}$ is called *nice* if for every $\epsilon > 0$ there are a 
>  positive integer $k$ and $k$ bounded intervals $I\_1,I\_2,...,I\_k$ such that 
>  $X \subset I\_1 \cup I\_2 \cup \cdots \cup I\_k$ and 
>  $\sum\limits\_{j=1}^k |I\_j|^{\epsilon} < \epsilon$.  
> 
>  Prove that there exist sets $X,Y \subset [0,1]$, both of them *nice*, such that $X+Y = [0,2]$,
>  where $X+Y:=\{x+y\mid x\in X,y\in Y\}.$
> 
> 
> 


This problem was posted some time ago at [math.SE](https://math.stackexchange.com/questions/1314672/two-sets-x-y-subset-0-1-such-that-xy-0-2).  

It's from an Iberoamerican contest for undergraduate students.  

The *nice* condition is stronger than Lebesgue measure zero, since there is a $\epsilon$ in the exponent.  

No solution was received. 
It was suggested the use of continued fractions whose digits are bounded, but I cannot see how this can be useful (I cannot prove or disprove that a set of such continued fractions are nice).  

Any suggestion is appreciated. Thanks in advance.



Sets $X,Y \subset [0,1]$, stronger than being measure $0$, such that $X+Y = [0,2]$
Following up on Bjørn's suggestion:


Let $S\subset\mathbb{N}$ be a set such that both $S$ and $\mathbb{N}\setminus S$ have lower density zero. For example, let $S=\bigcup\_{n\in\mathbb{N}} [(2n)!,(2n+1)!)$. Set
$$
X' = \{ \sum\_n a\_n 2^{-n}: n\in S, a\_n\in\{0,1\}\},
$$
$$
Y' = \{ \sum\_n a\_n 2^{-n}: n\notin S, a\_n\in\{0,1\}\}.
$$
In other words, $X'$ are the points in $[0,1]$ whose binary digit expansion has zeros at all places in $S$, and $Y'$ are the points in $[0,1]$ whose binary digit expansion has zeros at all places in the complement of $S$.


Now $X'$ and $Y'$ are nice. For example, for $X'$ fix $\varepsilon>0$ and take $N$ such that $s\_N:=|S\cap \{1,\ldots,N\}|\le \varepsilon N/2$. Then $X'$ can be covered by $2^{s\_N}\le 2^{\varepsilon N/2}$ intervals of length $2^{-N}$. 


It is clear from the definitions that $X'+Y'=[0,1]$. Take $X=X'\cup (1-X')$, $Y=Y'\cup (1-Y')$ to finish.


**Final remark**. The definition of *nice* is almost the same as having zero Hausdorff dimension. If the set $X$ is compact, it is exactly the same. The example in this answer is a rather standard one of sets of zero dimension which in some sense behave as if they were large.


