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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.
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.

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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.

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Inspired by Hausdorff dimension, we can try to let $X$ consist of numbers whose decimal expansion is of the form $$ 0.x_100x_4x_5x_60000\dots $$ and $Y$ consist of the "complementary" numbers: $$ 0.0x_2x_3000x_7x_8x_9x_{10}00000\dots $$ Adjust the lengths of the alternating segments as needed (so maybe $0.x_1000x_5x_6x_7x_8x_900\dots$) (Also adjust to cover $[0,2]$ instead of $[0,1]$.)

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