Two sets $X,Y \subset [0,1]$ such that $X+Y=[0,2]$

A set $X\subset \mathbb{R}$ is called nice if, for every $\epsilon > 0$, there are a positive integer $k$ and some 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 question is from an iberoamerican exam for undergraduate students (ciim 2010).

An attempt to solve this problem can be found at aops but it doesn't seem to be complete or correct. Any help is welcome.

Solution 1:

The middle thirds Cantor Set doesn't work since at stage $N$ there are $2^N$ intervals of length $\frac{1}{3^N}$. So for some $\epsilon$ the Hausdorff dimension of this Cantor set, the series is more than $0$:

$$ \left(\frac{2}{3^{\epsilon }}\right)^N =1 \longrightarrow \epsilon = \frac{\log 2}{\log 3} $$

It is also the limit set of the two functions $T_1(x) = \frac{x}{3}$ and $T_2(x) = \frac{x}{3} + \frac{2}{3}$.

If we remove successively smaller percentages, we can get Cantor sets of positive Lebesgue measure. The Smith-Volterra-Cantor set is nowhere-dense and has Lebesgue measure $\frac{1}{2}$.

These disasters come up when you try to show $\int_a^b f'(x) \, dx = f(b) - f(a)$ for certain trig series.

Considering this is like the Putnam exam, let's try to take "aggressive" Cantor sets. Instead of removing the middle $\frac{1}{3}$ at each stage, let's remove the middle $1-\frac{1}{N}$. Now after stage $N$ there are $2^N$ sets, but these have measure:

$$ 2^N \cdot \left(\frac{1}{N!}\right)^\epsilon \approx \left(\frac{2 e^\epsilon}{N}\right)^N \to 0$$

and this is true for all $\epsilon$. Using iterated functions we are using $T_{1,N}(x) = \frac{x}{N}$ and $T_{2,N}(x) = \frac{x}{N} + (1-\frac{1}{N})$. Here are some notes on dimension theory.

Here a lot of work has gone into constructing nice sets, but we need two nice sets $X,Y\subset [0,1]$ with $X + Y = [0,2]$. Necessarily $0,1 \in X \cap Y$.

It is certainly true that $C + C = [0,2]$ I don't think this is true for the set I have constructed.

Another route might be to try building "Cantor sets" using continued fractions whose digits are bounded or avoid a certain number.

$$ A_k = \{ [a_1, \dots, a_k, \dots] : 0 \leq a_i < k \} \subset \mathbb{Q} $$

These sets may be sparse enough and have the sum property you desire.