Suppose $(\mathbb{R}, \Omega, \mu)$ is Lebesgue measure space. Given an uncountable nowhere dense set $A \subset \mathbb{R}$ such that $\mu(A) = 0$, is it true that $A+A = \{a + b \in \mathbb{R}: a,b \in A\}$ has positive measure, i.e. $\mu(A+A) > 0$?
This is true for the Cantor set since $C + C = [0,2]$. Would this be true in general?
No. Given a base $b$ and a subset $S\subset \{0,\dots,b-1\}$, let $$C_b(S)=\big\{x\in[0,1)\colon x\text{ has a representation in base $b$ with digits in } S\big\}.$$ For example, $S_2(\{0,1\})=[0,1)$ and $S_3(\{0,2\})$ is $C\setminus\{1\}$, where $C$ is the Cantor set.
Consider $A=S_4(\{0,1\})$; then $A+A=S_4(\{0,1,2\})$. Because of the surjection $A\to[0,1]$ given by interpreting a base-$4$ string as binary, $A$ is uncountable. Similarly to the proof that the Cantor set is nowhere dense and has Lebesgue measure zero, both $A$ and $A+A$ satisfy these conditions.
