Is there a meager or measure zero set $A\subset \mathbb R$ with $A+A=A$ and $A-A=\mathbb R$? I am using sumset notation $$A+A=\{a+b:a,b\in A\}$$ $$A-A=\{a-b:a,b\in A\}$$ In more algebraic terms, $A$ must be an additive submonoid that generates $\mathbb R$ as a group. (At least, assuming $0\in A,$ which is harmless.)
It is a classic exercise to construct a meager measure zero set $A$ such that $A-A=\mathbb R.$ For example, take $A$ to be the set of numbers with no $9$ in any decimal expansion. But this doesn't satisfy $A+A=A.$
For any discontinuous additive function $f:\mathbb R\to\mathbb R$ the set $A=\{x:f(x)\geq 0\}$ satisfies the conditions except it only has inner measure zero. I want the outer measure to be zero.