Sums of sets of lower density 0 We say that a set $A\subseteq \mathbb{N}$ has lower density 0 if $$\text{lim inf}_{n\to\infty}\frac{|A\cap\{1,\ldots,n\}|}{n} = 0.$$
Given $A,B\subseteq \mathbb{N}$ we set $A+B = \{a+b: a\in A, b\in B\}$. 
Are there $A, B\subseteq \mathbb{N}$ with lower density 0, but $A+B$ does not have lower density 0?
 A: Let $A$ be the set of all natural numbers having digit $0$ in every odd-numbered place (counting from the decimal point), and let $B$ be the set of all numbers having a $0$ in every even-numbered place. Then $A$ and $B$ have density $0$, while $A+B=\mathbb N\setminus(A\cup B)$ has density $1$.
A: It is relatively easy to prove that the set of perfect squares has asymptotic density equal to $0$. Then either the set $Q_2 := \{x^2+y^2: x,y \in \mathbf N\}$ has positive lower asymptotic density, and we're done, or the lower density of $Q_2$ is zero, and then we just consider that $\mathbf N = Q_2 + Q_2$ (by Lagrange's four-squares theorem).
By the way, it follows, e.g., from E. Landau, Über die Einteilung der positiven ganzen Zahlen in vier Klassen nach der Mindeszahl der zu ihrer additiven Zusammensetzung erforderlichen Quadrate, Arch. Math. Phys. 13 (1908), 305-312 that the asymptotic density of $Q_2$ is actually zero, but this is more than what you need to answer the question in the OP.
A: Partition $\mathbb N$ into two sets $S$ and $T$, each having lower density $0$ and upper density $1$; for example, by taking
$$S=[1!,2!)\cup[3!,4!)\cup[5!,6!)\cup[7!,8!)\cup\cdots,$$
$$T=[2!,3!)\cup[4!,5!)\cup[6!,7!)\cup[8!,9!)\cup\cdots.$$
Now define
$$A=\{n\in\mathbb N:n+1\in S\}\cup\{1\},$$
$$B=\{n\in\mathbb N:n+1\in T\}\cup\{1\}.$$
Then each of the sets $A,B$ has lower density $0$ (and upper density $1$), and $A+B=\mathbb N\setminus\{1\}.$
