If $\mathcal{A}\subseteq\mathbb{N}$ is a set with asymptotic density $\mathrm{d}(\mathcal{A}):= \lim_{n\to\infty} |\mathcal{A}\cap[1,n]|/n = 1$, does $\mathcal{A}+\mathcal{B}$ always contains all but finitely many natural numbers, for every asymptotic bases $\mathcal{B}$?
I came up with this question by reading this survey on sumsets by I. Ruzsa. On Sections 4.6 and 4.7 (pp 77-81) he describes two theorems on how the (Schnirelmann's) density of an addition of a basis $\mathcal{B}$ to a set with positive lower asymp. density $\mathcal{A}$ can be bounded below in terms of the order of $\mathcal{B}$ and $\underline{\mathrm{d}}(\mathcal{A})$. Nonetheless, having $\mathrm{d}(\mathcal{A}+\mathcal{B})=1$ does not say anything about having all large integers or not!
I was able to prove it true when $\underline{\mathrm{d}}(\mathcal{B})>0$, and it goes by considering $n_0\in\mathbb{N}:\forall n\geqslant n_0$ holds $$|\mathcal{A}\cap[1,n]|>(1-\underline{\mathrm{d}}(\mathcal{B}))n,$$
thus, for every large $n$, there must be some $b\in\mathcal{B}: n-b\in\mathcal{A}$.
The case $\underline{\mathrm{d}}(\mathcal{B})=0$ seem a bit more elusive, though.