A subset $A$ of $\mathbb{N}$ is said to be of full density if:
$$\lim_{n\rightarrow \infty} \frac{|A\cap [1,n]|}{n}=1.$$
Suppose there exists a function $g:\mathbb{N}\rightarrow\mathbb{R}^{\geq 0}$, and a family $(A_i)_{i\in\mathbb{N}}$ of full-density subsets such that $g(k)< \epsilon_j$ for all $k\in A_j$ and $\epsilon_j\rightarrow 0$ as $j\rightarrow \infty$.
Is it necessarily true that there exists a full density subsequence of $\mathbb{N}$ along which $g$ converges to $0$?
(This would be a handy lemma for my work. My intuition suggests that this is true, but a construction is elusive to me at the moment.)
Here is a proof.
For $S\subseteq\mathbb N$ and $n\in\mathbb N$ define $d_n(S)=\frac{|S\cap[1,n]|}n.$
Define $B_j=\{k\in\mathbb N:g(k)\ge\epsilon_j\}\subseteq\mathbb N\setminus A_j;$ since $A_j$ has density one, $B_j$ has density zero.
Define integers $1=N_1\lt N_2\lt N_3\lt\cdots$ so that $d_n(B_1\cup\cdots\cup B_j)\le\frac1j$ for all $n\ge N_j$.
The set $$B=\bigcup_{j\in\mathbb N}(B_j\cap[N_j,\infty))$$ has density zero, since $N_j\le n\lt N_{j+1}\implies d_n(B)\le\frac1j.$
Thus the set $A=\mathbb N\setminus B$ has density one, and $g$ converges to $0$ along $A$ because $$k\in A,k\ge N_j\implies g(k)\lt\epsilon_j.$$