Sparse sets $S$ of integers such that $\frac{1}{x^2}\sum_{s\in S, s\le x}s\to 0$

Question

Let $A$ be a set of positive integers such that $$ \lim_{x\to \infty}\frac{1}{x^2}\sum_{a\in A \cap [1,x]}a=0. $$ Fix now a set $B$ of positive integers such that $|B\cap [1,x]| \le |A \cap [1,x]|$ for all $x\ge 1$.

Question. Is it true that $$ \lim_{x\to \infty}\frac{1}{x^2}\sum_{b \in B \cap [1,x]}b=0\,\,\,? $$

Answer 1

Yes, under these assumptions we have

$$\lim_{x\to \infty} \frac{1}{x^2}\sum_{b \in B \cap [1,x]} b=0\,.$$

For if we had

$$\limsup_{x\to \infty} \frac{1}{x^2} \sum_{b \in B \cap [1,x]} b = \lambda > 0\,$$

there would be an increasing sequence $(x_n)$ such that $B \cap [1,x_n]$ has more than $\frac{\lambda}{2}x_n$ elements. Then

$$\sum_{a \in A \cap [1,x_n]} a \geqslant \sum_{k = 1}^{\lfloor \lambda x_n/2\rfloor} k > \frac{\lambda^2 x_n^2}{9}$$

for all large enough $n$, so

$$\limsup_{x\to \infty} \frac{1}{x^2}\sum_{a \in A \cap [1,x]} a \geqslant \frac{\lambda^2}{9} > 0\,.$$