Let $S\subseteq\mathbb{Z}^d$ be a finite set. Then there is a positive integer $N$ such that for any $A\subseteq[N]^d$ with $|A|\ge\delta N^d$ there is a translate $S'=S+a$, with $a\in[\mathbb{Z}]^d$ such that $|S'\cap A|\ge \frac{\delta}{2}|S|$
I know I have to use the so called "averaging argument" to obtain a contradiction, like $|A|<\delta N^d$, for example supposing the thesis is not true. However I'm not able to obtain a contradiction. I am not sure where to use he arbitrariety of $N$. Any help would be really appreciated.
Let $N$ be chosen later, and let $A\subset [N]^d$ with $\lvert A\rvert \geq \delta N^d$. Let $S_a=(S+a)\cap A$, and consider the average of $\lvert S_a\rvert$,
$$ \sum_{a\in \mathbb{Z}^d} \lvert S_a\rvert = \sum_{a\in\mathbb{Z}^d}\sum_{b\in A}\sum_{s\in S}1_{s+a=b}.$$
If we change the order of summation to bring the sum over $A$ and $S$ outside this is clearly $\lvert A\rvert\lvert S\rvert$.
On the other hand, note that $S_a$ is empty unless $a\in A-S$, and so if $S\subset [-M,M]^d$ then the sum over $a$ can actually be restricted to those $a\in [-M,N+M]^d$.
Therefore, if we choose $N$ such that $(N+2M)^d\leq 2N^d$ (which is true for large enough $N$, since both $M$ and $d$ are fixed), the left-hand side is actually a sum over at most $2N^d$ different $a$. By averaging therefore there exists some $a$ such that
$$ \lvert S_a\rvert \geq \frac{\lvert A\rvert\lvert S\rvert}{2N^d}\geq \frac{\delta}{2}\lvert S\rvert.$$