The following theorem can be found in Hardy-Wright (Theorem 459), except that they state it only for $d=2$. Do you know of a reference where the proof of this general statement is written?


**Theorem:** *Let $d\ge2$ be an integer. Let $F$ be a bounded subset of $\Bbb R^d$. For every positive real number $r$, denote by $F(r)$ the set of points $x$ of $\Bbb Z^d$ such that $x\over r$ belongs to $F$. Assume that the cardinality of $F(r)$ divided by $r^d$ converges to some non-zero limit when $r$ goes to infinity. Then, when $r$ goes to infinity, the cardinality of the set of $(x\_1,\ldots,x\_d)$'s in $F(r)$ such that $\operatorname{GCD}(x\_1,\ldots,x\_d)=1$ is equivalent to $r^d/\zeta(d)$ when $r$ goes to infinity.*


Thank you very much for your time!



Reference request: probability that d numbers are coprime
Let $F$ be a bounded subset of $\mathbb R^d$ with $d \geq 2$. We define $F\_r := rF \cap \mathbb Z^d$ for any real number $r>0$ and assume that the limit
$$
 \mathcal V(F) := \lim\_{r \to + \infty} \frac{|F\_r|}{r^d}
$$
exists (for convex subsets, this is the Lebesgue volume of $F$). We rewrite the proof of Theorem 459 of Hardy–Wright so that it yields the following more general result.


*If $\mathcal V(F)$ is well-defined, then we have*
$$
 \lim\_{r \rightarrow + \infty} \frac{\left| \left\{ x \in F\_r, \operatorname{gcd}(x\_1, \cdots, x\_d) = 1 \right\} \right|}{r^d} = \frac{\mathcal V(F)}{\zeta(d)}.
$$


**Proof.**
We can and will assume that $0 \notin F$, which will not change any of the limits.
We also fix $N$ such that $F \subset [-N,N]^d$.


For every rational $r>0$, let $f(r) = \left| \left\{ x \in F\_r, \operatorname{gcd}(x\_1, \cdots, x\_d) = 1 \right\} \right|$. As $0 \notin F$, $|F\_r| = f(r)=0$ when $r<1/N$ and $f(r) \leq |F\_r| \leq (2rN+1)^d \leq (3rN)^d$ for all $r \geq 1/N$, so $|F\_r| \leq (3rN)^d$ in all cases. For any point $x$ of $F\_r$, there is a unique integer $k \in \mathbb N$ such that the gcd of the coordinates of $x$ is $k$, and then $x/k$ contributes to $f(r/k)$. Consequently (the right-hand side being in fact a finite sum)
$$
 |F\_r| = \sum\_{k=1}^{+ \infty} f(r/k).
 $$
By Möbius inversion, we then get
$$
 f(r) = \sum\_{k=1}^{+ \infty} \mu(k) |F\_{r/k}|.
$$


The sum of $\mu(k)/k^d$ converges absolutely towards $1/\zeta(d)$ as $d \geq 2$, so
$$
 \frac{f(r)}{r^d} - \frac{\mathcal V(F)}{\zeta(d)} = \sum\_{k=1}^{+ \infty} \frac{\mu(k)}{k^d} \left( \frac{|F\_{r/k}|}{(r/k)^d} - \mathcal V(F) \right).
$$
Let $\varepsilon>0$. By definition of $\mathcal V(F)$, we fix $n\_0$ such that if $r/k \geq n\_0$, $\left| \frac{|F\_{r/k}|}{(r/k)^d} - \mathcal V(F) \right| \leq \varepsilon$, which gives the inequality
$$
 \sum\_{k=1}^{\lfloor r/n\_0 \rfloor} \frac{1}{k^d} \left| \frac{|F\_{r/k}|}{(r/k)^d} - \mathcal V(F) \right| \leq \zeta(d) \varepsilon.
$$


On the other hand, the bounds on $|F\_{r/k}|$ give
$$
 \sum\_{k > \lfloor r/n\_0 \rfloor} \frac{1}{k^d} \left| \frac{|F\_{r/k}|}{(r/k)^d} - \mathcal V(F) \right| \leq \left((3N)^d + \mathcal V(F)\right) \times \frac{ (\lfloor r/n\_0 \rfloor)^{1-d} }{d-1}.
$$
Hence, for $r$ large enough, the absolute value of $\left|\frac{f(r)}{r^d} - \frac{\mathcal V(F)}{\zeta(d)}\right|$ is smaller than $2 \zeta(d) \varepsilon$, which proves the desired convergence. $\blacksquare$


*Acknowledgements.* This post greatly benefitted from exchanges with Samuel Le Fourn.


