3
$\begingroup$

I asked this question on MSE a week ago and it gave me a tumbleweed badge :-)

Let $\Lambda$ be a lattice in $\mathbb R^n$, with covolume $\Gamma$. Moreover, let $S$ be a bounded (Lebesgue-)measurable subset of $\mathbb R^n$ and for each $\alpha > 0$ define $\alpha S := \{\alpha \mathbf x : \mathbf x \in S\}$.

Under which (reasonable) hypothesis on $S$ is it true that $$\lim_{\alpha \to +\infty} \frac{\#(\alpha S \cap \Lambda)}{\alpha^n \Gamma} = \mu(S) \; ?$$ Here $\mu$ is the Lebesgue measure on $\mathbb R^n$.

Thank you for any suggestion/reference.

NOTE 1: As pointed out by Igor Rivin, at the top of the 9th page of

I. Kapovich, I. Rivin, P. Schupp, and V. Shpilrain, Densities in free groups and $\mathbb{Z}^k$, Visible Points and Test Elements, Mathematical Research Letters 14 (2007), pp. 263--284.

it is written that for $\Lambda = \mathbb{Z}^n$ and $S$ bounded, open, and with smooth piecewise boundary the claim is well-known to be true.

[2]:

Improve this question
$\endgroup$
1
  • $\begingroup$ In the case that $S$ is a polytope, much more precise statements are known. Search for "Ehrhart theory". $\endgroup$
    – user41593
    Commented Dec 22, 2015 at 21:16

1 Answer 1

Reset to default
4
$\begingroup$

You want something like "Riemann integrable", Lebesgue-measurable is clearly too weak (since if your lattice is the usual lattice, but your set is some nice set with all the rational points missing, you are screwed). For the argument, see the end of this paper (Kapovich, Rivin, Schupp, Shpilrain).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Could you please be more precise? Since the paper is quite involved. $\endgroup$
    – user40023
    Commented Dec 21, 2015 at 13:08
  • $\begingroup$ See the proof of proposition 2.3... $\endgroup$
    – Igor Rivin
    Commented Dec 21, 2015 at 13:40

You must log in to answer this question.