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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]:

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  • $\begingroup$ In the case that $S$ is a polytope, much more precise statements are known. Search for "Ehrhart theory". $\endgroup$ – Emanuele Tron Dec 22 '15 at 21:16
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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).

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  • $\begingroup$ Could you please be more precise? Since the paper is quite involved. $\endgroup$ – user40023 Dec 21 '15 at 13:08
  • $\begingroup$ See the proof of proposition 2.3... $\endgroup$ – Igor Rivin Dec 21 '15 at 13:40

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