Expected number of lattice points in $S$ is like $S$'s volume

Question

Hel${}$lo everyone,

I think I remember a theorem going something like this:

$L$ is a lattice in $\mathbb{R}^n$ whose base cell has area $a$. $S$ is a bounded open set in $\mathbb{R}^n.$ Take an element $x$ uniform on $L$'s base cell. Now \begin{equation} E[\text{# of points in } S\cap (L+x)]=vol(S)/a. \end{equation}

I imagine the proof might go something like this:

Call $\Lambda$ the base cell of $L$ centered about 0. Notice the expectation in question equals the sum of probabilities: \begin{align}\sum_{l\in L} P(l+x\in S)=\sum_{l\in L}\frac{1}{a}vol((\Lambda+l)\cap S)=vol(S)/a.\end{align}

Does anyone know if any of this is right or what this is called?

Thank${}$s!

Answer 1

Trivial by rearranging the integral. Identify a measurable fundamental region for $L$ centered about $0$ as $B$ and take $S$ to be a bounded, measurable set.

Now it is almost immediate that \begin{equation} \mathbb{E}_{U\sim \mathrm{Uniform}(B)} \left[\ |(U+L)\cap S|\ \right] = \frac{\mathrm{vol}\ S}{\mathrm{vol}\ B}. \end{equation}

Indeed,

\begin{align} \mathbb{E}_{U\sim \mathrm{Uniform}(B)} \left[\ |(U+L)\cap S|\ \right] &= \frac{1}{\mathrm{vol}\ B}\int_B |(U+L)\cap S| du \\ &= \frac{1}{\mathrm{vol} B}\sum_{\ell \in L} \int_B\mathbf{1}_{S}(u+\ell) du \\ &= \frac{1}{\mathrm{vol} B}\sum_{\ell \in L} \mathrm{vol}(B+\ell)\cap S \\ &= \frac{\mathrm{vol} S}{\mathrm{vol} B} \end{align} where the last equality follows since $(B+\ell)_\ell$ is a tiling of space.

What I was probably thinking of was the nontrivial Minkowski-Hlawka-Siegel theorem. From Zamir's book on lattices, for each dimension n>1 there is a random ensemble of lattices with unit determinant so that for any bounded measurable set S, the expected number of nonzero lattice points in S equals the volume of S.