Theorem 21.7.3.831AEEB7084F Let $K\subset M_{\mathbb R}$ be a convex body, let $a_ k\in M_{\mathbb R}$ be arbitrary, and let $r_ k\geq 0$ satisfy $r_ k/k\to 0$. Then for every continuous function $f\colon M_{\mathbb R}\to \mathbb R$,
\[ \lim _{k\to \infty }\frac1{k^ n} \sum _{m\in (kK)^{r_ k}\cap (M+a_ k)}f(m/k) =\int _ K f(p)\, \mathrm dp. \]
In particular $k^{-n}\# (kK\cap M)\to \operatorname {vol}(K)$. Here the exponent $n$ and the measure are those of the ambient space, also when $K$ has empty interior. The convergence is uniform with respect to the translations $a_ k$.
Proof.
For $n=0$ every set in question is the single point, and both sides equal its value under $f$. Suppose that $n>0$, and let $Q=[0,1)^ n$ be the half-open unit cube in the chosen coordinates. The cells
\[ \frac mk+\frac1k Q,\qquad m\in M+a_ k, \]
partition $M_{\mathbb R}$; each has volume $k^{-n}$, and its points lie within distance $c/k$ of its anchor $m/k$, where $c=\sqrt n$. Let $U_ k$ be the union of the cells whose anchors $m/k$ occur in the sum, that is, with $m/k\in K^{r_ k/k}$. Then
\[ K_{c/k}\subset U_ k\subset K^{(r_ k+c)/k}. \]
For the first inclusion, if $p\in K_{c/k}$, the anchor of the cell containing $p$ lies within $c/k$ of $p$, hence in $K$, so the cell is selected. For the second, every point of a selected cell lies within $c/k$ of its anchor, which lies within $r_ k/k$ of $K$. Since $K_{c/k}\subset K\subset K^{(r_ k+c)/k}$ and $(r_ k+c)/k\to 0$, proposition 21.7.2 gives $\operatorname {vol}(U_ k\mathbin \triangle K)\to 0$, and the volumes $\operatorname {vol}(U_ k)$ stay bounded.
For large $k$ the selected cells lie in the fixed compact set $K^1$. Let $\omega _ f$ be the modulus of continuity of $f$ on $K^1$. Since $k^{-n}f(m/k)$ is the integral of the constant $f(m/k)$ over the cell of $m$, we obtain
\[ \left|k^{-n}\sum _ m f(m/k)-\int _{U_ k}f(p)\, \mathrm dp\right| \leq \omega _ f(c/k)\operatorname {vol}(U_ k)\longrightarrow 0. \]
Since $f$ is bounded on $K^1$, the symmetric-difference estimate gives $\int _{U_ k}f\, \mathrm dp\to \int _ Kf\, \mathrm dp$. Taking $f=1$, $r_ k=0$ and $a_ k=0$ proves the counting formula. All the estimates depend only on the size of the cells, and not on their position, so they hold uniformly in the translations $a_ k$.
$\square$
Pome
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