PomeLemma 19.7.18.7887065F3C7C The above defined $d_{\mathrm{sg}}$ is a pseudometric on $\hat{\mathcal{S}}(C)$.
19.7.16 Generalities on semigroups3A76517E84BF
Write $\hat{\mathcal{S}}(C)$ for the set of subsets of $C\cap \mathbb {Z}^{n+1}$ and $\mathcal{S}(C)$ for the set of subsemigroups $S\subseteq C\cap \mathbb {Z}^{n+1}$. For each $k\in \mathbb {N}$ and $S\in \hat{\mathcal{S}}(C)$, we write
\[ S_ k\coloneqq \left\{ x\in \mathbb {Z}^ n: (x,k)\in S \right\} . \]
Note that $S_ k$ is a finite set by our assumption on $C$.
We introduce a pseudometric on $\hat{\mathcal{S}}(C)$ as follows:
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19.7.17
\begin{equation} d_{\mathrm{sg}}(S,S')\coloneqq \varlimsup _{k\to \infty } k^{-n} \left(|S_ k|+|S'_ k|-2|(S\cap S')_ k|\right). \end{equation}
Proof. Only the triangle inequality needs to be argued. Take $S,S',S''\in \hat{\mathcal{S}}(C)$. We claim that for any $k\in \mathbb {N}$,
\[ |S_ k|+|S'_ k|-2|S_ k\cap S'_ k|+|S''_ k|+|S'_ k|-2|S''_ k\cap S'_ k|\geq |S_ k|+|S''_ k|-2|S_ k\cap S''_ k|. \]
From this the triangle inequality follows. To argue the claim, we rearrange it to the following form:
\[ |S'_ k|-|S_ k\cap S'_ k| \geq |S'_ k\cap S''_ k|-|S_ k\cap S''_ k|, \]
which is obvious. $\square$
Given $S,S'\in \hat{\mathcal{S}}(C)$, we say $S$ is equivalent to $S'$ and write $S\sim S'$ if $d_{\mathrm{sg}}(S,S')=0$. This is an equivalence relation by 19.7.18.
Lemma 19.7.19.680A0890C1EE Given $S,S',S''\in \hat{\mathcal{S}}(C)$, we have
\[ d_{\mathrm{sg}}(S\cap S'',S'\cap S'')\leq d_{\mathrm{sg}}(S,S'). \]
In particular, if $S^ i,S'^ i\in \hat{\mathcal{S}}(C)$ ($i\in \mathbb {N}$) and $S^ i\to S$, $S'^ i\to S'$, then
\[ S^ i\cap S'^ i\to S\cap S'. \]
Proof. Observe that for any $k\in \mathbb {N}$,
\[ |S_ k\cap S_ k''|-|S_ k\cap S_ k'\cap S_ k''|\leq |S_ k|-|S_ k\cap S_ k'|. \]
The same holds if we interchange $S$ with $S'$. It follows that
\[ |S_ k\cap S_ k''|+|S_ k'\cap S_ k''|-2|S_ k\cap S_ k'\cap S_ k''|\leq |S_ k|+|S'_ k|-2|S_ k\cap S_ k'|. \]
The first assertion follows.
Next we compute
\[ \begin{split} d_{\mathrm{sg}}(S^ i\cap S'^ i,S\cap S')\leq & d_{\mathrm{sg}}(S^ i\cap S'^ i,S^ i\cap S')+d_{\mathrm{sg}}(S^ i\cap S',S\cap S')\\ \leq & d_{\mathrm{sg}}(S'^ i,S')+d_{\mathrm{sg}}(S^ i,S) \end{split} \]
and the second assertion follows. $\square$
The volume of $S\in \mathcal{S}(C)$ is defined as
\[ \operatorname{vol}S\coloneqq \lim _{k\to \infty } (ka)^{-n}|S_{ka}|=\varlimsup _{k\to \infty } k^{-n}|S_ k|, \]
where $a$ is a sufficiently divisible positive integer. The existence of the limit and its independence from $a$ both follow from the more precise result [Theorem 2, KK12], a special case of which can be stated as follows: Theorem 19.7.20 (Kaveh–Khovanskii).85641516C8CA Let $S\in \mathcal{S}(C)$. Then, for every sufficiently divisible positive integer $a$, the limit
\[ \lim _{k\to \infty }(ka)^{-n}|S_{ka}| \]
exists, is independent of $a$, and is equal to $\varlimsup _{k\to \infty } k^{-n}|S_ k|$. If moreover $S$ generates $\mathbb {Z}^{n+1}$ as an Abelian group, and if $C(S)$ denotes the closed convex cone generated by $S\cup \{ 0\} $, then
\[ \lim _{k\to \infty } k^{-n}|S_ k| = \operatorname{vol}\{ x\in \mathbb {R}^ n:(x,1)\in C(S)\} . \]
In particular, for every convex body $K\subseteq \mathbb {R}^ n$,
\[ \lim _{k\to \infty } k^{-n}\# \left(kK\cap \mathbb {Z}^ n\right)=\operatorname{vol}K. \]
The same assertion holds with $\mathbb {Z}^ n$ replaced by any lattice, provided the Lebesgue measure is normalized by that lattice.
Lemma 19.7.21.DCB8E8464FB6 Let $S,S'\in \mathcal{S}(C)$. Then
\[ |\operatorname{vol}S-\operatorname{vol}S'|\leq d_{\mathrm{sg}}(S,S'). \]
Proof. Choose a sufficiently divisible positive integer $a$. Taking the limsup in the definition of $d_{\mathrm{sg}}$ along $k=ma$ gives
\[ d_{\mathrm{sg}}(S,S')\geq \operatorname{vol}S+\operatorname{vol}S'-2\operatorname{vol}(S\cap S'). \]
It follows that $\operatorname{vol}S-\operatorname{vol}S'\leq d_{\mathrm{sg}}(S,S')$ and $\operatorname{vol}S'-\operatorname{vol}S\leq d_{\mathrm{sg}}(S,S')$. $\square$
We define $\overline{\mathcal{S}}(C)$ as the closure of $\mathcal{S}(C)$ in $\hat{\mathcal{S}}(C)$ with respect to the topology defined by the pseudometric $d_{\mathrm{sg}}$. By 19.7.21, $\operatorname{vol}\colon \mathcal{S}(C)\rightarrow \mathbb {R}$ admits a unique $1$-Lipschitz extension to
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19.7.22
\begin{equation} \operatorname{vol}\colon \overline{\mathcal{S}}(C)\rightarrow \mathbb {R}. \end{equation}
Lemma 19.7.23.EC9CD63CAB3A Suppose that $S,S'\in \overline{\mathcal{S}}(C)$ and $S\subseteq S'$. Then
\[ \operatorname{vol}S\leq \operatorname{vol}S'. \]
Proof. Take sequences $S^ j,S'^ j$ in $\mathcal{S}(C)$ such that $S^ j\to S$, $S'^ j\to S'$. By 19.7.19, after replacing $S^ j$ by $S^ j\cap S'^ j$, we may assume that $S^ j\subseteq S'^ j$ for each $j$. Then our assertion follows easily. $\square$
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