PomeLemma 19.7.29.B63FC87B8DDF Suppose that $S^ i\in \mathcal{S}'(C)$ is a decreasing sequence such that
\[ \lim _{i\to \infty }\operatorname{vol}S^ i>0. \]
Then there is $S\in \mathcal{S}'(C)$ such that $S^ i\to S$.
Lemma 19.7.29B63FC87B8DDF
Proof. By 19.7.27, we may replace $S^ i$ by its regularization and assume that $S^ i=C(S^ i)\cap \mathbb {Z}^{n+1}$. We define
\[ S=\left( \bigcap _{i=1}^{\infty }C(S^ i) \right) \cap \mathbb {Z}^{n+1}. \]
The convex bodies $\Delta (S^ i)$ form a decreasing sequence and satisfy
\[ \lim _{i\to \infty }\operatorname{vol}\Delta (S^ i)=\lim _{i\to \infty }\operatorname{vol}S^ i>0. \]
By continuity from above for Lebesgue measure,
\[ \operatorname{vol}\bigcap _ i\Delta (S^ i)=\lim _{i\to \infty }\operatorname{vol}\Delta (S^ i)>0. \]
Set
\[ K\coloneqq \bigcap _{i=1}^{\infty }C(S^ i). \]
The cone $K$ is full-dimensional. As noted above, $K\cap \mathbb Z^{n+1}$ generates $\mathbb Z^{n+1}$ as an Abelian group; hence $S=K\cap \mathbb Z^{n+1}$ lies in $\mathcal S'(C)$. Moreover, the closed convex cone generated by $S$ is exactly $K$. Therefore
\[ \Delta (S)=\{ x:(x,1)\in K\} =\bigcap _{i=1}^{\infty }\{ x:(x,1)\in C(S^ i)\} =\bigcap _{i=1}^{\infty }\Delta (S^ i). \]
By corollary 19.2.8 and theorem 19.2.5, we can compute the distance
\[ d_{\mathrm{sg}}(S,S^ i)=\operatorname{vol}S^ i-\operatorname{vol}S=\operatorname{vol}\Delta (S^ i)-\operatorname{vol}\Delta (S), \]
which tends to $0$ by construction. $\square$
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