Lemma 18.28.62.922E6C36BCC8 Let $\varphi \in \mathrm{PSH}(X,\theta )$. Let $(K,v)$ be a weighted subset of $X$. Let $\mu \in \mathrm{BM}(K,v)$. Then
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18.28.63
\begin{equation} \lim _{k\to \infty } \big ( \mathcal{L}_{k,\varphi }(N^ k_{v,K}) - \mathcal{L}_{k,\varphi }(N^ k_{v,\mu })\big ) = 0\, . \end{equation}
Proof.
By (18.28.41) and (18.28.44), for every $\epsilon >0$ and all $k\geq 1$,
\[ N^ k_{v,\mu }\leq N^ k_{v,K}\leq C_\epsilon \mathrm{e}^{\epsilon k}N^ k_{v,\mu }. \]
The corresponding unit balls therefore satisfy inclusions with radii differing by at most the factor $C_\epsilon \mathrm{e}^{\epsilon k}$. Taking logarithmic volumes and multiplying by $n!k^{-n-1}$ gives
\[ \left|\mathcal{L}_{k,\varphi }(N^ k_{v,K})-\mathcal{L}_{k,\varphi }(N^ k_{v,\mu })\right| \leq \frac{2n!}{k^{n+1}} h^0\left(X,L^ k\otimes T\otimes \mathcal{I}(k\varphi )\right)(\log C_\epsilon +\epsilon k). \]
The right-hand side has limsup at most $2\epsilon \operatorname{vol}(\theta _\varphi )$ by theorem 18.1.46. Letting $\epsilon \to 0+$ proves (18.28.63).
$\square$
Pome
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