Proposition 16.21.3.5ADA09DD223E Suppose that $\{ \theta \} $ is big, let $u\in \mathrm{PSH}(X,\theta )$, and put $w=P_\theta [u]$. Then
and $\mu _ w$ is carried by $\{ w=0\} $.
5ADA09DD223EProposition 16.21.3.5ADA09DD223E Suppose that $\{ \theta \} $ is big, let $u\in \mathrm{PSH}(X,\theta )$, and put $w=P_\theta [u]$. Then
and $\mu _ w$ is carried by $\{ w=0\} $.
Proof. Put $w_ k=P_\theta (u+k,V_\theta )$ for integers $k$; these are potentials by proposition 16.11.6. By corollary 16.13.5,
for the second inequality we used $w_ k\leq V_\theta $ and $w_ k\leq w\leq V_\theta $. Since $V_\theta \leq 0$, the sets $\{ u+k\leq V_\theta \} $ decrease to the pole set of $u$, which is $\mu _ u$-null, and the first term tends to zero in total variation by dominated convergence. By proposition 16.21.2, $w_ k$ increases almost everywhere to $w$, so $\mu _{w_ k}\to \mu _ w$ weakly by corollary 13.8.9. Integrating a continuous function $\chi \geq 0$ and letting $k\to \infty $, we obtain $\int _ X\chi \, \mu _ w\leq \int _ X\chi \mathbf1_{\{ w=V_\theta \} }\, \mu _{V_\theta }$, and lemma 16.12.1 gives the asserted inequality. Finally, a potential lies below $0$ quasi-everywhere if and only if it is nonpositive, so $V_\theta =P_\theta (0)$, and theorem 16.13.4 shows that $\mu _{V_\theta }$ is carried by $\{ V_\theta =0\} $. Hence $\mu _ w$ is carried by $\{ w=V_\theta =0\} $. $\square$
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