Theorem 15.9.1A15C1A7CB4F8

Proof. The canonical representative of the limit also lies between $\varphi -C$ and $\varphi +C$. It suffices to show that every subsequence has a further subsequence converging in capacity. We may therefore assume that $I_ j\leq 2^{-j}$ and that $u_ j\to u$ almost everywhere. For $N\geq j$ define

All these envelopes lie between $\varphi -C$ and $\varphi +C$. The contact theorem and ordered contact give an inequality for the finite rooftops. Partition the contact set of $p_{j,N}$ according to the first index $\ell $ for which $p_{j,N}=u_\ell $. On the corresponding piece, the measure of $p_{j,N}$ is bounded by that of $u_\ell $, and $\mathrm{e}^{p_{j,N}}=\mathrm{e}^{u_\ell }$. Therefore

Only positive measures enter this argument, so it applies also when their total masses are infinite.

As $N\to \infty $, $p_{j,N}$ decreases to $p_ j$; the limit is the envelope of the infimum because every competitor lies below every finite rooftop. The sequence $p_ j$ increases, and its canonical limit $p$ is $\theta $-plurisubharmonic. By lemma 5.29.19, both monotone limits hold in capacity. Applying lemma 13.14.8 to the bounded nonnegative weights $|\mathrm{e}^{p_{j,N}}-\mathrm{e}^ u|$, first in $N$ and then in $j$, we obtain

The inequalities $p_ j\leq u_\ell $ for all $\ell \geq j$ imply $p\leq u$, first almost everywhere and then for the canonical representatives. Also $p\sim u\sim \varphi $. Off the pole sets, $\mathrm{e}^ p<\mathrm{e}^ u$ exactly on $\{ p<u\} $, so $\mathrm{MA}^{\mathrm{ext}}_\theta (p)(\{ p<u\} )=0$. Since $P_{\delta \theta }(\varphi )-C$ is a competitor for $P_{\delta \theta }(p)$, the latter is not identically $-\infty $. Theorem 15.2.2, applied with coefficient zero, gives $p\geq u$, and hence $p=u$.

Finally, by Hartogs' lemma and the almost-everywhere convergence, $U_ j=(\sup _{\ell \geq j}u_\ell )^*$ decreases to $u$. The inequalities $p_ j\leq u_ j\leq U_ j$, together with the convergence in capacity of both monotone sequences, prove the claim. $\square$