Theorem 10.31.56AC45579AB2E

Proof. We follow the proofs of lemma 15.37.13, theorem 15.37.19 and check their ingredients in the present background. The envelope identity and Hartogs' lemma give $u_ j\leq \varphi +C$. Finite rooftops and stretching envelopes of members of $\mathcal E_\varphi $ stay in $\mathcal E_\varphi $ by proposition 16.28.64, lemma 16.28.63, and their masses are at least $\ell $ by (10.31.1). These facts replace the corresponding ones for semipositive backgrounds.

After a common translation, assume $u_ j,u\leq \varphi \leq 0$. Using lemma 5.44.2, extract once and for all a subsequence along which $u_ j\to u$ almost everywhere for smooth volume and

For $p_{j,N}=P_\theta (u_ j,\ldots ,u_ N)$, partitioning the contact set according to the first obstacle touched gives

In the proof of the extraction lemma, replace $\kappa $ by $\ell $. For each fixed $B\geq 1$, the stretched rooftops $r_{j,N}=Q_ B(p_{j,N},\varphi )$ satisfy

The exponential test (15.37.15) therefore bounds $\sup _ Xr_{j,N}$ from below, independently of $N$, once $j$ is sufficiently large. The construction combining the tail with the finitely many earlier terms, with $B=2b$, uses only convex combinations and finite rooftops that are not identically $-\infty $. It provides every defining competitor for

Since $h\leq u\leq \varphi $, upward stability gives $u\in \mathcal E_\varphi $.

For the continuity statements, put $p_ j=P_\theta (\inf _{k\geq j}u_ k)$ and $p=(\lim _ jp_ j)^*$. The monotone limits $p_{j,N}\downarrow p_ j$ and $p_ j\uparrow p$ have the common lower bound $h\not\equiv -\infty $, so they hold in capacity by lemma 5.29.19. All the limits lie in $\mathcal E_\varphi $, and $p\leq u$. Applying lemma 13.14.8 twice to (10.31.6) gives

Hence $\mu _ p(\{ p<u\} )=0$, and zero domination (theorem 13.19.1 with coefficient zero) gives $p=u$. The upper envelopes $(\sup _{k\geq j}u_ k)^*$ decrease to $u$ by Hartogs' lemma, and squeezing the $u_ j$ between them and $p_ j$ proves convergence in capacity.

Finally, the fixed finite dominating measure gives

uniformly in $j$, and theorem 13.15.23 proves weak convergence to $\mu _ u\leq \nu $. Starting with an arbitrary subsequence, we obtain a further subsequence with both convergences and the same limit. Hence the conclusions hold for the original sequence. $\square$