Theorem 10.31.114AC0DDA39172

Proof. Start with any subsequence and extract a further subsequence along which the integrals in the hypothesis are at most $2^{-j}$ and $u_ j\to u$ almost everywhere for smooth volume. The envelope identity and Hartogs' lemma give a uniform bound $u_ j,u\leq \varphi +C$. For $N\geq j$ let $p_{j,N}=P_\theta (u_ j,\ldots ,u_ N)$. Partitioning the contact set according to the first obstacle touched gives

Apply proposition 10.31.8 with $w=u$, in its one-sided form. It bounds $\sup _ Xp_{j,N}$ from below, independently of $N$. Consequently the decreasing sequence has a limit

that is not identically $-\infty $, and the same proposition puts $p_ j$ in $\mathcal E_\varphi $. The increasing sequence $p_ j$ has a regularized limit $p$, which belongs to $\mathcal E_\varphi $ by upward stability, since $p_1\leq p\leq u\preceq \varphi $. The inequality $p\leq u$ follows from the almost-everywhere convergence of the extracted sequence.

Both $p_{j,N}\downarrow p_ j$ and $p_ j\uparrow p$ hold in capacity by lemma 5.29.19, since $p_1\not\equiv -\infty $ is a common lower bound. Apply lemma 13.14.8 to (10.31.12), with the bounded weights $\min (1,|p_{j,N}-u|)$ and then $\min (1,|p_ j-u|)$. These weights converge in capacity: discard the deep sublevel sets of $p_1$ and $u$, and use that $\min (1,|x-y|)$ is Lipschitz on finite values. The result is

Hence $\mu _ p(\{ p<u\} )=0$, and zero domination gives $p=u$. The upper envelopes $(\sup _{k\geq j}u_ k)^*$ decrease to $u$. Squeezing the extracted $u_ j$ between these envelopes and $p_ j$ proves convergence in capacity. Since the initial subsequence was arbitrary, the conclusion holds for the original sequence. $\square$