10.31.7 Relative moments985DB0A9A096

Proof. We first treat the case $u_ j\leq w$. Since $\overline{\operatorname {vol}}_\theta (w)=M$, we may choose $w'\sim w$ with $\mu _{w'}(X)>M-\ell /2$. Translate $w'$ so that $w'\leq w$, and put $p_ j=P_\theta (u_ j,w')$. These rooftops belong to $\mathcal E_\varphi $. Contact support and ordered contact imply

Indeed, the integrand vanishes on the contact part $\{ p_ j=w'\} $, and on the remaining contact part $p_ j=u_ j<w'\leq w$.

We shall use the following estimate for maxima. If $a\in \mathcal E_\varphi $, $a\leq w'$, and $t>0$, then

To see this, put $m=\max (a,w'-t)$. It has the ordinary singularity type of $w'$, so $\mu _ m(X)\leq M$. On $\{ a>w'-t\} $, locality gives $\mu _ m=\mu _ a$. On the complementary set, where $m=w'-t$, ordered contact gives $\mu _ m\geq \mu _{w'}$. Integrating over the two sets proves (10.31.10).

Suppose that a subsequence had $\sup _ Xu_ j\to -\infty $; then also $\sup _ Xp_ j\to -\infty $. Splitting the mass of $p_ j$ at the level $w'-s$, $s>0$, and using (10.31.9) and (10.31.10), we obtain

For fixed $s$, the middle term tends to zero as $j\to \infty $, since $w'$ is finite off a pluripolar set and $\mu _{w'}$ is finite and non-pluripolar. Letting $j\to \infty $ and then $s\to \infty $ gives $\ell \leq M-\mu _{w'}(X)<\ell /2$, a contradiction. This proves the first assertion.

Now suppose $u_ j\to u$ in $L^1$. Then $u\leq w\preceq \varphi $. Fix $b>1$ and suppose that $Q_ b(u,w)\equiv -\infty $. Since $w-w'$ is bounded, translating competitors by a constant shows that $Q_ b(u,w')\equiv -\infty $ as well. Keep the same $p_ j$ and set

We claim that $\sup _ Xh_ j\to -\infty $. Otherwise, since $h_ j\leq p_ j\leq w'$, theorem 5.25.2 would give a subsequence converging in $L^1$ to some $h\not\equiv -\infty $. Passing to the limit in

first almost everywhere and then everywhere by lemma 5.24.3, would give a competitor for $Q_ b(u,w')$, a contradiction.

On the contact set of $h_ j$ one has $p_ j=b^{-1}h_ j+(1-b^{-1})w'$. Thus $h_ j\leq w'-bs$ implies $p_ j\leq w'-s$ there. Contact support and (10.31.9) therefore imply

Applying (10.31.10) with $a=h_ j$ and $t=bs$ on the complementary set, we obtain

Let first $j\to \infty $ and then $s\to \infty $. The choice of $w'$ again gives $\ell <\ell /2$, a contradiction. Therefore $Q_ b(u,w)\not\equiv -\infty $ for every $b>1$. The resulting convex inequalities and theorem 13.17.1, proposition 13.17.2 give

Letting $b\to \infty $ shows that the upper volumes of $u$ and $\varphi $ are equal; hence $u\in \mathcal E_\varphi $.

For the last assertion put $q_ j=P_\theta (u_ j,w)$. Contact gives

The case already treated bounds $\sup _ Xq_ j$ from below, and therefore also $\sup _ Xu_ j\geq \sup _ Xq_ j$. If $u_ j\to u$ in $L^1$, then, since $q_ j\leq w$, a subsequence of the $q_ j$ converges in $L^1$ to some $q\not\equiv -\infty $, and the case already treated gives $q\in \mathcal E_\varphi $. Since $q\leq u\leq \varphi +C$, upward stability gives $u\in \mathcal E_\varphi $ as well. $\square$

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. $\square$