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
C1417B0EF5E7
10.31.9
\begin{equation} \int _ X(w'-p_ j)\, \mu _{p_ j} \leq \int _ X(w-u_ j)\, \mu _{u_ j}\leq K. \end{equation}
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
525AD3E9DFE4
10.31.10
\begin{equation} \mu _ a(\{ a>w'-t\} ) \leq \mu _{w'}(\{ a>w'-t\} )+M-\mu _{w'}(X). \end{equation}
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
\[ \ell \leq \mu _{p_ j}(X) \leq \frac Ks+\mu _{w'}(\{ p_ j>w'-s\} )+M-\mu _{w'}(X). \]
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
\[ h_ j=Q_ b(p_ j,w')\in \mathcal E_\varphi . \]
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
\[ b^{-1}h_ j+(1-b^{-1})w'\leq p_ j\leq u_ j, \]
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
\[ \mu _{h_ j}(\{ h_ j\leq w'-bs\} )\leq b^ nK/s. \]
Applying (10.31.10) with $a=h_ j$ and $t=bs$ on the complementary set, we obtain
\[ \ell \leq b^ nK/s+ \mu _{w'}(\{ h_ j>w'-bs\} )+M-\mu _{w'}(X). \]
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
\[ (1-b^{-1})^ n M\leq \overline{\operatorname {vol}}_\theta (u)\leq M. \]
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
\[ \int _ X(w-q_ j)\, \mu _{q_ j} \leq \int _ X(w-u_ j)_+\, \mu _{u_ j}\leq K. \]
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.
$\square$
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