Lemma 17.4.3369EDE8DEEC5

Proof. By proposition 10.3.1 the suprema $\sup _ Xq_ k$ converge to $\sup _ Xq$, so they are bounded. Hence the tail envelopes $a_ j$ are $\theta $-psh, by theorem 5.10.2 applied in charts after adding smooth local potentials of $\theta $, and they decrease as $j$ increases. A subsequence of $(q_ k)$ converges to $q$ almost everywhere, so $a_ j\geq q$ almost everywhere, hence everywhere, since both functions are $\theta $-psh (section 16.11).

Conversely, let $x\in X$ and $\varepsilon >0$, let $h$ be a smooth local potential of $\theta $ on a coordinate ball $\Omega $ around $x$, and let $M>q(x)+h(x)$ be a real number. By upper semicontinuity there is a closed coordinate ball $K\subset \Omega $ around $x$ on which $q+h<M$ and $|h-h(x)|<\varepsilon $. The plurisubharmonic functions $q_ k+h$ converge to $q+h$ in $L^1_{\mathrm{loc}}(\Omega )$, so Hartogs' lemma in the form of theorem 5.10.4, applied with the constant function $M$ on $K$, gives $q_ k+h<M$ on $K$ for all large $k$. For such $k$ we have $q_ k<M-h(x)+\varepsilon $ on $K$. Since $K$ is a neighborhood of $x$, it follows that $a_ j(x)\leq M-h(x)+\varepsilon $ for all large $j$. Letting $\varepsilon \downarrow 0$ and $M\downarrow q(x)+h(x)$, or $M\to -\infty $ if $q(x)=-\infty $, we obtain $\lim _ ja_ j(x)\leq q(x)$. Thus $a_ j\downarrow q$.

Finally, if $q_ k\leq \psi $ for all $k$ with $\psi $ upper semicontinuous, then $\sup _{k\geq j}q_ k\leq \psi $, so $a_ j\leq \psi $ for every $j$, and therefore $q=\lim _ ja_ j\leq \psi $. $\square$