Lemma 13.7.3274B700496A8

Proof. Fix a compact set $K$ and let $M$ be an upper bound for all $u_ k$ and $u$ on $K$. On $[-\infty ,M]$, the function $\max \{ s,-C\} $ is a Lipschitz function of $\arctan s$: it is constant for $\arctan s\leq \arctan (-C)$, and the derivative of the tangent is bounded on the compact interval $[\arctan (-C),\arctan M]$. Hence extended capacity convergence implies capacity convergence of the truncations. Conversely, $|\arctan s-\arctan \max \{ s,-C\} |\leq \pi /2-\arctan C$ for all $s$, so that

where we also used $|\arctan s-\arctan t|\leq |s-t|$ for finite $s,t$. Choosing $C$ with $2(\pi /2-\arctan C)<\delta /2$, the set $\{ d_*(u_ k,u)>\delta \} $ is contained in $\{ |\max \{ u_ k,-C\} -\max \{ u,-C\} |>\delta /2\} $, which proves the equivalence.

Adding the same bounded real function $h$ to $s$ and $t$ changes $d_*$ by at most a bounded factor: since $1+\sigma ^2\leq 2(1+h^2)(1+(\sigma +h)^2)$, the formula $\arctan t-\arctan s=\int _ s^ t\mathrm d\sigma /(1+\sigma ^2)$ gives $d_*(s+h,t+h)\leq 2(1+h^2)d_*(s,t)$. Now let $(W_ i,h_ i)$ and $(W'_ l,h'_ l)$ be two choices of a finite cover of $X$ by charts with bounded images and of smooth local potentials $h_ i$ of $\theta $ on $W_ i$ and $h'_ l$ on $W'_ l$, and suppose that $u_ k+h_ i\to u+h_ i$ in extended capacity on every $W_ i$. Fix $l$, a compact set $K\subset W'_ l$ and $\delta >0$, and cover $K$ by finitely many compact sets $K_ i\subset W_ i\cap W'_ l$. The smooth function $h'_ l-h_ i$ is bounded on $K_ i$, say by $H_ i$, so the inequality just proved, applied with $h=h'_ l-h_ i$, gives

By lemma 13.7.1, the outer capacity of the left side, computed in the chart $W'_ l$, is at most a constant times the outer capacity of the right side, computed in $W_ i$, which tends to zero. By subadditivity of outer capacity, the outer capacity of $K\cap \{ d_*(u_ k+h'_ l,u+h'_ l)>\delta \} $ in $W'_ l$ tends to zero. Thus extended capacity convergence does not depend on the choice of the cover or of the local potentials.

If $u_ k$ decreases to $u$, then for every $C$ the truncations decrease to the bounded plurisubharmonic function $\max \{ u,-C\} $, so they converge in capacity by proposition 7.5.6; the equivalence just proved gives extended capacity convergence.

For the global assertion, put $m_ k=\sup _ Xu_ k$ and $v_ k=u_ k-m_ k$. By proposition 10.3.1, $\int _ X|v_ k|\, \omega ^ n\leq M_0$ for a constant $M_0$. Suppose that $m_ k\to +\infty $ along a subsequence, to which we pass. By Markov's inequality, the $\omega ^ n$-volume of $\{ u_ k<m_ k/2\} =\{ v_ k<-m_ k/2\} $ is at most $2M_0/m_ k$, which tends to zero. On the other hand, let $(W,z)$ be one of the charts of definition 13.7.2, with smooth local potential $h$ of $\theta $, and let $K\subset W$ be a compact set of positive volume, with $|h|\leq H$ and $u+h\leq M$ on $K$. On $K\cap \{ u_ k\geq m_ k/2\} $ we have $u_ k+h\geq m_ k/2-H$, so for large $k$ this set is contained in $K\cap \{ d_*(u_ k+h,u+h)>\delta _0\} $, where $\delta _0=(\pi /2-\arctan M)/2$. The outer capacity in $z(W)$ of the image of the latter set tends to zero, and hence so does its Lebesgue measure, by proposition 7.4.3(3). Since Lebesgue measure in the chart and the $\omega ^ n$-volume are comparable on $K$, the volume of $K\cap \{ u_ k\geq m_ k/2\} $ tends to zero, whereas by the first observation it tends to the volume of $K$, which is positive. This contradiction shows that the suprema are bounded above. $\square$