Theorem 13.19.6C0F29EF85157

Proof. Step 1: the rooftops. Adding a constant to $u$ changes neither $p$ nor the family $(p_ m)_{m>0}$, apart from a shift of the parameter $m$. We may therefore assume $u\leq 0$. Then

where the increasing limit is understood after upper semicontinuous regularization. Thus $p_ m\sim u$, and $p_ m$ and $p$ are locally bounded off $S$. By theorem 13.19.5, $p$ is an ordinary model with the same volumes as $u$. By lemma 5.29.19, $p_ m\to p$ in capacity. On compact subsets of $X\setminus S$ the potentials have a common bounded range, so all bounded local products converge there, by the bounded product theory of section 5.27.

Step 2: full measures. Let $v\in \mathrm{PSH}(X,\omega )$ be locally bounded off $S$, and put $v_ k=\max (v,-k)$. On each chart choose a smooth $q$ with $\mathrm{dd}^{\mathrm{c}}q\geq \omega $, so that the functions $v_ k+q$ are plurisubharmonic and decrease to $v+q$. Theorem 5.27.18 then gives full weak limits in every degree,

Indeed, $\mathrm{dd}^{\mathrm{c}}v_ k=\mathrm{dd}^{\mathrm{c}}(v_ k+q)-\mathrm{dd}^{\mathrm{c}}q$, and the powers can be expanded in the closed positive powers of $\mathrm{dd}^{\mathrm{c}}(v_ k+q)$ and smooth forms. The local limits patch together because the approximating bounded currents are global. The limits are currents of order zero with finite mass on the compact manifold $X$, and $\Phi _ v$ is positive. Stokes' theorem for the bounded potentials $v_ k$ gives $\int _ XW_ n(v)=0$. Consequently

The measure $\Phi _ v$ equals $\mathrm{MA}_\omega (v)$ off $S$. Every closed positive power of degree less than $n$ in this expansion has coefficient measures vanishing on $S$, by corollary 5.34.3. Thus only the top closed power contributes to $\mathbf1_ S\Phi _ v$. This identifies $\mathbf1_ S\Phi _ v$ with $\mathbf1_ S(\mathrm{dd}^{\mathrm{c}}(v+q))^ n_{\mathrm{full}}$ and proves independence of $q$. These assertions are local on each chart and do not require a single chart containing $S$. Finally, $J(v)\leq v_+(\omega )$, since every truncation $v_ k$ has mass at most $v_+(\omega )$.

Step 3: convergence of the lower-degree powers. We need local trace bounds for the closed powers of $p_ m$ that are uniform in $m$. Fix a relatively compact coordinate neighborhood and write $\kappa =\mathrm{dd}^{\mathrm{c}}|z|^2$. Choose $B>0$ with $B\kappa \geq \omega $ there, and $A>1$ with $A\omega -B\kappa \geq \kappa $. For $p_{m,k}=\max (p_ m,-k)$, the closed current $T_{m,k}=B\kappa +\mathrm{dd}^{\mathrm{c}}p_{m,k}$ is positive and

Expanding $Q_{m,k}^ n$ with the smooth positive difference $Q_{m,k}-T_{m,k}\geq \kappa $ gives, for $1\leq j\leq n$,

The last bound holds because $Q_{m,k}=A(\omega +\mathrm{dd}^{\mathrm{c}}(p_{m,k}/A))$ and $p_{m,k}/A$ is a bounded $\omega $-plurisubharmonic function. Letting $k\to \infty $ gives trace bounds, independent of $m$, for every full closed power of $\mathrm{dd}^{\mathrm{c}}(p_ m+B|z|^2)$.

For $j<n$, every subsequential weak limit of these closed positive $(j,j)$-currents puts no mass on $S$, again by corollary 5.34.3. Off $S$ the bounded capacity convergence identifies the limit with the corresponding full power of $\mathrm{dd}^{\mathrm{c}}(p+B|z|^2)$, which also puts no mass on $S$. Hence the limit is unique. Expanding back in the smooth correction and using a finite coordinate cover yields

Hence (13.19.7) gives $J(p_ m)\to J(p)$.

Step 4: the residual measures. Let $R=\mathbf1_ S\Phi _ u$. Since $p_ m-u$ is bounded, the equality case of lemma 5.27.19, applied on each chart after a common smooth correction, gives $\mathbf1_ S\Phi _{p_ m}=R$ for every $m$. The total masses $J(p_ m)$ are bounded by $v_+(\omega )$, so along any sequence $m_ i\to \infty $ a subsequence of $\Phi _{p_{m_ i}}$ converges weakly to a measure $\eta $. Since $\Phi _{p_ m}-R$ is positive, so is $\eta -R$. Off $S$, $\eta $ equals $\mathrm{MA}_\omega (p)$, by the bounded convergence of Step 1, and its total mass is $\lim _ mJ(p_ m)=J(p)$ by Step 3. The measure $\Phi _ p$ has the same total mass and the same restriction to $X\setminus S$, so

On the other hand, $u\leq p$, so the inequality in lemma 5.27.19 gives $\mathbf1_ S\Phi _ p\leq R$ on each chart. Together with the previous inequality, this forces $\Phi _ p(S)=R(S)$, hence $\mathbf1_ S\Phi _ p=R$, which is the asserted equality of residual measures. It also gives $\mathbf1_ S\eta =R$, since $\eta -R\geq 0$ and $\eta (S)=R(S)$. As $\eta $ and $\Phi _ p$ also agree off $S$, we conclude that $\eta =\Phi _ p$. Every weak subsequential limit of $\Phi _{p_ m}$ is therefore $\Phi _ p$, which proves the convergence $\Phi _{p_ m}\to \Phi _ p$. $\square$