Theorem 7.11.24D25577E2BC26

Proof. We estimate the capacities of the sets where the truncation $\max \{ U_ j,-T\} $ lies far below or far above $u_ T$ (Steps 1 and 2) and deduce convergence in capacity (Step 3). The estimate for $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ then follows from the convergence of the truncated measures (Step 4). Fix $T>0$ and write $a_ j=\max \{ U_ j,-T\} $ and $b=u_ T$. These plurisubharmonic functions take values in $[-T,0]$, and $a_ j\to b$ in $L^1_{\mathrm{loc}}(B)$ because truncation is $1$-Lipschitz. By theorem 7.11.11, $\varepsilon _ j=\int _ B|U_ j-u|\, \mathrm d\mu _ j\to 0$.

Step 1. The sets where $a_ j<b-\delta $. Let $\delta >0$. The set $\{ a_ j<b-\delta /2\} $ is relatively compact in $B$, because $b\leq 0$ while $a_ j$ tends to zero at $\partial B$. Applying lemma 7.10.1 with $s=t=\delta /2$ and then Chebyshev's inequality, we obtain

The weight $(b-a_ j)_+$ takes values in $[0,T]$, and it is at most $|U_ j-u|$, since truncation does not increase distances. By (7.11.20) with $p=2$, $\| (\mathrm{dd}^{\mathrm{c}}a_ j)^ n-\mu _ j\| _{\mathrm{TV}}\leq 2E_2T^{-2}$, where $\mu _ j=(\mathrm{dd}^{\mathrm{c}}U_ j)^ n$. Hence

and therefore $\operatorname {Cap}_ B(\{ a_ j<b-\delta \} )\leq 2^{n+1}\delta ^{-(n+1)}(\varepsilon _ j+2E_2/T)$.

Step 2. The sets where $a_ j>b+\delta $. Let $L\subset B$ be compact. By theorem 7.5.2, for every $\sigma >0$ there is an open set $G\subset B$ with $\operatorname {Cap}_ B(G)<\sigma $ such that $b$ is continuous on $B\setminus G$. We claim that $L\cap \{ a_ j>b+\delta \} \subset G$ for all sufficiently large $j$. Let $z\in L\setminus G$. Since $b$ is upper semicontinuous, $b<b(z)+\delta /3$ on a closed ball $\overline{B(z,r)}\subset B$. Hartogs' lemma (theorem 5.10.4), applied to the sequence $a_ j$ with the constant majorant $b(z)+\delta /3$, gives $a_ j<b(z)+\delta /3$ on $\overline{B(z,r)}$ for all large $j$. Since $b$ is continuous on $B\setminus G$, after shrinking $r$ we also have $b>b(z)-\delta /3$ on $(L\setminus G)\cap B(z,r)$. Hence $a_ j-b<2\delta /3$ on $(L\setminus G)\cap B(z,r)$ for all large $j$. As $L\setminus G$ is compact, finitely many such balls cover it, which proves the claim. Consequently $\varlimsup _ j\operatorname {Cap}_ B(L\cap \{ a_ j>b+\delta \} )\leq \sigma $ for every $\sigma >0$, that is, $\operatorname {Cap}_ B(L\cap \{ a_ j>b+\delta \} )\to 0$.

Step 3. Convergence in capacity. Let $K\subset B$ be compact, and apply Step 2 to a compact set $L\subset B$ with $K\subset \operatorname{Int}L$. The bounded functions $a_ j-b$ are quasi-continuous by theorem 7.5.2 and the remark following definition 7.5.1. Therefore lemma 7.5.4, together with Steps 1 and 2 applied with $\delta /2$ in place of $\delta $, gives

It remains to remove the truncation. Outside $\{ U_ j<-T\} \cup \{ u<-T\} $ we have $a_ j=U_ j$ and $b=u$. Hence both sets $K\cap \{ U_ j>u+\delta \} $ and $K\cap \{ U_ j<u-\delta \} $ in definition 7.5.5 are contained in

By proposition 7.11.22 with $p=2$, applied in part (1) to $U_ j$, whose energy is at most $E_2$, and in part (2) to the sequence $(U_ j)$,

and the capacity of the open set $\{ U_ j<-T\} $ equals its outer capacity. By subadditivity of the outer capacity, the upper limits as $j\to \infty $ of the outer capacities of the two sets in definition 7.5.5 are bounded by the sum of three terms, each of which tends to zero as $T\to \infty $. Since these upper limits do not depend on $T$, they vanish. Thus $U_ j\to u$ locally in capacity.

Step 4. The truncated limit measure. Fix $T>0$ again. By Step 3 and (7.11.17), the uniformly bounded functions $a_ j$ converge to $b$ locally in capacity, so $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n\to (\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ weakly on $B$ by theorem 7.5.8. By lemma 7.11.19, each $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n$ has total mass $\mu _ j(B)=m$. Testing with functions $\chi \in C_ c(B)$, $0\leq \chi \leq 1$, and letting $\chi $ increase to one, we see that $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ has total mass at most $m$. Moreover, (7.11.20) gives

for every real $\chi \in C_ c(B)$ with $|\chi |\leq 1$. Since $\mu _ j\to \mu $ weakly by proposition 7.11.5, letting $j\to \infty $ yields the same bound with $(\mathrm{dd}^{\mathrm{c}}u_ T)^ n$ and $\mu $ in place of $(\mathrm{dd}^{\mathrm{c}}a_ j)^ n$ and $\mu _ j$. Both limit measures are finite, so by the remark following the definition of the total variation distance, the supremum over such $\chi $ equals $\| (\mathrm{dd}^{\mathrm{c}}u_ T)^ n-\mu \| _{\mathrm{TV}}$. This proves (7.11.25). $\square$