Theorem 7.11.1178EB0192D174

Proof. If $m=0$, then $\mu =\mu _ j=0$ and there is nothing to prove. We assume that $m>0$, and use the cubes $Q\in \mathcal Q_ j$, the polydiscs $P(z,r)$ and the estimate (7.11.6) of proposition 7.11.5; recall that $K$ denotes the support of $\mu $.

Step 1. Convergence against the fixed measure $\mu $. By theorem 7.11.2 with $p=2$ and the uniform bound for $e_2(U_ j)$ in proposition 7.11.5, we have $S=\sup _ j\int _ BU_ j^2\, \mathrm d\mu <\infty $. Fix $T>0$. The truncations $\max \{ U_ j,-T\} $ are plurisubharmonic with values in $[-T,0]$, and they converge to $\max \{ u,-T\} $ in $L^1_{\mathrm{loc}}(B)$, since $s\mapsto \max \{ s,-T\} $ is $1$-Lipschitz. The measure $\mu $ satisfies the hypothesis of lemma 7.11.9, because $\mu \le (\mathrm{dd}^{\mathrm{c}}w)^ n$, so the truncations converge in $L^1(\mu )$. Since $|a^2-b^2|\le 2T|a-b|$ for $a,b\in [-T,0]$, it follows that $\int \max \{ u,-T\} ^2\, \mathrm d\mu \le S$. Letting $T\to \infty $, monotone convergence gives $\int u^2\, \mathrm d\mu \le S$; in particular $u\in L^2(\mu )$, and $u$ is finite $\mu $-almost everywhere. For every measurable $v\le 0$ we have $\int _ B|v-\max \{ v,-T\} |\, \mathrm d\mu \le T^{-1}\int _ Bv^2\, \mathrm d\mu $, because $0\le \max \{ v,-T\} -v\le v^2/T$. Applying this to $v=U_ j$ and $v=u$, and using the convergence of the truncations in $L^1(\mu )$, we get $\varlimsup _ j\int _ B|U_ j-u|\, \mathrm d\mu \le 2S/T$, and letting $T\to \infty $ proves the first convergence in (7.11.12).

Step 2. Cube averages. For $z\in K$ let $Q_ j(z)$ be the cube of $\mathcal Q_ j$ containing $z$, and write

for the average over this cube of a function $v$ integrable on it. If $v$ is $\mu _ j$-integrable, then $v$ is integrable on every cube $Q$ with $\mu (Q)>0$; on a cube with $\mu (Q)=0$ on which $v$ is not integrable we set $\mathcal A_ jv=0$, which does not affect integrals against $\mu $. Since $\mu $ is carried by $K$, the definition of $\mu _ j$ gives

For a plurisubharmonic function $v\le 0$ on $B$ and $z\in K$, the Cauchy–Schwarz inequality for the probability measure $\mathrm dV/|Q_ j(z)|$ on $Q_ j(z)$ and (7.11.6) with $p=2$ give

if $v(z)=-\infty $, the right side is infinite. This applies in particular to the possibly unbounded function $u$.

Fix $z\in K$ with $u(z)>-\infty $ and $\varepsilon >0$, and put $c=u(z)+2\varepsilon $. By upper semicontinuity of $u$ and Hartogs' lemma (theorem 5.10.4), there is a neighborhood of $z$ on which $u\le c$ and $U_ j\le c$ for all large $j$. Since the polydiscs $P(z,4\sqrt2\, h_ j)$ shrink to $z$, for $v=u$ or $v=U_ j$ and all large $j$ the function $v-c$ is plurisubharmonic and nonpositive on $P(z,4\sqrt2\, h_ j)$. Then (7.11.6) with $p=1$, applied to $v-c$, gives $\mathcal A_ j(c-v)(z)\le D_{n,1}(c-v(z))$, while $\mathcal A_ jv(z)\le c$ because $v\le c$ on $Q_ j(z)$. Thus, for all large $j$,

Consequently $\mathcal A_ ju(z)\to u(z)$, and $\mathcal A_ jU_ j(z)\to u(z)$ whenever $U_ j(z)\to u(z)$. Indeed, in both cases the lower bound tends to $u(z)+2\varepsilon (1-D_{n,1})$ as $j\to \infty $, the upper bound is $u(z)+2\varepsilon $, and $\varepsilon >0$ is arbitrary.

We use the following elementary fact: if Borel functions $F_ j$ on $K$ converge to zero $\mu $-almost everywhere and are bounded in $L^2(\mu )$, then $\int _ K|F_ j|\, \mathrm d\mu \to 0$. Indeed, for $\varepsilon >0$ the Cauchy–Schwarz inequality gives

and $\mu (\{ |F_ j|>\varepsilon \} )=\int _ K\mathbf1_{\{ |F_ j|>\varepsilon \} }\, \mathrm d\mu \to 0$ by dominated convergence; we let first $j\to \infty $ and then $\varepsilon \downarrow 0$.

By (7.11.14) and Step 1, the sequences $(\mathcal A_ jU_ j-u)$ and $(\mathcal A_ ju-u)$ are bounded in $L^2(\mu )$. Since $u$ is finite $\mu $-almost everywhere, we have shown that $\mathcal A_ ju\to u$ $\mu $-almost everywhere, so $\mathcal A_ ju\to u$ in $L^1(\mu )$ by the elementary fact. By the first convergence in (7.11.12), every subsequence of $(U_ j)$ has a further subsequence $(U_{j_ k})$ with $\sum _ k\int _ B|U_{j_ k}-u|\, \mathrm d\mu <\infty $. By the monotone convergence theorem, $\sum _ k|U_{j_ k}-u|$ is $\mu $-integrable, hence finite $\mu $-almost everywhere, so $U_{j_ k}\to u$ $\mu $-almost everywhere. By the pointwise convergence proved above, applied along this subsequence, $\mathcal A_{j_ k}U_{j_ k}\to u$ $\mu $-almost everywhere, and hence in $L^1(\mu )$. Consequently the whole sequence $\mathcal A_ jU_ j$ converges to $u$ in $L^1(\mu )$. The function $U_ j-u$ is $\mu _ j$-integrable, since $u$ is locally integrable and $\mu _ j$ has a bounded density with compact support. Therefore (7.11.13) gives

Step 3. The absolute difference. Put $g_ j=\mathcal A_ j((U_ j-u)_+)$ on $K$. Let $z\in K$ with $u(z)>-\infty $, and let $c$ be as in Step 2. For all large $j$ we have $U_ j\le c$ and $u\le c$ on $Q_ j(z)$, so $(U_ j-u)_+\le c-u$ there, and hence $0\le g_ j(z)\le c-\mathcal A_ ju(z)$. Since $\mathcal A_ ju(z)\to u(z)$, letting $j\to \infty $ and then $\varepsilon \downarrow 0$ shows that $g_ j\to 0$ $\mu $-almost everywhere. The Cauchy–Schwarz inequality and (7.11.14) give

so the $g_ j$ are bounded in $L^2(\mu )$ by Step 1. The elementary fact of Step 2 gives $\int _ Kg_ j\, \mathrm d\mu \to 0$, and by (7.11.13) this integral equals $\int _ B(U_ j-u)_+\, \mathrm d\mu _ j$. Combining this with (7.11.15) and the identity $|U_ j-u|=2(U_ j-u)_+-(U_ j-u)$, we obtain the second convergence in (7.11.12).

Finally, since $(\mathrm{dd}^{\mathrm{c}}U_ j)^ n=\mu _ j$, we have $\int _ BU_ j(\mathrm{dd}^{\mathrm{c}}U_ j)^ n=\int _ B(U_ j-u)\, \mathrm d\mu _ j+\int _ Bu\, \mathrm d\mu _ j$. The first term tends to zero by the second convergence in (7.11.12), and by (7.11.13) and Step 2 the second term equals $\int _ K\mathcal A_ ju\, \mathrm d\mu $, which tends to $\int _ Bu\, \mathrm d\mu $. This proves the last assertion. $\square$