Proposition 7.11.51151DF7B32E4

Proof. If $m=0$, then $\mu _ j=0$ and $U_ j=0$ for all $j$, and all assertions hold. Assume that $m>0$, put $d=\operatorname {dist}(K,\mathbb C^ n\setminus B)>0$, and choose $j_0$ such that $h_ j<d/(6\sqrt{2n})$ for $j\geq j_0$; in the rest of the proof we consider only $j\geq j_0$. Since $\mu $ is a finite Borel measure, it is a Radon measure (proposition 2.2.2), so $\mu $ vanishes outside its support $K$. Hence a cube $Q$ with $\mu (Q)>0$ meets $K$, and as its diameter is $\sqrt{2n}\, h_ j<d/6$, it lies in the $(d/6)$-neighborhood of $K$, which is contained in $B$. These cubes are disjoint, each has volume $h_ j^{2n}$, and they lie in a bounded set; so there are only finitely many of them. Therefore $\mu _ j$ is a finite sum of measures with constant densities on cubes contained in $B$, so it has a bounded density and compact support in $B$, and $\mu _ j(B)=\sum _ Q\mu (Q)=m$. This proves (1).

For the weak convergence, let $g$ be a continuous function on $\mathbb C^ n$. Then

and on every cube with $\mu (Q)>0$ the integrand is bounded by the modulus of continuity of $g$ at $\sqrt{2n}\, h_ j$ on the closed $(d/6)$-neighborhood of $K$. Hence the difference tends to zero.

To prove (2), fix a cube $Q\in \mathcal Q_ j$ which meets $K$ and a point $z\in Q$, and put $\varrho =4\sqrt2\, h_ j$. Every point of $Q$ differs from $z$ by less than $\sqrt2\, h_ j$ in each complex coordinate, so $Q\subset P(z,\varrho /2)$. Moreover $P(z,\varrho )$ lies in the Euclidean ball of radius $4\sqrt{2n}\, h_ j$ about $z$, and $\operatorname {dist}(z,K)\le \sqrt{2n}\, h_ j$, so this polydisc lies in the $(5d/6)$-neighborhood of $K$ and hence in $B$. Let $v\le 0$ be a plurisubharmonic function on $P(z,\varrho )$; if $v(z)=-\infty $, there is nothing to prove, so we assume that $v(z)>-\infty $. The function $\zeta \mapsto v(z+\varrho \zeta )$ is plurisubharmonic and nonpositive on the unit polydisc (proposition 5.9.3), with the finite value $v(z)$ at the origin. The first estimate of proposition 5.10.8, with its constant $C_{n,p}$, and the change of variables $\zeta \mapsto z+\varrho \zeta $, whose Jacobian is $\varrho ^{2n}$, give

Since $|Q|=h_ j^{2n}$ and $(\varrho /h_ j)^{2n}=32^ n$, we obtain (7.11.6) with $D_{n,p}=32^ nC_{n,p}^ p$.

For (3), let $\varphi \in \mathcal E_0(B)$. Since $\varphi \le 0$ is plurisubharmonic on $B$, (7.11.6) applies to $v=\varphi $ at every point $z$ of every cube with $\mu (Q)>0$. Integrating over $z\in Q$ with respect to $\mu $ and summing over the cubes, we obtain

and (7.11.7) follows from (7.11.4).

For (4), note that $\mu _ j$ equals $f_ j\eta ^ n$ for a bounded density $f_ j\ge 0$, so the solution $U_ j$ exists by theorem 7.10.2, applied with zero boundary values. Since $U_ j$ is subharmonic (theorem 5.9.1) with boundary limits $0$, the maximum principle (proposition 3.4.7) gives $U_ j\le 0$, and its Monge–Ampère measure has finite mass $\mu _ j(B)=m$; hence $U_ j\in \mathcal E_0(B)$. Taking $\varphi =U_ j$ in (7.11.7), we get

If $e_ p(U_ j)>0$, dividing by $e_ p(U_ j)^{p/(n+p)}$ gives the asserted bound; if $e_ p(U_ j)=0$, there is nothing to prove.

Finally write $B=B(a,R)$ and $\rho (z)=|z-a|^2-R^2$. The function $\rho $ is bounded and plurisubharmonic on $B$, and $(\mathrm{dd}^{\mathrm{c}}\rho )^ n=\eta ^ n=n!\pi ^{-n}\, \mathrm dV$. Applying theorem 7.11.2 to the finite measure $(\mathrm{dd}^{\mathrm{c}}\rho )^ n$ on $B$, with $p=2$, and using the uniform bound for $e_2(U_ j)$, we obtain $S=\sup _ j\int _ BU_ j^2\, \mathrm dV<\infty $. Since $U_ j\le 0$, theorem 5.10.4 shows that every subsequence has a further subsequence converging in $L^1_{\mathrm{loc}}(B)$ to a plurisubharmonic function $u$; convergence to $-\infty $ is excluded by the $L^2$ bound. Exhausting $B$ by compact sets and passing to a diagonal subsequence, we may assume that $U_ j\to u$ almost everywhere on $B$, so Fatou's lemma gives $\int _ Bu^2\, \mathrm dV\le S$. For a compact set $L\subset B$, the Cauchy–Schwarz inequality gives

which is small when $|B\setminus L|$ is small. Since $U_ j\to u$ in $L^1(L)$, it follows that $U_ j\to u$ in $L^1(B)$ along this subsequence. In particular, if a subsequence of $(U_ j)$ converges in $L^1(B)$, its limit is represented by such a function $u$. Finally $u\le 0$ almost everywhere, hence everywhere, by the uniqueness assertion of theorem 5.9.1 applied to $\max \{ u,0\} $ and $0$. $\square$