7.11.8 Convergence of energy integrals6293A9FE1CA2

Proof. We show that $\varlimsup _ jv_ j=v$ $\nu $-almost everywhere (Step 1) and that $\int v_ j\, \mathrm d\nu \to \int v\, \mathrm d\nu $ (Step 2), and combine the two facts (Step 3). First, the bounds $-L\le v\le L$ hold almost everywhere: for every compact set $K\subset D$ we have $(|v|-L)_+\le |v-v_ j|$, so $\int _ K(|v|-L)_+\, \mathrm dV\le \| v_ j-v\| _{L^1(K)}\to 0$. They hold everywhere by the uniqueness assertion of theorem 5.9.1, applied on each connected component of $D$ to the plurisubharmonic functions $\max \{ v,-L\} $ and $v$, and to $\max \{ v,L\} $ and the constant $L$.

Step 1. The upper limit of $v_ j$. We show that

For $\ell \ge 1$ let $b_\ell =(\sup _{j\geq \ell }v_ j)^*$ be the upper semicontinuous regularization of the supremum. By theorem 5.10.2, these are plurisubharmonic functions with values in $[-L,L]$, and they decrease as $\ell $ increases. We claim that $b_\ell \downarrow v$ everywhere. Let $z\in D$ and $\varepsilon >0$. By upper semicontinuity, $v<v(z)+\varepsilon $ on a neighborhood of a closed ball $\overline{B'}\subset D$ centered at $z$, and Hartogs' lemma (theorem 5.10.4), applied on the connected component of $D$ containing $z$ with the constant majorant $v(z)+\varepsilon $, gives $v_ j<v(z)+\varepsilon $ on $\overline{B'}$ for all large $j$. Hence $\sup _{j\ge \ell }v_ j\le v(z)+\varepsilon $ on $B'$ for all large $\ell $, so $b_\ell (z)\le v(z)+\varepsilon $, and the decreasing limit of $b_\ell $ is at most $v$.

Conversely, $b_\ell \ge v_ j$ for all $j\ge \ell $, so $(v-b_\ell )_+\le |v-v_ j|$ for $j\ge \ell $; integrating over a compact set $K\subset D$ and letting $j\to \infty $, we find that $b_\ell \ge v$ almost everywhere on $K$, hence almost everywhere on $D$. The decreasing limit of the $b_\ell $ is bounded below by $-L$, so it is plurisubharmonic by theorem 5.10.1; it is at most $v$ and equals $v$ almost everywhere, hence everywhere by the uniqueness of plurisubharmonic representatives, applied on each connected component of $D$.

The sets $N_\ell =\{ \sup _{j\ge \ell }v_ j<b_\ell \} $ are Borel, and by corollary 7.6.2 they satisfy $\operatorname {Cap}_ D^*(N_\ell \cap K)=0$ for every compact $K\subset D$. By countable subadditivity of outer capacity (proposition 7.4.3), the union $N=\bigcup _\ell N_\ell $ has the same property, so $\nu (N)=0$ by hypothesis. For $z\notin N$ we have $\sup _{j\ge \ell }v_ j(z)=b_\ell (z)$ for every $\ell $, and therefore $\varlimsup _ jv_ j(z)=\lim _\ell b_\ell (z)=v(z)$. This proves (7.11.10). The argument applies to every sequence of plurisubharmonic functions with values in $[-L,L]$ that converges to $v$ in $L^1_{\mathrm{loc}}(D)$.

Step 2. Convergence of $\int v_ j\, \mathrm d\nu $. We first construct convex combinations of the $v_ j$ whose integrals converge to $\int v\, \mathrm d\nu $. Let $\mathcal C_\ell $ be the set of finite convex combinations of $v_\ell ,v_{\ell +1},\ldots $; its elements are plurisubharmonic functions with values in $[-L,L]$ (proposition 5.9.5). Let

Since $\mathcal C_{\ell +1}\subset \mathcal C_\ell $, the numbers $d_\ell $ increase, and they are bounded by $L\, \nu (D)^{1/2}$; let $d$ be their limit. Choose $r_\ell \in \mathcal C_\ell $ with

For $k\geq \ell $ the midpoint $(r_ k+r_\ell )/2$ belongs to the convex set $\mathcal C_\ell $, so its norm is at least $d_\ell $, and the parallelogram identity gives

The right-hand side tends to zero as $\ell \to \infty $, so we can choose $\ell _1<\ell _2<\cdots $ with $\| r_{\ell _{i+1}}-r_{\ell _ i}\| _{L^2(\nu )}\le 2^{-i}$. By the triangle inequality in $L^2(\nu )$ and the monotone convergence theorem, the function $g=\sum _ i|r_{\ell _{i+1}}-r_{\ell _ i}|$ satisfies $\| g\| _{L^2(\nu )}\le 1$. Hence $g<\infty $ $\nu $-almost everywhere, and at every point where $g$ is finite the sequence $(r_{\ell _ i})$ converges.

On the other hand, the $r_{\ell _ i}$ are plurisubharmonic with values in $[-L,L]$, and $r_\ell \to v$ in $L^1_{\mathrm{loc}}(D)$, since $\| r_\ell -v\| _{L^1(K)}\le \sup _{j\ge \ell }\| v_ j-v\| _{L^1(K)}$ for every compact $K\subset D$. By Step 1, applied to the sequence $(r_{\ell _ i})$, we have $\varlimsup _ ir_{\ell _ i}=v$ $\nu $-almost everywhere. Therefore $r_{\ell _ i}\to v$ $\nu $-almost everywhere, and dominated convergence, with the constant bound $L$ and the finite measure $\nu $, gives $\int r_{\ell _ i}\, \mathrm d\nu \to \int v\, \mathrm d\nu $.

We now show that $\int v_ j\, \mathrm d\nu \to \int v\, \mathrm d\nu $. Otherwise there are $c>0$ and a subsequence $(v_{j_ k})$ such that either $\int v_{j_ k}\, \mathrm d\nu \ge \int v\, \mathrm d\nu +c$ for all $k$, or $\int v_{j_ k}\, \mathrm d\nu \le \int v\, \mathrm d\nu -c$ for all $k$. The subsequence satisfies the hypotheses of the lemma, so the construction above, applied to it, provides convex combinations of its terms whose integrals converge to $\int v\, \mathrm d\nu $. This is impossible, since every convex combination of the $v_{j_ k}$ satisfies the same inequality as the $v_{j_ k}$ themselves.

Step 3. Conclusion. By Step 1, $\varlimsup _ jv_ j\le v$ everywhere, so $(v_ j-v)_+\to 0$ pointwise. These functions are bounded by $2L$, so $\int (v_ j-v)_+\, \mathrm d\nu \to 0$ by dominated convergence. Since

Step 2 proves the assertion.

Finally, suppose that $\nu \le (\mathrm{dd}^{\mathrm{c}}w)^ n$ with $w$ bounded and plurisubharmonic on $D$, and put $M=\sup _ Dw-\inf _ Dw$. If $M=0$, then $w$ is constant and $\nu =0$. Otherwise $(w-\inf _ Dw)/M$ is a competitor for $\operatorname {Cap}_ D$ (definition 7.4.2), whose Monge–Ampère measure is $M^{-n}(\mathrm{dd}^{\mathrm{c}}w)^ n$. Hence $\nu (A)\leq M^ n\operatorname {Cap}_ D(A)$ for every Borel set $A\subset D$, and $\operatorname {Cap}_ D(A)\le \operatorname {Cap}_ D^*(A)$ because $\operatorname {Cap}_ D$ is increasing. If $E$ is as in the hypothesis, then $\nu (E\cap K)=0$ for every compact $K\subset D$, and exhausting $D$ by countably many compact sets gives $\nu (E)=0$. $\square$

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. $\square$