18.28.19 Partial equilibrium measuresB13AC61C497D

Proof. Step 1. We first prove (18.28.22) when $\varphi = V_{\theta }$.

Let $S\subseteq X$ be a proper analytic subset, such that $V_{\theta }$ is locally bounded on $X\setminus S$. Such $S$ exists by 13.13.8. Fix holomorphic balls $B\Subset B'\Subset X\setminus S$.

By proposition 5.14.3 and 18.28.9, we can take an increasing sequence $(\eta _ j)_ j$ in $\mathrm{PSH}(X,\theta )$ such that

Note that $P_{\theta ,K}[V_{\theta }](v)\sim V_{\theta }$ by 18.28.15. Take $C>0$ so that

Then we may replace $\eta _ j$ by $\eta _ j\lor (P_{\theta ,K}[V_{\theta }](v)-C)$ and assume that $\eta _ j\sim V_{\theta }$ for each $j$. In particular, $\eta _ j$ is locally bounded on $X\setminus S$.

For each $j\geq 1$, let $\gamma _ j\in \mathrm{PSH}(X,\theta )$ be the balayage of $\eta _ j$ on $B$, defined in proposition 5.17.9. Note that $(\gamma _ j)_ j$ is increasing due to proposition 5.14.22.

For the first equality in (18.28.22), we may assume that $B\Subset X\setminus (S \cup K)$. It suffices to show that $\theta _{P_{\theta ,K}[V_{\theta }](v)}^ n$ does not charge $B$.

Since $B$ lies outside $K$ by assumption, we also have $\gamma _ j|_ K=\eta _ j|_ K \leq v$ for all $j\geq 1$. In particular, $\gamma _ j\leq P_{\theta ,K}[V_{\theta }](v)$.

It follows that $\gamma _ j \nearrow P_{\theta ,K}[V_{\theta }](v)$ almost everywhere as well. Testing against non-negative continuous functions supported in $B$ and using theorem 13.13.9, we find that $\theta ^ n_{P_{\theta ,K}[V_{\theta }](v)}$ does not charge $B$, as desired.

We next prove the second equality in (18.28.22). The proof is essentially the same as that of 13.13.21.

It suffices to show that $\theta ^ n_{P_{\theta ,K}[V_{\theta }](v)}$ does not charge the set $K\cap \{ P_{\theta ,K}[V_{\theta }](v)|_ K<v\} $.

Let $x\in (X\setminus S) \cap K$ be a point such that $P_{\theta ,K}[V_{\theta }](v)(x)<v(x)-\epsilon $ for some $\epsilon >0$. We may assume that $x\in B$ and $B'$ is small enough, so that

1. B2B18BBFA88D

   $P_{\theta ,K}[V_{\theta }](v)|_{\overline{B'}}<v(x)-\epsilon $.

2. D0D5ED334E0E

   $v|_{\overline{B'}\cap K}>v(x)-\epsilon $.

By 1, we have $\gamma _ j|_{\partial B}\leq v(x)-\epsilon $, hence by proposition 5.14.22, $\gamma _ j|_ B\leq v(x)-\epsilon $. By 2, we have $\gamma _ j|_{B\cap K}\leq v|_{B\cap K}$. On the other hand, $\gamma _ j|_{K\setminus B}=\eta _{j}|_{K\setminus B}\leq v|_{K\setminus B}$. Therefore, $\gamma _ j|_ K\leq v$. It follows again that $\gamma _ j\leq P_{\theta ,K}[V_{\theta }](v)$. Therefore, we conclude as in the previous paragraph that $\gamma _ j\nearrow P_{\theta ,K}[V_{\theta }](v)$. Thus, testing against non-negative continuous functions supported in $B$, we conclude that $\theta ^ n_{P_{\theta ,K}[V_{\theta }](v)}$ does not charge $B$ again by theorem 13.13.9.

Step 2. We handle the general case.

By corollary 18.28.17,

Now corollary 16.3.14 gives

where in the second inequality we have used Step 1. Hence (18.28.23) follows, and a fortiori (18.28.22) follows as well. $\square$

Proof. The first equation in (18.28.25) follows from 18.28.21. For the second, it suffices to observe that

by corollary 18.28.14. $\square$

Proof. We may assume that $\varphi $ is $\mathcal{I}$-model by replacing $\varphi $ by $P_{\theta }[\varphi ]_{\mathcal{I}}$.

Note that it suffices to prove (18.28.28) at $t=0$, which is equivalent to

It suffices to prove the right derivative at $0$ for an arbitrary $f$; applying the same argument to $-f$ then gives the left derivative. In other words, we need to prove that

We first reduce the proof of (18.28.30) to the case $f\leq 0$. Assume that this case is known. Choose $A\in \mathbb R$ such that $f-A\leq 0$ on $K$. Since for any $t>0$,

then we have

By (16.5.13), for any $t>0$,

So the limit in (18.28.30) exists and is equal to the left-hand side of (18.28.30) with $f-A$ in place of $f$ plus $A\int _ X\theta _{\varphi }^ n$. So (18.28.30) for $f-A$ implies the desired formula for $f$ in view of 18.28.15 and the monotonicity theorem theorem 13.13.11. Hence we may assume that $f\leq 0$.

Fix $t>0$. Set

By the comparison principle (16.5.6) and proposition 16.5.30, we find

By 18.28.21 and corollary 18.28.14, the measure $\theta _{\eta _0}^ n$ is carried by $K\cap \{ \eta _0=v\} $, and the measure $\theta _{\eta _ t}^ n$ is carried by $K\cap \{ \eta _ t=v+tf\} $. On $K\cap \{ \eta _0=v\} $, we have $\eta _ t\leq v+tf=\eta _0+tf$, hence

On $K\cap \{ \eta _ t=v+tf\} $, we have $\eta _0\leq v$, hence

It follows that

and

Thus, letting $t\to 0+$, we get the inequality

Similarly, for $t>0$, we have

Letting $t\to 0+$, we find

Here we used the following elementary continuity consequence:

for $t>0$, and hence $P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v+tf) \nearrow P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)$ almost everywhere when $t\to 0+$. Next we claim that

Choose a continuous extension $\tilde f\in C^0(X)$ of $f$. Since 18.28.21, corollary 18.28.14 show that all the measures appearing here are carried by $K$, the integrals over $K$ against $f$ are the same as the integrals over $X$ against $\tilde f$. Thus corollary 13.13.12 applies.

We conclude (18.28.29). $\square$

Proof. Step 1. First we construct $(v^-_ j)_ j$.

Let

Since $K$ is non-pluripolar, we have $C_{K,v} \in \mathbb {R}$ by 5.17.4. Set

Now define $\widetilde v \colon X \to \mathbb {R}$ as

Since $M_{K,v}\geq \sup _ K v$, the function $\widetilde v$ is lower semicontinuous. Therefore there exists an increasing and uniformly bounded sequence $(v^-_ j)_ j$ in $C^\infty (X)$ such that $v^-_ j \nearrow \widetilde v$ and

The right-hand side can clearly be bounded by a constant as in (18.28.32).

Step 2. We construct $(v_ j^+)_ j$.

Since

is usc, there exists a decreasing and uniformly bounded sequence $(v^+_ j)_ j$ in $C^\infty (X)$ such that $v^+_ j \searrow h$. We may guarantee that

Note that $\inf _ K v-1\leq h\leq C_{K,v} \lor (\inf _ K v-1)$. So (18.28.32) is satisfied.

Step 3. We finish the proof. Fix $\varphi \in \mathrm{PSH}(X,\theta )_{>0}$ and $\delta \in [0,1]$.

Note that 2 follows from corollary 18.28.14, 18.28.15, proposition 16.5.37 and 1. Let us prove 1.

Step 3.1. We prove 1 for $(v^-_ j)_ j$.

Observe that $P_{\theta + \delta \omega ,X}[\varphi ](v^-_ j)$ is increasing in $j\geq 1$, and

We claim that

Let $\eta \in \mathrm{PSH}(X,\theta +\delta \omega )$ be a potential such that $\eta \preceq \varphi $ and $\eta |_ K\leq v$. Take some $\epsilon >0$. Then, since $\eta $ is upper semicontinuous, $\widetilde v$ is lower semicontinuous and $\eta < \epsilon +\widetilde v$, compactness gives $j_0>0$ such that $\eta < \epsilon + v^-_ j$ for $j \geq j_0$. Hence,

for $j\geq j_0$. Therefore,

Letting $\epsilon \to 0+$, we find

Since $\eta $ is arbitrary, (18.28.33) follows.

Step 3.2. We prove 1 for $(v^+_ j)_ j$.

Note that $P_{\theta + \delta \omega ,X}[\varphi ](v^+_ j)$ is decreasing in $j$, and

In particular, $\chi \in \mathrm{PSH}(X,\theta +\delta \omega )_{>0}$, since $P_{\theta + \delta \omega ,K}[\varphi ](v)$ has the same singularity types as $P_{\theta + \delta \omega }[\varphi ]$ by 18.28.15. Observe that

for all $j\geq 1$, since $v_ j^+$ is continuous. Thus, $\chi \leq h$. On the other hand, $h|_ K\leq v$ quasi-everywhere. So in particular, $\chi |_{K}\leq v$ quasi-everywhere. On the other hand,

by 18.28.15. Hence, by corollary 18.28.13,

Therefore, (18.28.34) is an equality. $\square$

Proof. We only give the proof for the relative $\mathcal{I}$-envelopes; the proof for the relative $P$-envelopes is identical, replacing $P_{\theta }[\bullet ]_{\mathcal{I}}$ by $P_{\theta }[\bullet ]$ and using corollary 18.28.13 for the final candidate condition.

Set

By 18.28.15, $d_ S(\eta _ j,P_{\theta }[\varphi _ j]_{\mathcal{I}})=0$ and $d_ S(\eta ,P_{\theta }[\varphi ]_{\mathcal{I}})=0$ for all $j$. Moreover, theorem 16.9.49 gives

in case 1, and the same convergence holds in case 2 by corollary 16.9.12 and theorem 16.9.49. Consequently, $\eta _ j\xrightarrow {d_ S}\eta $ in both cases.

Assume first that $(\varphi _ j)_ j$ is decreasing. Then $(\eta _ j)_ j$ is decreasing and $\eta _ j\geq \eta $. Let $\eta _\infty \coloneqq \inf _ j \eta _ j$. The Monge–Ampère masses of $\eta _ j$ converge to the mass of $\eta $, by the preceding $d_ S$-convergence and theorem 16.9.30. Since $\eta _ j\geq \eta _\infty \geq \eta $, the monotonicity theorem theorem 13.13.11 gives

Hence $\eta _ j\xrightarrow {d_ S}\eta _\infty $ by corollary 16.9.14. The uniqueness of the $d_ S$-limit gives $d_ S(\eta _\infty ,\eta )=0$. Since $\eta _\infty \geq \eta $, proposition 16.9.6 implies $\eta _\infty \sim _ P \eta $. In particular, $\eta _\infty \preceq _{\mathcal{I}}\varphi $ by corollary 16.8.14. The condition $\eta _\infty \leq v$ on $K$ quasi-everywhere is inherited from the $\eta _ j$'s, so $\eta _\infty $ is a candidate for $P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)$. Thus $\eta _\infty \leq \eta $, and the reverse inequality was already observed. This proves the $\mathcal{I}$-part of 1.

Assume next that $\varphi _ j\nearrow \varphi $ almost everywhere. Then $(\eta _ j)_ j$ is increasing and $\eta _ j\leq \eta $. Put $\tilde\eta \coloneqq \operatorname *{sup}_{}{}\! {}^{*}_ j \eta _ j$. By corollary 18.28.11 and corollary 16.9.12, we have $\eta _ j\xrightarrow {d_ S}\tilde\eta $. The $d_ S$-convergence above therefore gives $d_ S(\tilde\eta ,\eta )=0$, hence $\tilde\eta \sim _ P \eta $ by proposition 16.9.6. In particular, $\tilde\eta \preceq _{\mathcal{I}}\varphi $ by corollary 16.8.14. Also $\tilde\eta \leq v$ on $K$ quasi-everywhere by proposition 5.14.5. Hence $\tilde\eta $ is a candidate for $P_{\theta ,K}[\varphi ]_{\mathcal{I}}(v)$, so $\tilde\eta \leq \eta $. Since $\eta _ j\leq \eta $ for all $j$, this proves $\eta _ j\nearrow \eta $ almost everywhere. $\square$