Theorem 13.13.24A60C61FA5696

Proof. Step 1. We first reduce to the case where $f$ is bounded.

Step 1.1. Reduce to the case where $f\leq 0$.

Take $C\in \mathbb {R}$ so that $P_{\theta }(f)\leq C$. Then

Thus we can replace $f$ by $\min \{ f,C\} $. Replacing $f$ by $f-\sup _ X f$, we reduce easily to the case where $f\leq 0$.

Step 1.2. Reduce to the case where $f$ is bounded. Assume that in this case, the theorem is known. We prove (13.13.25) for $f\leq 0$.

For each $j>0$, we have

where $f_ j=f\lor (-j)$.

Now fix $C>0$ and two open sets $G'\Subset G\Subset X\setminus \operatorname{nK}(\{ \theta \} )$. Fix a smooth function $\chi \colon X\rightarrow [0,1]$ so that $\chi |_{G'}\equiv 1$ and $\chi $ is supported in $G$.

Define

Then $U_ C$ is $\mathcal{F}$-open. It follows from proposition 13.11.4 1 that for any $j>0$,

In particular, thanks to (13.13.26),

Note that the $P_{\theta }(f_ j)\lor (V_{\theta }-C)$'s for various $j$ are uniformly bounded from below on $U_ C$, thanks to 13.13.8.

Next we claim that $(P_{\theta }(f_ j))_ j$ is decreasing and converges to $P_{\theta }(f)$.

The sequence is decreasing; let us compute its limit. In fact, $P_{\theta }(f_ j)\leq f_ j$ quasi-everywhere. It follows that $\inf _ j P_{\theta }(f_ j)\leq f$ quasi-everywhere and hence

The reverse inequality is trivial, and our assertion follows.

Since $P_{\theta }(f)\not\equiv -\infty $, we know that the set $\{ f=-\infty \} $ is pluripolar. It follows that $P_{\theta }(f_ j)-f_ j$ converges to the following function $g\colon X\rightarrow [-\infty ,\infty )$ in capacity:

Therefore,

Next we claim that

Since $g\leq 0$ quasi-everywhere, the left-hand side of (13.13.28) is non-negative. It suffices to prove the $\leq $ direction.

We wish to let $j\to \infty $ directly in (13.13.27), but since $U_ C$ is not open in general, this cannot be done directly. In the sequel, we shall slightly enlarge $U_ C$ to get an open set and then take the limit.

Fix $\epsilon >0$. We can find an open subset $W\Subset X\setminus \operatorname{nK}(\{ \theta \} )$ so that

For example, we can construct $W$ as follows. By example 13.12.5, we can find an open set $A\subseteq G$ so that $\operatorname{Cap}_{C^{-1}\theta }(A)<\epsilon $ and $P_{\theta }(f)-V_{\theta }$ is continuous on $G\setminus A$. Then $U_ C\cap (G\setminus A)$ is relatively open in $G\setminus A$, so we may choose an open set $W_0\subseteq G$ such that

Taking $W=W_0\cup A$, we get $U_ C\subseteq W$ and $W\setminus U_ C\subseteq A$, hence (13.13.29).

Thanks to (13.13.29) and (13.13.27), we have

Letting $j\to \infty $ and applying theorem 13.12.8, we find that

Since $\epsilon >0$ is arbitrary, (13.13.28) follows.

Now letting $C\to \infty $ in (13.13.28), we find

Since $G'\Subset X\setminus \operatorname{nK}(\{ \theta \} )$ is arbitrary, and $\operatorname{nK}(\{ \theta \} )\subseteq X$ is a proper analytic subset, we finally conclude

In other words, $P_{\theta }(f)=f$ almost everywhere with respect to $\theta _{P_{\theta }(f)}^ n$. This proves (13.13.25).

Step 2. We now assume that $-C'\leq f\leq 0$ for some $C'>0$.

For each $j>0$, take an open subset $V_ j\subseteq X$ so that

1. F0D9C03CCEDA

   $\operatorname{Cap}_{C'^{-1}\theta }(V_ j)\leq 2^{-j-1}$, and

2. AF1E921609BC

   $f|_{X\setminus V_ j}$ is continuous.

Without loss of generality, we may assume that the $V_ j$'s are decreasing. Take a continuous function $f_ j\colon X\rightarrow [-C',0]$ extending $f|_{X\setminus V_ j}$. This is always possible thanks to Tietze's extension theorem. For each $j>0$, we let

Then $h_ j$ is lower semicontinuous and $h_ j$ agrees with $f$ outside $V_ j$. Set

Then the set $\{ h\neq f\} $ is contained in the intersection of the $V_ j$'s and hence is a pluripolar set, thanks to theorem 13.12.4. In particular, $P_{\theta }(f)=P_{\theta }(h)$.

Now we can apply 13.13.21 to conclude that

for each $j>0$.

Fix two open subsets $G'\Subset G\Subset X\setminus \operatorname{nK}(\{ \theta \} )$. Note that $-C'\leq h_ j\leq 0$. In particular,

Hence,

Now we can apply the same arguments as in Step 1.2 to conclude that

Since $G'\Subset X\setminus \operatorname{nK}(\{ \theta \} )$, (13.13.25) follows. $\square$