Theorem 12.10.2330380C32B7A7

Proof. By theorem 11.1.7, every bounded $\omega $-plurisubharmonic function has mass $V$, and truncation therefore bounds every extended mass by $V$. For $f_ k=\min (f,k)$, integration of the bounded equation gives $c_\omega (f_ k)=V/\int _ Xf_ k\, \mathrm{d}V$, so the definition (15.28.1) gives $b_\omega (f)=V/M$. A threshold subsolution has mass at least $b_\omega (f)M=V$. Its mass is consequently $V$, and the nonnegative difference between its measure and $b_\omega (f)f\, \mathrm{d}V$ has mass zero.

Now let $u\in \mathrm{PSH}(X,\omega )$ have extended mass $V$. Normalize $u\leq 0$, put $\mu _ u=\mathrm{MA}^{\mathrm{ext}}_\omega (u)$, and fix $a>1$. For $u_ j=\max (u,-j)$, let $w_ j=P_\omega (au_ j)$. Comparison with constants gives $-aj\leq w_ j\leq 0$. The bounded contact theorem theorem 16.28.4 shows that $\mathrm{MA}_\omega (w_ j)$ is concentrated on the contact set $C_ j=\{ w_ j=au_ j\} $. Since the two functions are canonical representatives and $w_ j\leq au_ j$ almost everywhere, this inequality holds everywhere. Both functions are $a\omega $-plurisubharmonic, so positivity and proposition 13.14.6 give

For integers $j\geq N$, locality of the products on $\{ u>-N\} $ and the mass identity for bounded functions yield

Choose $N$ with $a^ n\mu _ u(\{ u\leq -N\} )<V$; this is possible because $\mu _ u$ is finite and puts no mass on the pole set of $u$. If $\sup _ Xw_ j<-aN$, the contact set $C_ j$ lies in $\{ u_ j<-N\} $, and the preceding inequalities contradict $\int _ X\mathrm{MA}_\omega (w_ j)=V$. Hence $\sup _ Xw_ j\geq -aN$ for $j\geq N$, and for smaller $j$ the bound $\sup _ Xw_ j\geq -aj$ suffices.

The envelopes $w_ j$ decrease with $j$, and by compactness of normalized quasi-plurisubharmonic functions this uniform lower bound for their suprema prevents their limit from being identically $-\infty $. Their limit $w\in \mathrm{PSH}(X,\omega )$ satisfies $w\leq au$, also at the poles of $u$. With $\delta =1/a$, the function $\delta w\in \mathrm{PSH}(X,\delta \omega )$ is therefore a competitor for $P_{\delta \omega }(u)$, so this envelope is not identically $-\infty $. Since $a>1$ was arbitrary, this holds for every $0<\delta <1$, and the Lelong estimate preceding the theorem completes the proof. The normalization $u\leq 0$ does not affect any conclusion. $\square$