PomeLemma 10.7.5.C789A06C06B4 Let $\psi $ be a quasi-plurisubharmonic function on $X$ that is smooth outside its pole set $A=\{ \psi =-\infty \} $. Then there are a Kähler form $\sigma \geq \omega /2$ on $X\setminus A$ and functions $\chi _ R\in C^\infty _ c(X\setminus A)$, $R\geq 1$, with $0\leq \chi _ R\leq 1$, $\chi _ R\to 1$ pointwise as $R\to \infty $, and $|\mathrm d\chi _ R|_\sigma \leq C/R$ for a constant $C$. This applies in particular when $\psi $ is a finite sum of quasi-plurisubharmonic functions with gentle analytic singularities and of quasi-plurisubharmonic functions that are smooth outside a point $x_ k$ and equal to $c_ k\log |z-z(x_ k)|^2$ plus a smooth function near $x_ k$, in local coordinates, with $c_ k>0$; then $A$ is the union of their pole sets.
Lemma 10.7.5C789A06C06B4
Proof. The set $A$ is a null set, since $\psi $ is locally integrable, and in particular $A\neq X$. Choose $b\geq 0$ with $\mathrm{dd}^{\mathrm{c}}\psi \geq -b\omega $, which is possible by lemma 6.14.4(4), and subtract a constant so that $\psi <-2b-2$; this is possible because $\psi $ is upper semicontinuous on the compact manifold $X$. Put $\rho =-\log (-\psi )$ on $X\setminus A$. The chain rule gives
Since $0<-1/\psi <1/(2b+2)$, the first term is at least $-\omega /2$, and since $\partial \rho =-\partial \psi /\psi $, the form $\sigma =\omega +\mathrm{dd}^{\mathrm{c}}\rho $ satisfies
It is a Kähler form on $X\setminus A$. The rank-one form $\frac{\mathrm i}{2\pi }\partial \rho \wedge \bar\partial \rho $ has, in a $\sigma $-unitary coframe, the single eigenvalue $|\partial \rho |_\sigma ^2/(2\pi )$, so the last inequality gives $|\partial \rho |_\sigma ^2\leq 2\pi $; hence $|\mathrm d\rho |_\sigma $ is bounded by a numerical constant.
The function $-\rho =\log (-\psi )$ tends to $+\infty $ at $A$, because $\psi $ is upper semicontinuous and equal to $-\infty $ on $A$, and its sublevel sets $\{ -\rho \leq t\} =\{ \psi \geq -\mathrm e^ t\} $ are closed in $X$, because $\psi $ is upper semicontinuous, and disjoint from $A$; so they are compact subsets of $X\setminus A$. Choose a smooth function $\chi $ on $\mathbb R$ equal to one on $(-\infty ,1]$ and to zero on $[2,\infty )$, with $0\leq \chi \leq 1$, and put $\chi _ R=\chi (-\rho /R)$. Then $\chi _ R$ has compact support in $X\setminus A$, equals one on $\{ -\rho \leq R\} $, and $|\mathrm d\chi _ R|_\sigma \leq \sup |\chi '|\, |\mathrm d\rho |_\sigma /R$. The same gradient bound shows that every path in $X\setminus A$ that eventually leaves every compact subset of $X\setminus A$ has infinite $\sigma $-length, since $-\rho $ tends to $+\infty $ along it, by the compactness of its sublevel sets.
For the last assertion, recall that a function $\varphi $ with gentle analytic singularities of type $(c,\mathfrak a)$ is smooth outside its pole set, where it equals $c$ times the logarithm of the smooth positive function $\exp (\varphi /c)$ (definition 6.14.12). Hence a finite sum as in the statement is quasi-plurisubharmonic and smooth outside the union of the pole sets of its terms, and this union is its pole set, since all terms are bounded above. $\square$
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