Theorem 9.11.26.D12D6E199E87 The right-hand side of (9.11.25) is the same for every resolution, including resolutions that modify regular points. It is a positive $\sigma $-finite Borel measure, gives zero mass to $X_{\mathrm{sing}}$ and to locally pluripolar sets, and may have infinite total mass; for bounded $u$ it is finite. Put $r_ n=\max (n-2,0)$. There are finite constants $C$ and, for each $q>r_ n$, $C_ q$, depending only on the fixed background data and, for $C_ q$, on $q$, such that for every nonpositive $u\in \mathrm{PSH}(X,\theta )$ and every $p>0$,
\[ \int _ X\mathrm{e}^{pu}\, \mathrm{MA}^{\mathrm{ext}}_{X,\theta }(u) \leq C(1+p^{-1})^{r_ n},\qquad \int _ X(1-u)^{-q}\, \mathrm{MA}^{\mathrm{ext}}_{X,\theta }(u) \leq C_ q. \]
The constants may be obtained on any one fixed resolution.
Proof.
Given two resolutions, choose a proper analytic subset $S\subset X$ containing $X_{\mathrm{sing}}$ and the images of both exceptional sets. Each map is an isomorphism over $X\setminus S$, and there the two measures are given by the same bounded local products, followed by truncation and exhaustion. On each resolution, the inverse image of $S$ is a proper analytic subset and has zero extended mass by theorem 13.14.3. Hence both pushforwards give $S$ zero mass. This proves independence, including for resolutions whose exceptional sets lie over regular points.
Positivity follows from positivity on $Y$. On $X\setminus S$ either resolution is a Borel isomorphism, so a countable exhaustion of $Y$ by sets of finite measure pushes forward to one on $X$; adjoining the null set $S$ proves $\sigma $-finiteness. A locally pluripolar subset of $X$ pulls back to a pluripolar set off the exceptional locus, and the exceptional locus itself has zero mass. This gives the asserted vanishing.
If $u$ is bounded, its pullback is bounded on the compact manifold $Y$, and local bounded-product estimates give finite total mass. For general nonpositive $u$, apply theorem 13.15.23 on $Y$ to the smooth background $\pi ^*\theta $. Pushforward and the identity $u\circ \pi $ identify the weighted integrals exactly, and the constants come from $\pi ^*\theta $ and a Hermitian metric on $Y$.
$\square$
Pome
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