Lemma 5.26.2.88C7A8DCE36E There are constants $\alpha >0$ and $C<+\infty $, depending only on $(X,\omega ,\mathrm{d}V)$, such that
5.26 Uniform exponential integrability5A19CE5C3CCE
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For a single quasi-plurisubharmonic function $v$ on $X$, Skoda's local integrability theorem, recalled in [SGPT], applies after local smooth corrections. Taking the smallest exponent over a finite cover gives $\mathrm{e}^{-a(v)v}\in L^1(X)$ for some $a(v)>0$. Normalized compactness makes the exponent and the bound for the integral uniform: otherwise, a convergent weighted sum of counterexamples would violate this qualitative statement.
Let $K_\omega $ be the constant of theorem 5.25.2 for the background $\omega $, so that every $v\in \mathrm{PSH}(X,\omega )$ with $\sup _ Xv=0$ satisfies
Proof. Suppose that no such pair exists. Set $a_ j=2^{-j}$ and choose normalized potentials $v_ j\in \mathrm{PSH}(X,\omega )$ with
The partial sums $V_ N=\sum _{j\leq N}a_ jv_ j$ decrease, since every $v_ j\leq 0$. They belong to $\mathrm{PSH}(X,\omega )$ by proposition 5.23.2, because $\sum _ ja_ j\leq 1$, and their $L^1$ norms are uniformly bounded by (5.26.1). By corollary 5.25.3, their limit $V$ is therefore $\omega $-plurisubharmonic. Since all summands are nonpositive, $V\leq a_ jv_ j$ for every $j$.
Fix $p>0$. For every $j\geq 1/p$,
Thus $\mathrm{e}^{-pV}$ is not integrable for any $p>0$, which contradicts the integrability of $\mathrm{e}^{-a(V)V}$ obtained above from Skoda's theorem. $\square$
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