Theorem 16.9.59E1576893E02B

Proof. Fix $\lambda '>\lambda >0$. We want to find $j_0>0$ so that for $j\geq j_0$, (16.9.60) holds.

Step 1. We first assume that $\varphi $ has analytic singularities.

Let $\pi \colon Y\rightarrow X$ be a log resolution of $\varphi $ and let $E_1,\ldots ,E_ N$ be all prime divisors in the polar locus of $\varphi $ on $Y$. Recall that by theorem 5.16.18, a non-zero local holomorphic function $f$ lies in the right-hand side of (16.9.60) if and only if for all $i=1,\ldots ,N$,

where $A_ X$ denotes the log discrepancy. Similarly, if $f$ lies in the left-hand side of (16.9.60), then for all $i=1,\ldots ,N$,

As Lelong numbers are continuous with respect to $d_ S$ by theorem 16.9.55, we can find $j_0>0$ so that when $j\geq j_0$, $\lambda '\nu (\varphi _ j,E_ i)\geq \lambda \nu (\varphi ,E_ i)$ for all $i$. In particular, (16.9.61) follows.

Step 2. We handle the general case.

By corollary 16.9.36, we are free to increase $\theta $ and assume that $\theta _{\varphi }$ is a Kähler current.

Take a quasi-equisingular approximation $(\psi _ k)_ k$ of $\varphi $ in $\mathrm{PSH}(X,\theta )$. The existence is guaranteed by theorem 5.18.11. Take $\lambda ''\in (\lambda ,\lambda ')$. Then by definition of quasi-equisingular approximations (cf. (5.18.9)), we can find $k>0$ so that

Observe that $\psi _ k\geq \varphi $ for all $k>0$. By proposition 16.9.42, we then find that for any $k>0$, we have $\varphi _ j\lor \psi _ k\xrightarrow {d_ S}\varphi \lor \psi _ k=\psi _ k$ as $j\to \infty $. By Step 1, we can find $j_0>0$ so that for $j\geq j_0$,

It follows that for $j\geq j_0$,

This completes the proof. $\square$