Fix $\Omega$ a Kähler form on $X$, and set $\omega_{j}:=\omega+\varepsilon_{j}\Omega$, where $\varepsilon_{j}>0$ decreases to $0$. We start by showing that ${\mathcal{E}}^{1}(X,\omega_{j})\subset L^{1}(\mu)$.

Fix $\varphi\in{\mathcal{E}}^{1}(X,\omega_{j})$. We can assume without loss of generality that $\sup_{X}\varphi=-1$. It follows from propositions 3.6 and 2.7 in [GZ 1] that there exists a constant $C=C(\omega,\Omega)>0$ independent of $j$ such that $Cap_{\omega_{j}}(\varphi<-t)\leq C/t$ for all $t>0$. Since $Cap_{\omega}(\cdot)\leq Cap_{\omega_{j}}(\cdot)$, the measure $\mu$ satisfies ${\mathcal{H}}(\alpha,A,\omega_{j})$. We infer

with an upper-bound which is independent of $j$.

The main result in [GZ 2] guarantees in this case that there exists a unique function $\varphi_{j}\in{\mathcal{E}}^{1}(X,\omega_{j})$ such that

where $\lambda_{j}=\int_{X}(\omega+\varepsilon_{j}\Omega)^{n}>1$ decreases to 1 as $j$ goes to infinity.

The normalization $\sup_{X}\varphi_{j}=-1$ implies that the sequence $(\varphi_{j})$ is relatively compact in $L^{1}(X)$ (see proposition 2.7 in [GZ 1]). Let $\varphi$ be a cluster point of $(\varphi_{j})$. Relabelling if neccessary, we assume $\varphi_{j}\rightarrow\varphi$ in $L^{1}(X)$. Note that $\varphi\in PSH(X,\omega)$ and $\sup_{X}\varphi=-1$ (by Hartogs’ lemma, see proposition 2.7, [GZ 1]). We are going to show that $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ and $\omega_{\varphi}^{n}=\mu$.

Set $\Phi_{j}:=(\sup_{l\geq j}\varphi_{l})^{*}$, where $u^{*}$ denotes the upper-semi-continuous regularization of $u$. Then $\Phi_{j}\in PSH(X,\omega_{j})$ with $\Phi_{j}\geq\varphi_{j}$, hence $\Phi_{j}\in{\mathcal{E}}^{1}(X,\omega_{j})$ (see proposition 3.2 in [GZ 2]), and $\Phi_{j}$ decreases towards $\varphi$. For $l\geq j$, we have

It follows therefore from an inequality due to J.-P.Demailly [Dem 1] that $(\omega_{j}+dd^{c}\Phi_{j})^{n}\geq\mu$. Now by (1) and lemma 7.2 in [GZ 2],

is uniformly bounded with respect to $j$ thanks to (1).

We infer from proposition 1.2 that $\varphi\in{\mathcal{E}}^{1}(X,\omega)$ and $(\omega_{j}+dd^{c}\Phi_{j})^{n}\rightarrow(\omega+dd^{c}\varphi)^{n}$. Thus $(\omega+dd^{c}\varphi)^{n}\geq\mu$, but these are two probability measures, whence $\mu=(\omega+dd^{c}\varphi)^{n}$. The uniqueness of $\varphi$ follows from Theorem 7.4, [GZ 2]. ∎