Set $\omega_{j}:=\omega+\varepsilon_{j}\Omega$. We can assume w.l.o.g. that $\varphi_{j}\leq\varphi\leq 0$. Set

Observe that, $K$ being fixed, $(\varphi_{j}^{K})_{j}$ is uniformly bounded and decreases towards $\varphi^{K}$ as $j$ goes to infinity. Therefore $(\omega_{j}+dd^{c}\varphi_{j}^{K})^{n}\rightarrow(\omega+dd^{c}\varphi^{K})^{n}$, by a classical result of E.Bedford and A.Taylor [BT 82]. Moreover the sequence of positive measures $(-\varphi_{j}^{K})(\omega_{j}+dd^{c}\varphi_{j}^{K})^{n}$ has uniformly bounded mass, since by lemma 7.2 in [GZ 2],

where $M:=\sup_{j}\int_{X}(-\varphi_{j})(\omega_{j}+dd^{c}\varphi_{j})^{n}<+\infty$.

Since $\varphi_{j}$ is u.s.c., a standard argument yields that any cluster point $\nu$ of the sequence $(-\varphi_{j}^{K})(\omega_{j}+dd^{c}\varphi_{j}^{K})^{n}$ satisfies $0\leq(-\varphi^{K})(\omega+dd^{c}\varphi^{K})^{n}\leq\nu$. In particular

is bounded from above uniformly with respect to $K$. Since $\varphi^{K}$ decreases towards $\varphi$, this shows $\varphi\in{\mathcal{E}}^{1}(X,\omega)$.

It remains to show that $(\omega_{j}+dd^{c}\varphi_{j})^{n}\rightarrow(\omega+dd^{c}\varphi)^{n}$. Since $(\omega_{j}+dd^{c}\varphi_{j}^{K})^{n}\rightarrow(\omega+dd^{c}\varphi^{K})^{n}$ for any fixed $K$, it is enough to get an upper bound on the mass of $(\omega_{j}+dd^{c}\varphi_{j}^{K})^{n}$ in $(\varphi_{j}\leq-K)$ which is uniform in $j$. This follows from Chebyshev inequality, namely

This yields the desired result. ∎

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]. ∎