We first show that $(\varphi_{j})$ converges to $\varphi$ in $L^{1}(\mu)$. Observe that the sequence $(\varphi_{j})$ is uniformly bounded: this follows from Theorem 2.1 since $(\omega+dd^{c}\varphi_{j})^{n}$ satisfies ${\mathcal{H}}(\alpha,A_{j},\omega)$, where $A_{j}=e^{t\sup_{X}\psi_{j}-c_{\psi_{j}}}A\leq e^{tM_{\mu}}A$ is bounded from above. It follows then from standard arguments that $\int_{X}\varphi_{j}d\mu\rightarrow\int_{X}\varphi d\mu$ (see e.g. the proof of lemma 5.2 in [Ce]).

Fix $\varepsilon>0$ and let $G$ be an open set of $X$ such that $\varphi$ is continuous on $X\setminus G$ and $Cap_{\omega}(G)\leq\varepsilon$ (see corollary 3.8 in [GZ 1]). By Hartogs’ lemma, $\varphi_{j}\leq\varphi+\varepsilon$ on the compact set $X\setminus G$, if $j\geq j_{\varepsilon}$. Observe that

$$\int_{X\setminus G}|\varphi-\varphi_{j}|d\mu\leq 2\varepsilon+\int_{X\setminus G}[\varphi-\varphi_{j}]d\mu\leq 3\varepsilon,$$

if $j\geq j^{\prime}_{\varepsilon}$. O the other hand since $\mu$ satisfies ${\mathcal{H}}(\alpha,A,\omega)$, we get

$$\int_{G}|\varphi-\varphi_{j}|d\mu\leq 2M\mu(G)\leq 2MA\varepsilon^{1+\alpha},$$

where $M=\sup_{j}||\varphi_{j}||_{L^{\infty}(X)}$. This shows that $||\varphi-\varphi_{j}||_{L^{1}(\mu)}\rightarrow 0$.

The proof for $(e^{t\psi_{j}})$ is similar: it suffices to note that the functions $u_{j}:=e^{t\psi_{j}-t\sup_{X}\psi_{j}}$ are $\omega$-psh and uniformly bounded. One can then apply the rest of the argument. ∎