Set $\psi_{j}:=\max(\psi,-j)\in PSH(X,\omega)\cap L^{\infty}(X)$. Observe that the probability measures $(\omega+dd^{c}\psi_{j})^{n}$ converge in $X\setminus E$ towards the measure $\nu=e^{t\psi}\mu$. Since $\nu(X)=\nu(X\setminus E)=1$, it follows that $(\omega+dd^{c}\varphi_{j})^{n}$ converges to $\nu$ on all of $X$.

Fix $\varepsilon>0$ and set $v_{\varepsilon}:=(\psi+\varepsilon v)/(1+\varepsilon)\in PSH(X,\omega)$, where $v\in PSH(X,\omega)$, $v\leq 0$, is such that $e^{v}$ is continuous and $(v=-\infty)=E$. It follows from lemma 2.2 that for all $s>0$,

$$Cap_{\omega}(\psi_{j}<-s-1)\leq\int_{(\psi_{j}<-s)}(\omega+dd^{c}\psi_{j})^{n}\leq\int_{(v_{\varepsilon}\leq-s/(1+\varepsilon))}(\omega+dd^{c}\psi_{j})^{n}.$$

Observe that $e^{v_{\varepsilon}}$ is continuous on $X$, hence the sublevel sets $(v_{\varepsilon}\leq c)$ are compact. We infer, letting $j\rightarrow+\infty$,

$$Cap_{\omega}(\psi<-s-1)\leq\int_{(v_{\varepsilon}\leq-s/(1+\varepsilon))}e^{t\psi}d\mu.$$

Letting $\varepsilon$ go to zero and using that $\mu(X)=1$ yields

$$Cap_{\omega}(\psi<-s-1)\leq\int_{(\psi<-s)}e^{t\psi}d\mu\leq e^{-s}.$$

Therefore the capacity of the sublevel sets of $\psi$ decreases fast as $s\rightarrow+\infty$, hence by lemma 6.2 in [GZ 2] we get $\psi\in{\mathcal{E}}^{1}(X,\omega)$. Since $e^{-t\varphi}(\omega+dd^{c}\varphi)^{n}\equiv e^{-t\psi}(\omega+dd^{c}\psi)^{n}$, it follows from proposition 4.3 that $\varphi\equiv\psi$. ∎