More precisely we are going to show the following: consider $\Omega$ an open subset of $X$, $v\in PSH^{-}(\Omega)$ and $P\subset\{v=-\infty\}$. Fix $0<\varepsilon<1/n$ and $V_{t}:=V_{G_{t},\omega}$ where $G_{t}=\{x\in\Omega\,/\,v(x)<-t\}$. Then

$$\varphi_{\varepsilon}(x):=\frac{1}{\varepsilon}\int_{1}^{+\infty}\frac{1}{t^{1+\varepsilon}}[V_{t}(x)-\sup_{X}V_{t}]dt$$

is a $\omega$-psh function such that $P\subset\{\varphi_{\varepsilon}=-\infty\}$.

Indeed since $G_{t}$ is open, we have $V_{t}\in PSH(X,\omega)$ and $V_{t}=0$ on $G_{t}$ (see proposition 3.6). Observe that $\varphi_{\varepsilon}$ is a sum of negative $\omega$-psh functions hence it is either identically $-\infty$ or a well defined $A\omega$-psh function with $A=\varepsilon^{-1}\int_{1}^{+\infty}t^{-(1+\varepsilon)}dt=1$. Recall that $-C+\sup_{X}V_{t}\leq\int_{X}V_{t}\omega^{n}\leq\sup_{X}V_{t}$ (proposition 1.7). Therefore $\int_{X}\varphi_{\varepsilon}\omega^{n}\geq-C$ hence $\varphi_{\varepsilon}\in PSH(X,\omega)$.

Fix $x\in\Omega$ such that $v(x)<-1$. Observe that $V_{t}-\sup_{X}V_{t}\leq 0$ with $V_{t}(x)=0$ if $x\in G_{t}$, i.e. when $|v(x)|>t$. Therefore

$$\varphi_{\varepsilon}(x)\leq-\frac{1}{\varepsilon}\int_{1}^{|v(x)|}\frac{\sup_{X}V_{t}}{t^{1+\varepsilon}}dt.$$

Recall now that $Cap_{\omega}$ is always dominated by $Cap_{BT}$ hence $Cap_{\omega}(G_{t})\leq C_{1}/t<1$ if $t$ is large enough. We infer from the previous proposition that

$$-\sup_{X}V_{t}\leq-[Cap_{\omega}(G_{t})]^{-1/n}\leq-C_{2}t^{1/n},$$

which yields

$$\varphi_{\varepsilon}(x)\leq\frac{-C_{2}}{\varepsilon}\int_{1}^{|v(x)|}\frac{dt}{t^{1+\varepsilon-1/n}}\leq-C_{3}|v(x)|^{1/n-\varepsilon}+C_{4}.$$

Note that $\varphi_{\varepsilon}(x)=-\infty$ whenever $v(x)=-\infty$ hence $P\subset\{\varphi_{\varepsilon}=-\infty\}$. ∎