(i)$\Longrightarrow$(ii). Let $u\in C^{0}(X)$. By (i) there exists a decreasing net $(\varphi_{j})$ of $\theta$-psh model functions converging pointwise to $P_{\theta}(u)$. Since $P_{\theta}(u)\leq u$, we see that the compact set $X$ is for each $\varepsilon>0$ the increasing union of the open sets $\{\varphi_{j}<u+\varepsilon\}$, hence $\varphi_{j}<u+\varepsilon$ for some $j$. Since $\varphi_{j}-\varepsilon$ is $\theta$-psh and dominated by $u$, we get $\varphi_{j}-\varepsilon\leq P_{\theta}(u)$ by definition of the envelope, which proves (ii).

(ii)$\Longrightarrow$(iii). Since the set of $\varphi\in\mathcal{D}(X)\cap\PSH(X,\omega)$ such that $\varphi\leq u$ is stable by max, (ii) shows that we can construct an increasing family $\varphi_{j}\in\mathcal{D}(X)\cap\PSH(X,\omega)$ converging pointwise to $P_{\theta}(u)$. But $P_{\theta}(u)-\varphi_{j}$ is usc for each $j$, and Dini’s lemma therefore shows that the convergence is uniform on $X$.

(iii)$\Longrightarrow$(i). Let $\varphi$ be a $\theta$-psh function. We first claim that for each $x\in X$ we have

Indeed, given $\varepsilon>0$ there exists $u\in C^{0}(X)$ such that $u\geq\varphi$ and $u(x)\leq\varphi(x)+\varepsilon$, simply because $\varphi$ is usc. Since $\varphi$ is $\theta$-psh we have $\varphi\leq P_{\theta}(u)$. By (iii) may then find $\psi\in\mathcal{D}(X)\cap\PSH(X,\omega)$ such that $P_{\theta}(u)\leq\psi\leq P_{\theta}(u)+\varepsilon$. We thus have $\psi\geq\varphi$ and $\psi(x)\leq\varphi(x)+2\varepsilon$, and the claim follows.

Now consider the set $I$ of all $\psi\in\mathcal{D}(X)\cap\PSH(X,\omega)$ such that $\psi>\varphi$ on $X$. Note that this condition implies that $\psi\geq\varphi+\varepsilon$ for some $\varepsilon>0$, since $\varphi-\psi$ is usc. We claim that $I$ is a directed set, which will conclude the proof. To see this, let $\psi_{1},\psi_{2}\in I$ and choose $\varepsilon>0$ such that $\min\{\psi_{1},\psi_{2}\}\geq\varphi+3\varepsilon$. We then also have $P(\min\{\psi_{1},\psi_{2}\})\geq\varphi+3\varepsilon$. By (iii) we find $\psi_{3}\in\mathcal{D}(X)\cap\PSH(X,\omega)$ such that $|\psi_{3}-P(\min\{\psi_{1},\psi_{2}\})|\leq\varepsilon$. Then $\varphi+\varepsilon\leq\psi_{3}-\varepsilon\leq\min\{\psi_{1},\psi_{2}\}$, which concludes the proof. ∎