We write here $V_{E}$ for $V_{E,\omega}$ since $\omega$ is fixed and no confusion can arise.

Let $E$ be an open subset of $X$. Observe that $V_{E}\leq 0$ on $E$, hence $V_{E}^{*}\leq 0$ on $E$ which is open. Therefore $V_{E}^{*}\leq V_{E}$, whence equality. This proves 1).

Let $w\in PSH(X,\omega)$, $w\leq 0$, and fix $P\subset\{w=-\infty\}$. Fix $E$ a Borel subset of $X$. Clearly $V_{E\cup P}\leq V_{E}$ hence $V_{E\cup P}^{*}\leq V_{E}^{*}$. Conversely let $\varphi\in PSH(X,\omega)$ be such that $\varphi\leq 0$ on $E$. Then $\forall\varepsilon>0$, $\psi_{\varepsilon}:=(1-\varepsilon)\varphi+\varepsilon w\in PSH(X,\omega)$ satisfies $\psi_{\varepsilon}\leq 0$ on $E\cup P$, hence $\psi_{\varepsilon}\leq V_{E\cup P}$. Letting $\varepsilon\rightarrow 0$ we infer $\varphi\leq V_{E\cup P}$ on $X\setminus P$, hence $\varphi\leq V_{E\cup P}^{*}$ on $X$. Thus $V_{E}^{*}\leq V_{E\cup P}^{*}$.

Let $E_{j}$ be an increasing sequence of subsets of $X$ and set $E=\cup_{j\geq 1}E_{j}$. Let $v:=\lim\searrow V_{E_{j}}^{*}$ (the limit is decreasing by 3.4.1). If $E$ is $PSH(X,\omega)$-polar then so are all the $E_{j}^{\prime}$s, hence $V_{E}^{*}\equiv+\infty=\lim V_{E_{j}}^{*}$. So let us assume $E$ is not $PSH(X,\omega)$-polar. Then $v\in PSH(X,\omega)$ since $v\geq V_{E,\omega}^{*}\not\equiv-\infty$ (see proposition 1.6.3). Observe that $v=0$ on the set $E\setminus N$, where $N=\cup_{j\geq 1}\{V_{E_{j}}<V_{E_{j}}^{*}\}$. The latter is called a negligible set. It follows from the local theory [5] together with theorem 5.2 that $N$ is $PSH(X,\omega)$-polar. Therefore $V_{E}^{*}\leq v\leq V_{E\setminus N}^{*}=V_{E}^{*}$ by 2).

Let $K_{j}$ be a decreasing sequence of compact subsets and set $K=\cap_{j}K_{j}$. Clearly $\lim\nearrow V_{K_{j}}\leq V_{K}$. Fix $\varepsilon>0$ and let $\varphi\in PSH(X,\omega)$ be such that $\varphi\leq 0$ on $K$. Then $\{\varphi<\varepsilon\}$ is an open set which contains all $K_{j}^{\prime}s$, for $j\geq j_{\varepsilon}$ large enough. Thus $\varphi-\varepsilon\leq 0$ on $K_{j}$, hence $\varphi-\varepsilon\leq\lim\nearrow V_{K_{j}}$. Taking the supremum over all such $\varphi^{\prime}$s and letting $\varepsilon\rightarrow 0$ yields the reverse inequality $V_{K}\leq\lim\nearrow V_{K_{j}}$. The conclusion on the convergence of the upper semi-continuous regularizations follows now from proposition 1.6.

It remains to prove 5). By Choquet’s lemma, there exists an increasing sequence $\varphi_{j}\in PSH(X,\omega)$ such that $\varphi_{j}\leq 0$ on $E$ and $V_{E}^{*}=(\sup_{j}\varphi_{j})^{*}$. Set $G_{j}:=\{\varphi_{j}<1/j\}$. This defines a decreasing sequence of open subsets containing $E$. Observe that $\varphi_{j}-1/j\leq V_{G_{j}}\leq V_{E}$, hence $\lim\varphi_{j}\leq\lim V_{G_{j}}\leq V_{E}$. Therefore $V_{E}^{*}=\lim V_{G_{j}}^{*}$. ∎