Upon translating by a constant we may assume that $0\geq\varphi\geq-C$. Let $\varepsilon>0$. If we choose $0<t\ll 1$ such that $t(C+1)\leq\varepsilon/2$ then we have

$$\nu\{\psi+\varepsilon<(1-t)\varphi+t\}\leq\nu\{\psi+\varepsilon/2<\varphi\}=0.$$

By Lemma 8.3 it follows that

$$\Capa_{\omega}\{\psi+\varepsilon<\varphi\}\leq t^{-n}\nu\{\psi+\varepsilon<(1-t)\varphi+t\}=0$$

(since $\MA(\psi+\varepsilon)=\nu$). But $\{\psi+\varepsilon<\varphi\}$ is open by continuity of $\varphi$, hence empty by Lemma 4.2. We have thus proved that $\varphi\leq\psi+\varepsilon$ on $X$ for all $\varepsilon>0$, and the result follows. ∎