Set $h(t):=h_{\varphi,\psi}(t):=E_{\omega}((1-t)\varphi+t\psi)$ for $0\leq t\leq 1$. Note that $h$ is a polynomial of degree at most $n$ when $\varphi$ and $\psi$ are model functions. By continuity of the energy along decreasing nets, the same is true in general. In particular, $h$ is differentiable on $[0,1]$.

Pick any decreasing sequence $(\psi_{j})_{j=1}^{\infty}$ of $\omega$-psh model functions converging to $\psi$. Note that $h_{\varphi^{\langle s\rangle},\psi_{j}}\to h_{\varphi,\psi}$ as polynomials when $s\to\infty$ and $j\to\infty$, hence $\frac{d}{dt}\big|_{t=0+}h_{\varphi^{\langle s\rangle},\psi_{j}}\to h^{\prime}(0+)$. Since (6.9) holds true for bounded functions by Proposition 6.3, it suffices to show

First, we have

by (6.6). By Lemma 6.7 the first term of the right hand side tends to $0$, and the second term converges to $\int\varphi\MA(\varphi)$ since $\MA(\varphi)$ puts no mass on $\{\varphi=-\infty\}$.

Second, for fixed $j$ we have $\lim_{s\to\infty}\int\psi_{j}\MA(\varphi^{\langle s\rangle})=\int\psi_{j}\MA(\varphi)$ since $\psi_{j}$ is continuous.

Finally, Lemma 2.23 yields $\lim_{j\to\infty}\int\psi_{j}\MA(\varphi)=\int\psi\MA(\varphi)$, completing the proof. ∎