Note that $\varphi+tf\geq\varphi-|t|\sup_{X}|f|$ implies $P_{\omega}(\varphi+tf)\geq\varphi-|t|\sup_{X}|f|$, hence $P_{\omega}(\varphi+tf)\in\mathcal{E}^{1}(X,\omega)$ for all $t$. We are going to show that

For all $t\in\mathbf{R}$. If $\varphi$ is continuous, then the result follows immediately from Theorem 7.2.

In general, let $(\varphi_{m})_{m=1}^{\infty}$ be a decreasing sequence of $\omega$-psh model functions converging to $\varphi$, see Proposition 4.5.

For each $t\in\mathbf{R}$ the sequence $(P_{\omega}(\varphi_{m}+tf))_{m=1}^{\infty}$ is a decreasing sequence of $\omega$-psh functions, and we claim that $\lim_{m}P_{\omega}(\varphi_{m}+tf)=P_{\omega}(\varphi+tf)$. Indeed let $\tilde{\varphi}_{t}=\lim_{m}P_{\omega}(\varphi_{m}+tf)$. Since $\varphi_{m}+tf\geq\varphi+tf$, we have $P_{\omega}(\varphi_{m}+tf)\geq P_{\omega}(\varphi+tf)$, hence $\tilde{\varphi}_{t}\geq P_{\omega}(\varphi+tf)$. Conversely, $\tilde{\varphi}_{t}\leq P_{\omega}(\varphi_{m}+tf)\leq\varphi_{m}+tf$ for all $m$, hence $\tilde{\varphi}_{t}\leq\varphi+tf$ and it follows $\tilde{\varphi}_{t}=P_{\omega}(\varphi+tf)$ as required.

We apply (7.1) to $\varphi_{m}$:

As $m\to\infty$, $E_{\omega}(P_{\omega}(\varphi_{m}+tf))$ and $E_{\omega}(\varphi_{m})$ decrease to $E_{\omega}\circ P_{\omega}(\varphi+tf)$ and $E_{\omega}(\varphi)$, respectively and by Proposition 6.9, $\int f\MA(P_{\omega}(\varphi_{m}+sf))$ converges to $\int f\MA(P_{\omega}(\varphi+sf))$ for each $s$. Finally (7.1) follows from (7.2) using dominated convergence in view of the upper bound $|\int f\MA(P_{\omega}(\varphi_{m}+sf))|\leq\sup_{X}|f|$ for all $m$ and all $s$. ∎