To prove (6.6), first assume $\psi=-t$, where $t\geq 1$. Pick $s\geq t$ so that $\varphi^{\langle t\rangle}=\max\{\varphi_{s},-t\}$, $\{\varphi>-t\}=\{\varphi_{s}>-t\}$, and

$$\one_{\{\varphi>-t\}}\MA(\varphi^{\langle t\rangle})=\one_{\{\varphi_{s}>-t\}}\MA(\varphi^{\langle t\rangle})=\one_{\{\varphi_{s}>-t\}}\MA(\varphi_{s})=\one_{\{\varphi>-t\}}\cdot\one_{\{\varphi>-s\}}\MA(\varphi_{s}),$$

where the second equality follows from Theorem 5.1. As $s\to\infty$, $\one_{\{\varphi>-s\}}\MA(\varphi_{s})(E)\to\MA(\varphi)(E)$ for any Borel set $E$, so the right hand side of the equation above converges to $\one_{\{\varphi>-t\}}\MA(\varphi)$.

Now consider $\varphi,\psi\in\mathcal{E}^{1}(X,\omega)$ and set $u=\max\{\varphi,\psi\}\in\mathcal{E}^{1}(X,\omega)$. Then

• •

  $\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(u)=\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(u^{\langle t\rangle})$ since $\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}\subseteq\{u>-t\}$;

• •

  $\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(u^{\langle t\rangle})=\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(\varphi^{\langle t\rangle})$ by (5.1) applied to $\varphi^{\langle t\rangle}$ and $\psi$, noticing the inclusion $\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}\subseteq\{\varphi^{\langle t\rangle}>\psi\}$;

• •

  $\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(\varphi^{\langle t\rangle})=\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(\varphi)$ by the previous step and the inclusion $\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}\subseteq\{\varphi^{\langle t\rangle}>-t\}$.

To summarize, we get

(6.8)

$$\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(u)=\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(\varphi).$$

Now

$\displaystyle\one_{\{\varphi^{\langle t\rangle}>\psi^{\langle t\rangle}\}}\MA(u)$

$\displaystyle=\one_{\{\varphi>-t\geq\psi\}}\MA(u)+\one_{\{\varphi>\psi>-t\}}\MA(u)$

As $t\to\infty$ the first term tends to $0$ since $\MA(u)$ puts no mass on the pluripolar set $\{\psi=-\infty\}$ (see Remark 6.5), and the second term converges to $\one_{\{\varphi>\psi>-\infty\}}\MA(u)=\one_{\{\varphi>\psi\}}\MA(u)$.

Thus the left-hand side of (6.8) tends to $\one_{\{\varphi>\psi\}}\MA(u)$ as $t\to\infty$. Similarly, the right-hand side tends to $\one_{\{\varphi>\psi\}}\MA(\varphi)$. Finally the comparison principle follows exactly as in the proof of Corollary 5.3. The proof is complete. ∎