Concavity follows from (6.4) and Proposition 2.21. Monotonicity is a consequence of (6.3), and the last equation follows from (6.2) since $(\omega+dd^{c}\varphi)^{j}\wedge\omega^{n-j}$ is a probability measure for each $j$ thanks to Proposition 2.19 and the normalization $\{\omega\}^{n}=1$. ∎

That $E_{\omega}$ is nondecreasing, concave and satisfies $E_{\omega}(\varphi+c)=E_{\omega}(\varphi)+c$ follows formally from Proposition 6.1 (using that $\PSH(X,\omega)\cap\mathcal{D}(X)$ is convex and invariant under addition of a constant).

Upper semicontinuity is also a direct consequence of these algebraic properties of $E_{\omega}$ and of Theorem 2.10. Indeed, pick $\varphi_{0}\in\PSH(X,\omega)$ and $t\in\mathbf{R}$ such that $E_{\omega}(\varphi_{0})<t$. We need to show that $E_{\omega}(\varphi)<t$ for $\varphi$ in a neighborhood $U$ of $\varphi_{0}$ in $\PSH(X,\omega)$. By definition, there exists $\psi_{0}\in\PSH(X,\omega)\cap\mathcal{D}(X)$ such that $\psi_{0}\geq\varphi_{0}$ and $E_{\omega}(\psi_{0})<t-\varepsilon$ for some $\varepsilon>0$. By Theorem 2.10, $U:=\{\varphi\in\PSH(X,\omega)\mid\sup_{X}(\varphi-\psi_{0})<\varepsilon\}$ is an open neighborhood of $\varphi_{0}$ in $\PSH(X,\omega)$. By (6.2) we have $E_{\omega}(\varphi)\leq E_{\omega}(\psi_{0})+\varepsilon<t$ for all $\varphi\in U$, which proves upper semicontinuity.

Finally, being usc and nondecreasing, $E_{\omega}$ is automatically continuous along decreasing nets. ∎

We may assume $\varphi\leq 0$. Set $\mu_{t}:=\MA(\varphi^{\langle t\rangle})$. Since (6.2) applies to bounded $\omega$-psh functions by Proposition 6.3, we get

$$E_{\omega}(\varphi^{\langle t/2\rangle})-E_{\omega}(\varphi^{\langle t\rangle})\geq\frac{1}{n+1}\int(\varphi^{\langle t/2\rangle}-\varphi^{\langle t\rangle})\mu_{t}=\frac{1}{n+1}\int_{0}^{t/2}\mu_{t}\left\{\varphi^{\langle t/2\rangle}-\varphi^{\langle t\rangle}\geq s\right\}\,ds\\ \geq\frac{1}{n+1}\int_{0}^{t/2}\mu_{t}\left\{\varphi^{\langle t/2\rangle}-\varphi^{\langle t\rangle}\geq t/2\right\}\,ds=\frac{t}{2(n+1)}\mu_{t}\left\{\varphi\leq-t\right\},$$

where $\mu_{t}=\MA(\varphi^{\langle t\rangle})$. Since $\lim_{t\to\infty}E_{\omega}(\varphi^{\langle t/2\rangle})=\lim_{t\to\infty}E_{\omega}(\varphi^{\langle t\rangle})=E_{\omega}(\varphi)$ by the continuity of $E_{\omega}$ along decreasing sequences, the proof is complete. ∎

We may assume $\sup_{X}|f|=1$. Pick $s\geq t$. The probability measures $\mu_{t}:=\MA(\varphi^{\langle t\rangle})$ and $\mu_{s}$ agree on $\{\varphi>-t\}$. Hence

$$\left|\int f\mu_{t}-\int f\mu_{s}\right|\leq(\mu_{t}+\mu_{s})\{\varphi\leq-t\}\leq\frac{1}{t}\left(\int-\varphi^{\langle t\rangle}\mu_{t}+\int-\varphi^{\langle s\rangle}\mu_{s}\right)\\ \leq\frac{n+1}{t}(|E_{\omega}(\varphi^{\langle t\rangle})|+|E_{\omega}(\varphi^{\langle s\rangle})|)\leq\frac{2(n+1)}{t}|E_{\omega}(\varphi)|.$$

The result follows by letting $s\to\infty$. ∎

Given $f\in\mathcal{D}(X)$, we have by definition that $\int f\MA(\varphi^{\langle t\rangle})\to\int f\MA(\varphi)$ as $t\to\infty$ and $\int f\MA(\varphi^{\langle t\rangle}_{j})\to\int f\MA(\varphi_{j})$ as $t\to\infty$ for every $j$. Moreover, Lemma 6.8 shows that the latter convergence is uniform in $j$. Since for each $t$ we have $\int f\MA(\varphi^{\langle t\rangle}_{j})\to\int f\MA(\varphi^{\langle t\rangle})$ as $j\to\infty$ by Theorem 3.1, the result follows. ∎

Pick $s,t>0$. Since (6.1) holds for bounded $\omega$-psh functions, we see using (3.1) that

$$-\infty<E\left(\frac{\varphi+\psi}{2}\right)\leq E\left(\frac{\varphi^{\langle t\rangle}+\psi^{\langle s\rangle}}{2}\right)\leq\frac{2^{-(n+1)}}{n+1}\int\psi^{\langle s\rangle}\MA(\varphi^{\langle t\rangle}).$$

Since $\psi^{\langle s\rangle}$ decreases to $\psi$ at any point of $X$, the right hand side converges to

$$\frac{2^{-(n+1)}}{n+1}\int\psi\MA(\varphi^{\langle t\rangle})\leq\frac{2^{-(n+1)}}{n+1}\int_{\{\varphi>-t\}}\psi\MA(\varphi^{\langle t\rangle})=\frac{2^{-(n+1)}}{n+1}\int_{\{\varphi>-t\}}\psi\MA(\varphi)$$

by monotone convergence. We obtain the desired estimate by letting $t\to\infty$. ∎

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. ∎

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

$$\lim_{j\to\infty}\lim_{s\to\infty}\int(\psi_{j}-\varphi^{\langle s\rangle})\MA(\varphi^{\langle s\rangle})=\int(\psi-\varphi)\MA(\varphi).$$

First, we have

	$\displaystyle\int\varphi^{\langle s\rangle}\,\MA(\varphi^{\langle s\rangle})$	$\displaystyle=\int_{\{\varphi\leq-s\}}(-s)\,\MA(\varphi^{\langle s\rangle})+\int_{\{\varphi>-s\}}\varphi\MA(\varphi^{\langle s\rangle})$
		$\displaystyle=\int_{\{\varphi\leq-s\}}(-s)\,\MA(\varphi^{\langle s\rangle})+\int_{\{\varphi>-s\}}\varphi\MA(\varphi)$

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. ∎