As in [BBGZ09], the strategy is to first use a variational argument going back to Alexandrov [Ale38] in order to produce a solution $\varphi\in\mathcal{E}^{1}(X,\omega)$.

Consider the functional $F_{\mu}:\PSH(X,\omega)\to[-\infty,+\infty[$ defined by

(8.3)

$$F_{\mu}(\varphi):=E_{\omega}(\varphi)-\int\varphi\mu.$$

We first claim that $F_{\mu}$ is usc on $\PSH(X,\omega)$. By Proposition 6.2, $E_{\omega}$ is usc so that it is sufficient to prove $\varphi\mapsto\int\varphi\mu$ is continuous on $\PSH(X,\omega)$. Pick a net $\varphi_{k}\to\varphi$ in $\PSH(X,\omega)$, i.e. $\varphi_{k}(x)\to\varphi(x)$ for any $x\in X^{\mathrm{div}}$. Since divisorial points are dense in $\Delta_{\mathcal{X}}$ by [JM10], and the family $\{\varphi_{k}|_{\Delta_{\mathcal{X}}}\}$ is equicontinuous by Theorem 2.9, it follows that $\varphi_{k}|_{\Delta_{\mathcal{X}}}\to\varphi|_{\Delta_{\mathcal{X}}}$ uniformly. Whence $\int\varphi_{k}\mu\to\int\varphi\mu$ since $\Delta_{\mathcal{X}}$ contains the support of $\mu$ by assumption.

Now write $\PSH_{0}(X,\omega):=\{\varphi\in\PSH(X,\omega)\mid\sup\varphi=0\}$, and observe that $F_{\mu}(\varphi+c)=F_{\mu}(\varphi)$ for any constant $c$ by Proposition 6.2, so that $\sup_{\PSH(X,\omega)}F_{\mu}=\sup_{\PSH_{0}(X,\omega)}F_{\mu}$. Since $F_{\mu}$ is usc, and $\PSH_{0}(X,\omega)$ is compact by Theorem 2.10, it actually attains its maximum. We can thus find $\varphi\in\PSH_{0}(X,\omega)$ such that

$$F_{\mu}(\varphi)=\sup_{\PSH(X,\omega)}F_{\mu}.$$

Clearly $E_{\omega}(\varphi)>-\infty$, so $\varphi\in\mathcal{E}^{1}(X,\omega)$. Let us show that $\MA(\varphi)=\mu$. Pick any model function $f\leq 0$ on $X$. For $t\in\mathbf{R}$, consider the function

$$h(t)=E_{\omega}\circ P_{\omega}(\varphi+tf)-\int(\varphi+tf)\mu.$$

In view of Corollary 7.3, $h(t)$ is differentiable at $t=0$ with derivative

$$h^{\prime}(0)=\int f\,\MA(\varphi)-\int f\,\mu.$$

But since $P_{\omega}(\varphi+tf)\leq\varphi+tf\leq 0$, it follows that $h(t)\leq F_{\mu}\circ P_{\omega}(\varphi+tf)\leq F_{\mu}(\varphi)=h(0)$ for all $t$. Thus $h$ has a local maximum at $t=0$, so $h^{\prime}(0)=0$, that is $\int f\mu=\int f\MA(\varphi)$. This implies $\MA(\varphi)=\mu$, as $f$ was an arbitrary model function.