Lemma 16.22.17420706BAE2F

Proof. Fix a smooth probability volume form $\mathrm dV$ on $X$, put $m=\int _ X\langle \theta _\psi ^ n\rangle $, and consider

The set $\mathcal S$ contains $\psi $, and $H$ takes values in $[H(\psi ),0]$ on it. We construct an increasing sequence in $\mathcal S$ that almost maximizes $H$ at every stage. Put $v_0=\psi $. Given $v_ k\in \mathcal S$, let

and choose $v_{k+1}\in \mathcal S$ with $v_{k+1}\geq v_ k$ and $H(v_{k+1})\geq s_ k-2^{-k}$. The regularized limit $\phi =(\sup _ kv_ k)^*$ is a potential with $\psi \leq \phi \leq 0$, and $\int _ X\langle \theta _\phi ^ n\rangle =m$ by corollary 13.8.9; thus $\phi \in \mathcal S$. Since $v_ k$ increases to $\phi $ almost everywhere, between the integrable bounds $\psi $ and $0$, dominated convergence gives $H(v_ k)\to H(\phi )$.

Let $w\in \mathcal S$ with $w\geq \phi $. Then $w\geq v_ k$ for every $k$, so

and letting $k\to \infty $ gives $H(w)\leq H(\phi )$. Since $w\geq \phi $, it follows that $w=\phi $ almost everywhere, hence everywhere. Now $P_\theta [\phi ]$ lies between $\phi $ and $0$, because $\phi \leq 0$ is a competitor for it, and it has mass $m$ by proposition 16.21.2. Thus $P_\theta [\phi ]\in \mathcal S$ and $P_\theta [\phi ]\geq \phi $, so $P_\theta [\phi ]=\phi $. Since $m>0$, $\phi $ is a model potential. $\square$