Theorem 13.13.43 (Domination principle).B798B54E6F2E Let $\varphi ,\psi \in \mathrm{PSH}(X,\theta )_{>0}$. Assume that
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\begin{equation} \exists \phi \in \mathrm{PSH}(X,\theta )\quad \textup{s.t.} \quad \varphi \preceq \phi , \quad \psi \preceq \phi ,\quad \int _ X \theta _{\varphi }^ n=\int _ X \theta _{\psi }^ n=\int _ X \theta _{\phi }^ n, \end{equation}
and
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\begin{equation} \int _{\{ \varphi <\psi \} } \theta _{\varphi }^ n=0. \end{equation}
Then $\varphi \geq \psi $.
Proof. Take the common less singular potential $\phi$ in the hypotheses as the reference potential. The inequalities $\varphi\preceq\phi$, $\psi\preceq\phi$ and the two mass equalities say exactly that $\varphi,\psi\in\mathcal E(X,\theta,\phi)$. The reference mass is positive, because it equals the positive mass of $\varphi$. The second hypothesis says that the Monge--Ampère measure of $\varphi$ vanishes on $\{\varphi<\psi\}$. The relative full-mass domination principle therefore gives $\varphi\geq\psi$. This identifies every hypothesis of the earlier proof; in particular the common full-mass reference cannot be dropped.
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