Proposition 16.3.9.EF5AC4B19C4D Let $\theta '=\theta +\mathrm{dd}^{\mathrm{c}}g$ for some $g\in C^{\infty }(X)$. Then for any $\varphi \in \mathrm{PSH}(X,\theta )$, we have $\varphi '\coloneqq \varphi -g\in \mathrm{PSH}(X,\theta ')$ and
Proposition 16.3.9EF5AC4B19C4D
\[ P_{\theta }[\varphi ]\sim P_{\theta '}[\varphi ']. \]
Proof. By symmetry, it suffices to show that
\[ P_{\theta }[\varphi ]\preceq P_{\theta '}[\varphi ']. \]
We may assume that $g\geq 0$ (after replacing $g$ by $g+C$ with $C\geq -\inf _ X g$, which leaves $\theta '$ unchanged and shifts $\varphi '$ by an additive constant). Then for any $\psi \in \mathrm{PSH}(X,\theta )$ with $\psi \preceq \varphi $ and $\psi \leq 0$, we set $\psi '\coloneqq \psi -g\in \mathrm{PSH}(X,\theta ')$. Then $\psi '\preceq \varphi '$ and $\psi '\leq 0$, so $\psi '\leq P_{\theta '}[\varphi ']$. Since $\psi $ is arbitrary, it follows that
\[ P_{\theta }[\varphi ]-\sup _ X g\leq P_{\theta }[\varphi ]-g\leq P_{\theta '}[\varphi '], \]
which gives $P_{\theta }[\varphi ]\preceq P_{\theta '}[\varphi ']$. $\square$
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