Corollary 16.8.33.4EE47D80CC25 Let $\Phi \in \mathrm{PSH}(X\times S,p_1^*\theta )$. Assume that for any $c\in \mathbb {R}$, $x\in X$ and $z\in S$, we have
\[ \Phi (x,z)=\Phi (x,z+\mathrm{i}c). \]
Then $\int _ X (\theta +\mathrm{dd}^{\mathrm{c}}\Phi _ z)^ n$ is independent of $z\in S$, where $\Phi _ z \in \mathrm{PSH}(X,\theta )$ is given by $\Phi _ z(x)=\Phi (x,z)$.
Proof.
Fix a Kähler form $\omega $ on $X$. We may regard $\Phi $ as an element of $\mathrm{PSH}(X\times S,p_1^*(\theta +\omega ))$. For any $0<a<b<1$, the restriction $(\Phi _ t)_{t\in [a,b]}$ is then a subgeodesic from $\Phi _ a$ to $\Phi _ b$ in the class $\theta +\omega $. The slices have positive mass in this class, so theorem 16.8.32 gives $\Phi _ a\sim _ P\Phi _ b$. By proposition 16.8.22, this directly implies
\[ \int _ X \theta _{\Phi _ a}^ n=\int _ X \theta _{\Phi _ b}^ n. \]
Since $\Phi $ is invariant under imaginary translations, $\Phi _ z$ depends only on $\operatorname{Re}z$, so the mass is independent of $z\in S$.
$\square$
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