PomeD26C76DEC922Proposition 12.15.2.D26C76DEC922 Let $\omega $ be a Hermitian form on $X$. The value $\mathcal V_\mu (\omega )$ is the same for all smooth positive probability measures $\mu $ if and only if
is independent of the smooth potential $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v>0$.
Proof. For a smooth potential $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v>0$, put
Then $\mu _ v$ is a smooth positive probability measure, and $v$ solves the normalized equation for $\mu _ v$ with scalar $M_ v$. Uniqueness of the scalar and theorem 12.2.3 give $\mathcal V_{\mu _ v}(\omega )=M_ v$. Hence, if the volume does not depend on the measure, neither does the mass. Conversely, suppose that all these masses equal one constant $M$. For an arbitrary smooth positive probability measure $\mu $, integrating the normalized equation gives $c_\omega (\mu )=M$, so $\mathcal V_\mu (\omega )=M$ for every $\mu $. $\square$
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