Lemma 15.30.240B884870886

Proof. Suppose first that $g$ is smooth. Since $\kappa $ is Kähler, theorem 14.2.1 on $Y$ gives a smooth $\phi _ g$ with $A_ g=\kappa +\mathrm{dd}^{\mathrm{c}}\phi _ g>0$ and $A_ g^ k=(g/M_ g)\kappa ^ k$; the scalar is determined by integration, since both sides have mass one. On $X$, since $h_0^{m+1}=0$ and $A_ g^{k+1}=0$ for dimensional reasons,

Uniqueness of the scalar gives $c_ H(Rg)=CV_0/M_ g$. For continuous $g>0$, choose smooth $g_ l>0$ converging uniformly to $g$. Since $g$ has a positive minimum, $(1-\varepsilon _ l)g\leq g_ l\leq (1+\varepsilon _ l)g$ for some $\varepsilon _ l\in (0,1)$ tending to zero. The order and homogeneity of the scalar in lemma 15.27.1 give

Since $M_{g_ l}\to M_ g$, the smooth case proves the formula. $\square$