Proposition 10.30.106436A9863EBA

Proof. Take $U$ from theorem 13.14.18. All its multiplier ideals at positive weights are trivial, so every divisorial order in $D_{m,Z}$ vanishes on every model. The self-trace identity gives $N_ Z(U)=0$, which is also the divisorial part of the zero potential. Both families therefore consist of the pullbacks of the class of $\omega $. Since $\omega $ is smooth and positive, its affine class is nef and realizes the family on $Y$ itself. The extended mass of $U$ is infinite by construction, whereas the smooth measure $\omega ^ n$ has finite mass because $Y$ is compact. $\square$