Proposition 18.26.127C9BBE3DA7BE

Proof. Fix $\pi $. By proposition 18.26.9, the measures $\mu _\pi (p_ j)$ converge in total variation to a positive measure $\lambda _\pi $. Put

This is a Borel set of Lebesgue measure zero. Every measure $\mu _\pi (p_ j)$ vanishes on its complement, so $\lambda _\pi (Y_\pi \setminus B_\pi )=0$. The polar currents are closed and positive by corollary 18.26.11, and their $\eta _\pi $-trace masses are at most $A_\pi $. By compactness of positive currents of bounded mass, a subsequence converges weakly. Its limit $R$ is closed and positive, and $R\wedge \eta _\pi ^{n-1}=\lambda _\pi $. Since positivity bounds all coefficient measures by the Hermitian trace, $R$ is carried by $B_\pi $. Any two subsequential limits have the same null carrier and the same trace, so they agree by lemma 5.46.4. As every subsequence has a further convergent subsequence, the whole sequence of currents converges weakly to this unique limit $R_\pi $.

Apply proposition 5.46.5 to $g$, with background $\pi ^*\omega $ and potential $p_ j\circ \pi $. It gives $g_*\operatorname {Pol}_{\pi '}(p_ j)=\operatorname {Pol}_\pi (p_ j)$. Proper pushforward is weakly continuous, since a smooth test form downstairs pulls back to a smooth test form upstairs, and passing to the limit proves compatibility. The limits are constructed separately in each coordinate, so no diagonal subsequence over $\mathscr R$ is needed.

To obtain one pole carrier, put $h_ m=\sum _{j=1}^ m2^{-j}p_ j$. These functions decrease, and they are $\omega $-plurisubharmonic because the coefficients have sum at most one and $\omega >0$. Normalized compactness, theorem 5.25.2, bounds their $L^1$ norms uniformly, so corollary 5.25.3 gives a limit $h\in \mathrm{PSH}(X,\omega )$. Since all summands are nonpositive, each pole set of a $p_ j$ is contained in the pole set of $h$. Hence $B_\pi \subset \{ h\circ \pi =-\infty \} $, which gives the last assertion. $\square$