Lemma 11.45.4688DA0FB893F

Proof. The one-sided ordinary class has the description

Indeed, $u\preceq \psi $ gives $\nu (u,D)\geq d$. Divisorial removal, lemma 5.38.1, shows that $\theta +\mathrm{dd}^{\mathrm{c}}u-d[D]\geq 0$. Thus the locally integrable difference $u-\psi $ has an $\eta $-plurisubharmonic representative $r$. The equality $u=\psi +r$ holds off $D$ by canonical equality and on $D$ because both sides equal $-\infty $. Conversely, adding $\psi $ to $r$ adds the positive current $d[D]$; since $r$ has a finite compact upper bound, the one-sided comparison follows. Off $D$ the currents agree, so finite-level locality identifies their extended products, including at any additional poles of $r$. Both measures give $D$ zero mass, proving the global equality. The bounded ordinary type corresponds exactly to bounded $r$, as in theorem 11.41.1.

The function $g_\sigma =\mathrm{e}^{\sigma \psi }f$ extends continuously by zero on $D$ and is positive elsewhere. Since $\eta >0$, theorem 14.19.3 gives a continuous solution $R_\sigma $ of

For this bounded density, the greatest solution of theorem 15.35.2 is bounded and coincides with $R_\sigma $ by corollary 14.19.12. Thus it dominates every $\eta $-plurisubharmonic exponential subsolution. The correspondence (11.45.6) transfers this order to the entire one-sided ordinary class and gives $s_\sigma =\psi +R_\sigma $. Bounded residual uniqueness gives uniqueness in the ordinary type. The bounded residual equation has finite mass, so the singular equation does too. $\square$