Proposition 20.3.62.65113AF2CE2F Suppose that $Z$ does not have the bounded mass property, that is, $v_+(\omega _ Z)=+\infty $ in the notation of definition 13.14.14, and that $A>\delta $. Then there is $a\in \mathrm{PSH}(Y,\Omega )$ such that
\[ a\sim _{\mathcal I}v,\qquad \int _ Y\mathrm{MA}^{\mathrm{ext}}_\Omega (v)<+\infty , \qquad \int _ Y\mathrm{MA}^{\mathrm{ext}}_\Omega (a)=+\infty . \]
Both potentials have the same divisorial Lelong numbers on every modification; when $D$ is the trace measure of subsection 20.3.37, these are the trace data computed in theorem 20.3.39. The product manifold $Y$ does not have the bounded mass property.
Proof.
Choose the nonpositive potential $H$ of theorem 13.14.18 on $Z$; it has infinite extended mass, and $\mathrm{e}^{-tH}$ is integrable for every $t>0$. Put $a=w\circ q+H\circ r$. The product formula and $A-\delta >0$ show that the mass of $a$ is infinite, while (20.3.61) gives the finite positive mass of $v$.
We prove equality of the multiplier ideals locally, for every $t>0$. Since $H\leq 0$, integrability against $\mathrm{e}^{-ta}$ implies integrability against $\mathrm{e}^{-tv}$, so $\mathcal I(ta)\subseteq \mathcal I(tv)$. Conversely, lemma 5.37.7 gives $\mathcal I(tv)=\mathcal I(tw)\mathcal O_ Y$. Local generators pulled back from $C$ are square integrable against $\mathrm{e}^{-tw}$, so by Fubini's theorem they are also square integrable against $\mathrm{e}^{-tw}\mathrm{e}^{-tH}=\mathrm{e}^{-ta}$; multiplication by bounded holomorphic functions and finite sums preserve this local integrability. Thus $\mathcal I(tv)\subseteq \mathcal I(ta)$, which proves equality at every positive weight. Theorem 5.37.3 then gives equality of all divisorial Lelong numbers after modification.
Finally, since $a$ has infinite mass, proposition 13.14.16 shows that $Y$ does not have the bounded mass property.
$\square$
Pome
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