Proposition 15.35.13.C791D4F48DEC Suppose $b_ h(f)=0$. Let $\lambda _ j>0$ decrease to zero, and let $u_ j$ be any solution of (15.35.1) at exponent $\lambda _ j$. Put
\[ M_ j=\sup _ Xu_ j,\qquad v_ j=u_ j-M_ j,\qquad c_ j=\mathrm e^{\lambda _ jM_ j}. \]
Then
\begin{gather*} c_ j\longrightarrow 0,\qquad \lambda _ jM_ j\longrightarrow -\infty ,\\ \| \mathrm{MA}^{\mathrm{ext}}_ h(u_ j)\| _{\mathrm{TV}} =\| \mathrm{MA}^{\mathrm{ext}}_ h(v_ j)\| _{\mathrm{TV}} \longrightarrow 0. \end{gather*}
Moreover, $\mathrm e^{\lambda _ ju_ j}\to 0$ uniformly on $X$, where the exponential is zero on the pole set, and $u_ j\to -\infty $ uniformly, that is, $\sup _ Xu_ j\to -\infty $.
Proof.
Set $g_ j=\mathrm e^{\lambda _ jv_ j}f$. By lemma 5.44.5, applied to the finite non-pluripolar measure $f\, \mathrm dV$, $g_ j\to f$ in $L^1$; each $g_ j$ is nonzero, since $v_ j$ is finite almost everywhere. Since the Monge–Ampère measure is unchanged by adding constants,
\[ \mathrm{MA}^{\mathrm{ext}}_ h(v_ j)=c_ jg_ j\, \mathrm dV, \qquad \text{hence}\qquad 0<c_ j\leq b_ h(g_ j) \]
by theorem 15.28.4. The upper semicontinuity of the threshold, theorem 15.32.1, gives $\limsup _ jc_ j\leq b_ h(f)=0$. Since $0\leq g_ j\leq f$, the total masses are at most $c_ j\| f\| _{L^1}$ and tend to zero. Moreover $\lambda _ jM_ j=\log c_ j\to -\infty $, and $0\leq \mathrm e^{\lambda _ ju_ j}\leq \mathrm e^{\lambda _ jM_ j}=c_ j$. Finally, once $\lambda _ j<1$ and $\log c_ j<0$, we have $M_ j=\lambda _ j^{-1}\log c_ j\leq \log c_ j$, so $M_ j\to -\infty $.
$\square$
Pome
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