Corollary 13.15.18.5804300415EA Assume $v_+(\omega )=+\infty $ and put $V_\omega =\int _ X\omega ^ n$. Fix a real number $b>V_\omega $ and a sequence $0<\varepsilon _ j<1$ tending to zero. The smooth potentials $w_ j$ of corollary 13.14.23, chosen with mass $b$, converge to zero in capacity. For every $T>0$,
\[ \left\| \mathbf1_{\{ w_ j\geq -T\} }(\omega +\mathrm{dd}^{\mathrm{c}}w_ j)^ n-\omega ^ n \right\| _{\mathrm{TV}}\longrightarrow 0, \qquad \int _{\{ w_ j<-T\} }(\omega +\mathrm{dd}^{\mathrm{c}}w_ j)^ n\longrightarrow b-V_\omega . \]
After passage to a subsequence, there is a positive measure $\sigma $ of mass $b-V_\omega $ such that
\[ (\omega +\mathrm{dd}^{\mathrm{c}}w_ j)^ n\rightharpoonup \omega ^ n+\sigma , \qquad \mathbf1_{\{ w_ j<-T\} }(\omega +\mathrm{dd}^{\mathrm{c}}w_ j)^ n\rightharpoonup \sigma \quad \text{for every }T>0. \]
Proof.
The construction in corollary 13.14.23 gives $w_ j=s_ j u_ j$, where $u_ j$ is smooth and $\omega $-plurisubharmonic, $\sup _ Xu_ j=0$ and $0<s_ j<\varepsilon _ j$. The sublevel estimate (5.29.4) gives, for every fixed $\delta >0$ and all sufficiently large $j$,
\[ \operatorname {Cap}_\omega (\{ |w_ j|>\delta \} ) =\operatorname {Cap}_\omega (\{ u_ j<-\delta /s_ j\} ) \leq Cs_ j/\delta \longrightarrow 0. \]
Put $\mu _ j=(\omega +\mathrm{dd}^{\mathrm{c}}w_ j)^ n$. Fix $T>0$ and write $E_{j,T}=\{ w_ j\geq -T\} $. The lower bound in corollary 13.14.23 gives
\[ \mathbf1_{E_{j,T}}\mu _ j\geq \lambda _{j,T}:=(1-\varepsilon _ j)^ n\mathbf1_{E_{j,T}}\omega ^ n. \]
Since $w_ j\to 0$ in $L^1$, the smooth volume of $X\setminus E_{j,T}$ tends to zero, and hence $\lambda _{j,T}\to \omega ^ n$ in total variation. For every $p>0$, theorem 13.15.9 gives $\int _ X\mathrm{e}^{pw_ j}\, \mathrm{d}\mu _ j\to V_\omega $, and therefore
\[ \limsup _ j\mu _ j(E_{j,T}) \leq \mathrm{e}^{pT}\lim _ j\int _ X\mathrm{e}^{pw_ j}\, \mathrm{d}\mu _ j =\mathrm{e}^{pT}V_\omega . \]
Letting $p\downarrow 0$ and combining with the lower bound, we obtain $\mu _ j(E_{j,T})\to V_\omega $. Positivity now yields
\[ \begin{split} \| \mathbf1_{E_{j,T}}\mu _ j-\omega ^ n\| _{\mathrm{TV}} & \leq \mu _ j(E_{j,T})-\lambda _{j,T}(X) +\| \lambda _{j,T}-\omega ^ n\| _{\mathrm{TV}}\\ & \longrightarrow 0. \end{split} \]
Subtracting this shallow mass from $\mu _ j(X)=b$ proves the assertion about the mass on $\{ w_ j<-T\} $.
The positive measures $\mu _ j$ have fixed mass on the compact manifold $X$, so a subsequence converges weakly to a measure $\nu $ of mass $b$. The inequality $\mu _ j\geq (1-\varepsilon _ j)^ n\omega ^ n$ gives $\nu \geq \omega ^ n$. Set $\sigma =\nu -\omega ^ n$. For each $T>0$, subtracting the total-variation limit of the shallow restrictions from $\mu _ j\rightharpoonup \nu $ gives the last assertion along the same subsequence, for every $T$.
$\square$
Pome
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