Corollary 18.26.18.2D7E515584AA Fix $H=\chi h$, $S=K^ n$, $C>0$ and $M>0$ from the construction in proposition 18.26.17, and put $A=\int _ X\omega ^ n$. Choose $0<t_0<1/C$. For $0<t\leq t_0$ set
\[ u_ t=tH,\qquad p_ t=P_\omega [u_ t], \]
and define $u_0=p_0=0$ separately. Both singularity volumes of $p_ t$ equal
\[ V(p_ t)=A-Mt^ n,\qquad 0\leq t\leq t_0. \]
For all $s,t\in [0,t_0]$,
\[ d_\omega (p_ s,p_ t)=M|s^ n-t^ n|, \qquad \delta _\omega ^{\mathrm{pol}}(p_ s,p_ t)=0. \]
Thus this family, with the restriction of $d_\omega $, is a complete ordered interval, isometric to $[0,Mt_0^ n]$. Every $p_ t$ with $t>0$ is not an ideal model and has trivial multiplier ideals. For any prescribed $0<\delta <A$, one can choose $t_0>0$ so that the whole interval lies in $\mathcal M_\delta (\omega )$.
Proof.
Fix $t>0$. Let $u_{t,k}=\max (tH,-k)$ and $\Theta _ k=\omega +\mathrm{dd}^{\mathrm{c}}u_{t,k}$. These bounded closed positive currents satisfy, for $1\leq j\leq n$,
\[ \int _ X\Theta _ k^ j\wedge \omega ^{n-j}=A. \]
Indeed, subtract $\omega ^ j$, write the difference as a telescoping sum of mixed products, and integrate $\mathrm{dd}^{\mathrm{c}}\bigl (u_{t,k}\sum _{i=0}^{j-1}\Theta _ k^ i\wedge \omega ^{j-1-i}\bigr )\wedge \omega ^{n-j}$ by parts for bounded potentials.
Choose an open neighborhood $U$ of $S$ on which $\chi =1$. There $\Theta _ k=\omega +t\mathrm{dd}^{\mathrm{c}}h_{k/t}$, so positivity gives
\[ t^ j(\mathrm{dd}^{\mathrm{c}}h_{k/t})^ j\wedge \omega ^{n-j} \leq \Theta _ k^ j\wedge \omega ^{n-j}. \]
For the finitely many $1\leq j<n$, take a common subsequence along which the local closed positive currents $(\mathrm{dd}^{\mathrm{c}}h_{k/t})^ j$ converge weakly on $U$, with limits $R_ j$; the inequalities above give the required local bounds on their traces. Along a further subsequence, the global measures $\Theta _ k^ n$ and the local top-degree measures also converge weakly. By lemma 18.26.15, the latter have mass $M$ and supports eventually contained in a fixed compact subset of $U$ and converging to $S$. Their extensions by zero therefore converge, along a further subsequence, to a measure $\rho $ supported on $S$ with $\rho (S)=M$.
Since $S$ has Hausdorff dimension zero, corollary 5.34.3 shows that every $R_ j$, $j<n$, has coefficient measures vanishing on $S$. The smooth measure $\omega ^ n$ also gives $S$ zero mass. On $U$ we expand
\[ \Theta _ k^ n=\sum _{j=0}^ n\binom nj t^ j(\mathrm{dd}^{\mathrm{c}}h_{k/t})^ j\wedge \omega ^{n-j}. \]
If $\eta $ is the chosen global weak limit, this expansion gives $\eta (S)=t^ n\rho (S)=Mt^ n$, while $\eta (X)=A$. Off $S$, the function $u_ t$ is locally bounded and its truncations eventually agree with it on each compact set, so there $\eta $ agrees with $\mathrm{MA}_\omega (u_ t)$. The non-pluripolar measure gives $S$ zero mass, which proves
\[ \int _ X\mathrm{MA}_\omega (u_ t)=A-Mt^ n. \]
This argument uses only a weak subsequential limit in top degree; it does not define an unbounded Bedford–Taylor product on $S$.
The current inequality $\omega +\mathrm{dd}^{\mathrm{c}}u_ t\geq (1-Ct)\omega $ gives $A-Mt^ n\geq (1-Ct_0)^ nA>0$. Invariance of non-pluripolar mass under bounded changes in the closed background and theorem 13.19.5 now give the formula for both singularity volumes of the normalized ordinary model $p_ t$. Since $H\leq 0$, monotonicity of the ordinary envelope gives $p_ s\geq p_ t$ for $s\leq t$, and the zero endpoint satisfies the same order. Thus $\max (p_ s,p_ t)=p_ s$, and the formula for ordered pairs in (13.22.2) gives $d_\omega (p_ s,p_ t)=M(t^ n-s^ n)$. Hence $p_ t\mapsto Mt^ n$ is an isometry onto $[0,Mt_0^ n]$, and the interval is complete.
The proof of proposition 18.26.17 applies to every $0<t\leq t_0$: all first polar currents vanish on all modifications, and all multiplier ideals are trivial. The polar currents also vanish for $p_0=0$, so the polar comparison vanishes on the family. For $t>0$, the strict volume deficit makes $p_ t\ne 0$, and its trivial ideals then show that it is not an ideal model. Finally, choose $t_0<1/C$ so small that $(1-Ct_0)^ nA>\delta $; the uniform lower bound for the volumes puts every $p_ t$ in $\mathcal M_\delta (\omega )$.
$\square$
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