Theorem 15.30.6.AB3D52EC73EF For every nonzero nonnegative $g\in L^1(Y,\kappa ^ k)$, with $M_ g=\int _ Yg\kappa ^ k$, one has
\[ b_ H(Rg)=\frac{c_0}{M_ g}. \]
There is $v\in \mathrm{PSH}(Y,\kappa )$ with $\sup _ Yv=0$ and $\int _ Y\mathrm{MA}^{\mathrm{ext}}_\kappa (v)=1$ whose pullback $u$ satisfies
\[ \mathrm{MA}^{\mathrm{ext}}_ H(u) =\frac{c_0}{M_ g}Rg\, \mathrm dV,\qquad \int _ X\mathrm{MA}^{\mathrm{ext}}_ H(u)=c_0. \]
If $v_-(h_0)=0$, then $v_-(H)=0$ and $H$ is not closed.
Proof.
We first compute the scalars of bounded densities, then use the closed factor's mass bound to identify the equation for an unbounded density.
Step 1: bounded densities. If $g$ is bounded and nonzero, theorem 14.19.2 on $Y$ gives a bounded continuous $v$ with $\mathrm{MA}_\kappa (v)=c_\kappa (g)g\kappa ^ k$. Closed Stokes gives $c_\kappa (g)=1/M_ g$. The bounded product expansion gives
\[ \mathrm{MA}_ H(v)=C h_0^ m\wedge \mathrm{MA}_\kappa (v) =\frac{c_0}{M_ g}Rg\, \mathrm dV. \]
Uniqueness of the bounded normalization scalar therefore gives $c_ H(Rg)=c_0/M_ g$, including densities with zeros.
Step 2: the threshold. For general $g\in L^1$, the original truncations on $X$ satisfy
\[ R\min (g,j/A)\leq \min (Rg,j)\leq R\min (g,j/a). \]
By Step 1 and lemma 15.27.1,
\[ \frac{c_0}{\int _ Y\min (g,j/a)\kappa ^ k} \leq c_ H(\min (Rg,j)) \leq \frac{c_0}{\int _ Y\min (g,j/A)\kappa ^ k}. \]
Both bounds tend to $c_0/M_ g$.
Step 3: the exact equation. By proposition 15.28.7, $b_\kappa (g)=1/M_ g$. Thus theorem 15.28.4 on $Y$ gives $v\in \mathrm{PSH}(Y,\kappa )$ with supremum zero and
\[ \mathrm{MA}^{\mathrm{ext}}_\kappa (v)\geq (g/M_ g)\kappa ^ k. \]
For $v_ j=\max (v,-j)$, closed Stokes gives $\int _ Y\mathrm{MA}_\kappa (v_ j)=1$. The defining non-pluripolar truncations consequently give $\int _ Y\mathrm{MA}^{\mathrm{ext}}_\kappa (v)\leq 1$. The prescribed lower measure has mass one, so equality holds both in mass and in the measure inequality.
The pullback $u$ is $H$-plurisubharmonic, and $\max (u,-j)$ is the pullback of $v_ j$. Multiplying the bounded product expansion by the pullback of $\mathbf1_{\{ v>-j\} }$ and passing to the defining non-pluripolar limit gives
\[ \mathrm{MA}^{\mathrm{ext}}_ H(u) =C h_0^ m\wedge \mathrm{MA}^{\mathrm{ext}}_\kappa (v) =\frac{c_0}{M_ g}Rg\, \mathrm dV. \]
Its mass is $c_0$. Finally, when $v_-(h_0)=0$, pulling back bounded potentials from $X_0$ whose masses tend to zero proves $v_-(H)=0$ by the same expansion. Closedness of $h_0$ would contradict that vanishing by Stokes' theorem, so $H$ is not closed.
$\square$
Pome
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