Lemma 10.25.9BCA9F26FE4F4

Proof. For a twice differentiable function $G$ of $L=-\log |z|^2$, differentiation away from zero gives

On the inner disk this yields

Both densities are integrable at zero. Both profiles are locally integrable, and their distributional Hessians have no additional atom at zero. To see this, integrate twice by parts outside $\{ |z|=\varepsilon \} $: the boundary terms against a smooth test function are bounded by constant multiples of $\varepsilon |G(-\log \varepsilon ^2)|$ and $|G'(-\log \varepsilon ^2)|$, and both tend to zero for the two profiles. Thus (10.25.10) holds in the sense of distributions across the center. On the cutoff annulus all derivatives are bounded. Hence, for a sufficiently large common $C$,

is a positive measure, smooth and strictly positive away from zero, with no atom at zero. Pairing $\mathrm{dd}^{\mathrm{c}}v$ with the constant test function gives $\int _ E\alpha _ v=C\int _ E\omega _ E$.

Let $Y=E^{n-1}$, let $\pi _ Y:X\to Y$ be the projection onto the remaining factors, and let $\theta _ Y$ be $C$ times the sum of the $n-1$ flat forms of $Y$. On $X\setminus D$ the smooth wedge product is

for $n=1$ this means just $\alpha _ v$, with $Y$ a point. The right side defines a finite product measure on $X$ that gives $D$ zero mass. By theorem 13.14.3, the non-pluripolar measure agrees with this product off $D$ and gives the pluripolar set $D$ zero mass, so the two global measures are equal. Integrating the product proves the full-mass identity. Since $\theta >0$, its envelope is $V_\theta =0$ (section 5.40). The bounded profile therefore has minimal ordinary singularities, whereas $v_{\mathrm c}$ is not bounded below by $0$ minus a constant. $\square$