Theorem 15.37.28D23EB7EFA5E3

Proof. For the reference zero, $P_{\epsilon \theta }(0)=0$ for every $\epsilon >0$. Letting $\epsilon \downarrow 0$ in theorem 16.28.70 therefore gives

Positive exponent. By lemma 15.24.4, write $\mu =f\nu $ with $\nu =\mathrm{MA}_\theta (\psi )$, where $\psi \in \mathrm{PSH}(X,\theta )$ is bounded and $0\leq f\in L^1(\nu )$. For positive integers $j$, let $f_ j=\min (f,j)$. These densities are bounded, and the measures $f_ j\nu $ are nonzero. By theorem 15.10.1 there are bounded $u_ j\in \mathrm{PSH}(X,\theta )$ with

These bounded potentials belong to $\mathcal E_0$ by proposition 16.28.62. Since $f_ j$ increases with $j$, exponential order (corollary 15.37.11) shows that $u_ j$ decreases; in particular $u_ j\leq u_1$. On the other hand, since $f_ j\leq f$,

so the suprema of the $u_ j$ have a common lower bound. Hence the decreasing limit $u$ is a $\theta $-plurisubharmonic function, not identically $-\infty $, and $u_ j\to u$ in $L^1(X)$. Moreover

Applying theorem 15.37.19 with reference zero, we obtain $u\in \mathcal E_0$ and weak convergence of the Monge–Ampère measures to $\mathrm{MA}^{\mathrm{ext}}_\theta (u)$. The right-hand sides converge in total variation to $\mathrm{e}^{\lambda u}f\nu $, since the functions $\mathrm{e}^{\lambda u_ j}f_ j$ converge $\nu $-almost everywhere and are bounded by $\mathrm{e}^{\lambda \sup _ Xu_1}f\in L^1(\nu )$. This proves the equation, and corollary 15.37.11 gives uniqueness.

Zero exponent. For each positive integer $j$, let $u_ j\in \mathcal E_0$ be the solution just constructed with exponent $1/j$. To bound the normalization, we compare with the solution $s\in \mathcal E_0$ for exponent one, so that $\mathrm{MA}^{\mathrm{ext}}_\theta (s)=\mathrm{e}^ s\mu $. Set $H=\max (0,\sup _ Xs)$ and $w_ j=s+(j-1)H$. Then $w_ j/j\geq s$, and

Exponential order, now with exponent $1/j$, gives $u_ j\leq w_ j$. Put

Then $\sup _ Xv_ j=0$ and $0<c_ j\leq \mathrm{e}^ H$, because $M_ j\leq \sup _ Xw_ j\leq jH$. By (15.37.29),

so $c_ j\geq v_-(\theta )/\mu (X)>0$. After extraction, theorem 5.25.2 gives $v_ j\to v$ in $L^1(X)$, where $v\not\equiv -\infty $ and $\sup _ Xv=0$, and $c_ j\to c>0$.

By lemma 5.44.5, $a_ j=\mathrm{e}^{v_ j/j}\to 1$ in $L^1(\mu )$. Consequently

Applying theorem 15.37.19 again, we obtain $v\in \mathcal E_0$ and identify the limit as $\mathrm{MA}^{\mathrm{ext}}_\theta (v)$; thus $\mathrm{MA}^{\mathrm{ext}}_\theta (v)=c\mu $. The scalar assertion of corollary 15.37.11, applied in both directions to two solutions, proves that $c$ is unique. $\square$