Theorem 10.31.44D5758568A543

Proof. We first take $\lambda =1$ and suppose that

Corollary 10.31.28 gives $Q\in \mathcal E_\varphi $ such that

is a finite nonzero measure. Consider

The family contains $Q$. For $w\in \mathcal T$, exponential order (corollary 13.19.3) gives $w\leq Q$. Also

These bounds prevent the collapse of any sequence in $\mathcal T$.

Let

and choose $w_ j\in \mathcal T$ whose weighted masses tend to $\alpha $. Extract an $L^1$ limit $w\not\equiv -\infty $. Since

the fixed-dominator theorem theorem 10.31.5 gives $w\in \mathcal E_\varphi $, convergence in capacity, and $\mu _{w_ j}\rightharpoonup \mu _ w$. Put $\nu _ j=\mathrm{e}^{-w_ j}\mu _{w_ j}$. After a further extraction, these finite measures converge weakly to a measure $\nu $, and

We claim that $\mu _ w=\mathrm{e}^ w\nu $. By lemma 5.44.2,

Let $f$ be continuous. Since $\nu _ j\leq \mu $, replacing $\mathrm{e}^{w_ j}$ by $\mathrm{e}^ w$ in $\int _ Xf \mathrm{e}^{w_ j}\, \mathrm{d}\nu _ j$ changes the integral by an amount tending to zero. For the remaining integral, approximate the fixed bounded measurable function $f \mathrm{e}^ w$ in $L^1(\mu )$ by continuous functions. Since $\nu _ j,\nu \leq \mu $, the approximation error is uniform in $j$, while the integrals of each continuous approximation converge by weak convergence. Thus

This proves the claim. The measure $\nu $ does not charge the poles of $w$, so $\mathrm{e}^{-w}\mu _ w=\nu $. Hence $w\in \mathcal T$ attains the maximum $\alpha $.

Suppose that $\mu -\nu $ were nonzero. Then corollary 10.31.28, applied to $\mu -\nu $, would give $v\in \mathcal E_\varphi $ with

In particular $\mu _ v\leq \mathrm{e}^{\sup _ Xv}\mu \leq A \mathrm{e}^{\sup _ Xv}\sigma $, so the bounded-carrier hypothesis of theorem 10.31.38 holds. That theorem, with $\lambda =1$, gives $a\in \mathcal E_\varphi $ with

The right-hand side lies between $\nu _0$ and $\mu $, so $a\in \mathcal T$, but its mass exceeds $\alpha $. This contradicts maximality. Therefore $\nu =\mu $, and $w$ solves the equation.

Now let $\mu $ be any nonzero finite non-pluripolar measure. The bounded-carrier decomposition lemma 15.24.4, applied to $\omega $, writes

Set $f_ k=\min (f,k)$ and $\mu _ k=f_ k\sigma $. These measures are nonzero, and the first part gives $u_ k\in \mathcal E_\varphi $ with $\mu _{u_ k}=\mathrm{e}^{u_ k}\mu _ k$. Since $\mu _ k$ increases with $k$, exponential order shows that $u_ k$ decreases. The uniform bound

shows that the decreasing limit $u$ is not identically $-\infty $. Moreover $\mu _{u_ k}\leq \mathrm{e}^{\sup _ Xu_1}\mu $, so fixed-dominator continuity gives $u\in \mathcal E_\varphi $ and weak convergence of these measures. On the right-hand side, $\mathrm{e}^{u_ k}f_ k\to \mathrm{e}^ u f$ in $L^1(\sigma )$ by dominated convergence, with dominating function $\mathrm{e}^{\sup _ Xu_1}f$. Thus $\mu _ u=\mathrm{e}^ u\mu $.

For general $\lambda >0$, apply the result for exponent one in the background $\lambda \theta $, with model $\lambda \varphi $ and datum $\lambda ^ n\mu $, and divide the resulting potential by $\lambda $. The scaling of the relative class and of the Monge–Ampère measure gives the equation. Uniqueness follows by applying corollary 13.19.3 to two solutions in both directions. $\square$