Theorem 15.35.29016C2906071

Proof. Since $f_ k\leq f_{k+1}$, lemma 14.19.28 gives $u_{k+1}\leq u_ k$.

We first show that every exponential subsolution $v$ lies below each $u_ k$. Let $t>0$. On the set $\{ u_ k<v-t\} $,

Since $u_ k$ is bounded, $P_{h/2}(u_ k)\geq \inf _ Xu_ k$ is not identically $-\infty $, and $v-t\preceq u_ k$. Strict domination, theorem 15.2.2, therefore gives $u_ k\geq v-t$. Letting $t\downarrow 0$ gives $u_ k\geq v$.

In particular $s\leq u_ k\leq u_1$ for every $k$, so the decreasing limit $u_\lambda \geq s$ is not identically $-\infty $ and belongs to $\mathrm{PSH}(X,h)$. By lemma 5.29.19, $u_ k\to u_\lambda $ in capacity. Choose a constant $A\geq \sup _ Xu_1$. The nonpositive functions $u_ k-A$ satisfy

and $\mathrm{e}^{\lambda A}f_ k\to \mathrm{e}^{\lambda A}f$ in $L^1(\mathrm{d}V)$. Theorem 15.34.2, applied with $\mu =\mathrm{d}V$ and constant exponents, shows that $\mathrm{MA}^{\mathrm{ext}}_ h(u_\lambda )$ is finite and that $u_\lambda $ solves (15.35.1). Finally, every exponential subsolution lies below all $u_ k$, hence below $u_\lambda $, which is itself a subsolution. $\square$