Proposition 15.35.17B1974D837A1A

Proof. Let $K=K_ a$ be the constant of the lemma with $\tau =a$, and let $\lambda _ j>a$ decrease to $a$. By the lemma, the functions

decrease and satisfy $V_ j\geq U_ a$. Their limit $V$ therefore belongs to $\mathrm{PSH}(X,h)$ and satisfies $V\geq U_ a$. Dominated convergence between the integrable bounds $U_ a$ and $V_1$ gives $V_ j\to V$ in $L^1$, and lemma 5.29.19 gives convergence in capacity. Since the added constants tend to zero, $U_{\lambda _ j}\to V$ in both senses.

The nonpositive functions $U_{\lambda _ j}-K$ satisfy

and $\mathrm e^{\lambda _ jK}f\to \mathrm e^{aK}f$ in $L^1$. Theorem 15.34.2 therefore gives $\mathrm{MA}^{\mathrm{ext}}_ h(V)=\mathrm e^{aV}f\, \mathrm dV$, together with total-variation convergence of the measures $\mathrm{MA}^{\mathrm{ext}}_ h(U_{\lambda _ j})$. Hence $V$ is a solution at $a$, so $V\leq U_ a$ by maximality, while $V\geq U_ a$ by construction. Thus $V=U_ a$. Since every sequence of exponents converging to $a$ from above has a decreasing subsequence, this proves the assertions. $\square$