Proposition 15.28.6D42D2088B421

Proof. The family is nonempty by theorem 15.28.4. By theorem 5.25.2, an $L^1$ limit of members of $\mathcal S_ f$ is $h$-plurisubharmonic with supremum zero, and lemma 15.28.3, with $g_ j=g=bf$, shows that it satisfies the measure inequality. Thus $\mathcal S_ f$ is a closed subset of the compact set of normalized $h$-plurisubharmonic functions, hence compact. It is also stable under finite maxima: if $s_1,s_2\in \mathcal S_ f$, then $\max (s_1,s_2)$ has supremum zero, and proposition 13.14.6, applied on the sets where the maximum equals $s_1$ or $s_2$, gives $\mathrm{MA}^{\mathrm{ext}}_ h(\max (s_1,s_2))\geq bf\, \mathrm{d}V$.

By Choquet's lemma there is a sequence $(s_ i)_{i\geq 1}$ in $\mathcal S_ f$ with $(\sup _ is_ i)^*=\Phi _ f$. Fix $s_0\in \mathcal S_ f$. The finite maxima $t_ N=\max (s_0,s_1,\ldots ,s_ N)$ belong to $\mathcal S_ f$, increase with $N$, and satisfy $s_0\leq t_ N\leq 0$; their regularized limit is $\Phi _ f$. Hence $t_ N\to \Phi _ f$ almost everywhere and, by dominated convergence, in $L^1$. Lemma 15.28.3 gives $\mathrm{MA}^{\mathrm{ext}}_ h(\Phi _ f)\geq bf\, \mathrm{d}V$, and $s_0\leq \Phi _ f\leq 0$ forces $\sup _ X\Phi _ f=0$. Thus $\Phi _ f\in \mathcal S_ f$, and by construction it dominates every member of $\mathcal S_ f$.

Finally, suppose $v_{\lambda _ j}\to v$ in $L^1$ with $\lambda _ j\downarrow 0$. The measures $\mathrm{MA}^{\mathrm{ext}}_ h(v_{\lambda _ j})=\mathrm{e}^{\lambda _ ju_{\lambda _ j}}f\, \mathrm{d}V$ have densities converging in $L^1$ to $bf$, by corollary 15.35.3. Applying lemma 15.28.3 to these densities and potentials shows that $v\in \mathcal S_ f$. $\square$