Lemma 5.27.19F7D352207205

Proof. Fix a compact set $E\subset S$ and $\varepsilon >0$, and put $w_ c=\max \{ b-c,(1+\varepsilon )a\} $ for $c>0$. Since $a=-\infty $ on the compact set $E$, upper semicontinuity gives a neighborhood of $E$ on which $a<-(C+c)/\varepsilon $ and $a\leq b+C$. Off $S$, on this neighborhood, $b-c-(1+\varepsilon )a\geq -C-c-\varepsilon a>0$, while on $S$ the second branch of the maximum equals $-\infty $. Thus $w_ c=b-c$ near $E$, and locality on open sets gives $(\mathrm{dd}^{\mathrm{c}}w_ c)^ n_{\mathrm{full}}(E)=(\mathrm{dd}^{\mathrm{c}}b)^ n_{\mathrm{full}}(E)$.

As $c\to \infty $, the functions $w_ c$ decrease to $(1+\varepsilon )a$. Theorem 5.27.18 applies to every $w_ c$ with the same set $S$, so their full measures converge weakly to $(1+\varepsilon )^ n(\mathrm{dd}^{\mathrm{c}}a)^ n_{\mathrm{full}}$. Since the mass of a compact set is upper semicontinuous under weak convergence of positive measures, we obtain

Here we apply the semicontinuity after multiplying the measures by a fixed nonnegative cutoff function with compact support, equal to one near $E$; both masses on $E$ are finite by the theorem. Letting $\varepsilon \downarrow 0$ and applying inner regularity to all compact sets $E\subset S$ gives the asserted inequality of measures. If $a-b$ is bounded near $S$, then also $\{ b=-\infty \} =S$, so the roles of $a$ and $b$ can be interchanged, and equality follows. $\square$