Proposition 15.8.429A74E6E8A18

Proof. Using the infimum construction in the proof of proposition 16.28.3, take continuous $f_ j\uparrow f$ with common bounds $m\leq f_ j\leq M$, and shift $\rho $ below $m$. The envelopes $p_ j=P_\theta (f_ j)$ increase and satisfy

Their regularized limit $q=(\lim _ jp_ j)^*$ is therefore $\theta $-plurisubharmonic. Outside a pluripolar set it equals the pointwise limit, which is at most $f$; hence $q\leq p=P_\theta (f)$.

We prove the reverse inequality without using a domination theorem. Let $u$ be any competitor for $P_\theta (f)$, and let $u_ t\leq f$ be the potentials given by lemma 5.40.5, which satisfy this inequality everywhere. For fixed $t,\eta >0$, the sets

are open, increase with $j$, and cover $X$. By compactness, $u_ t\leq f_ j+\eta $ everywhere for all large $j$. Thus $u_ t-\eta $ is a competitor for $p_ j$, and $q\geq u_ t-\eta $. Letting first $\eta \downarrow 0$ and then $t\downarrow 0$, we obtain $q\geq u$ almost everywhere, hence everywhere by lemma 5.24.3. Taking the supremum over all competitors $u$ gives $q\geq p$, so $q=p$.

For fixed $k$, the set $U_ k=\{ p<f_ k\} $ is open. If $j\geq k$, then $p_ j\leq p<f_ k\leq f_ j$ on $U_ k$, so theorem 15.8.1 gives $\mathrm{MA}_\theta (p_ j)=0$ there. On every relatively compact open subset of $X\setminus Z$ the functions $p_ j$ have common bounds. After a common smooth correction and a finite expansion, the local Bedford–Taylor continuity along increasing sequences shows that their measures converge to $\mathrm{MA}_\theta (p)$ there. Since the measure of an open set is at most the lower limit of the measures of the approximating sequence, $\mathrm{MA}_\theta (p)$ vanishes on $U_ k\setminus Z$. The union of these sets is $\{ p<f\} \setminus Z$, and the extended measure does not charge the pluripolar set $Z$. $\square$