Lemma 17.7.67B309BFF1A4D

Proof. Denote by $U^\alpha $ the complexification of $(u_ t^\alpha )$. By proposition 17.7.4, $u_ t^\alpha \leq (1-t)u_0^\alpha +tu_1^\alpha $, so the functions $U^\alpha $ are uniformly bounded above. Their regularized upper envelope $U$ is therefore $p^*\theta $-psh (theorem 5.10.2, applied in charts after adding smooth local potentials of $p^*\theta $), and it is invariant under imaginary translations, because the family is and these translations are homeomorphisms. Thus $U$ is the complexification of the family of its slices $U_ t$, which by lemma 17.7.3 are $\theta $-psh and have convex fibers $t\mapsto U_ t(x)$.

We next reduce to a countable subfamily. Cover $X$ by finitely many coordinate balls $D_\beta $ on which $\theta $ has smooth potentials $h_\beta $ (lemma 6.14.2). By theorem 5.10.2, applied to the plurisubharmonic functions $U^\alpha +h_\beta \circ p$ on $D_\beta \times S$ and to the functions $u_0^\alpha +h_\beta $ and $u_1^\alpha +h_\beta $ on $D_\beta $, there are countable sets $J_\beta \subset A$ such that, on $D_\beta \times S$ and on $D_\beta $ respectively, the regularized suprema over $\alpha \in J_\beta $ of $U^\alpha $, of $u_0^\alpha $ and of $u_1^\alpha $ equal $U$, $v_0$ and $v_1$. The countable set $J=\bigcup _\beta J_\beta $ then satisfies $(\sup _{\alpha \in J}U^\alpha )^*=U$, $(\sup _{\alpha \in J}u_0^\alpha )^*=v_0$ and $(\sup _{\alpha \in J}u_1^\alpha )^*=v_1$, since on each $D_\beta \times S$, respectively $D_\beta $, these regularized suprema lie between the regularized suprema over $J_\beta $ and over $A$.

Let $v_ t=(\sup _\alpha u_ t^\alpha )^*$; we show that $U_ t=v_ t$. Since $U_ t$ is upper semicontinuous on $X$ and dominates every $u_ t^\alpha $, we have $v_ t\leq U_ t$. For $a<t<b$, the inequality $u_ t^\alpha \leq \frac{b-t}{b-a}u_ a^\alpha +\frac{t-a}{b-a}u_ b^\alpha \leq \frac{b-t}{b-a}v_ a+\frac{t-a}{b-a}v_ b$ and upper semicontinuity of the right side give $v_ t\leq \frac{b-t}{b-a}v_ a+\frac{t-a}{b-a}v_ b$. Thus $t\mapsto v_ t(x)$ is convex for every $x$. In the same way, the bound $u_ t^\alpha \leq (1-t)u_0^\alpha +tu_1^\alpha $ gives $v_ t\leq (1-t)v_0+tv_1$. Fix one member $\alpha _0$ of the family and let $w_0=\inf _ tu_ t^{\alpha _0}$, a $\theta $-psh potential by proposition 17.7.4. Since $w_0\leq v_ t\leq U_ t$ and $U$ is bounded above, all $v_ t(x)$ and $U_ t(x)$ are finite on the set where $w_0$ is finite, which has full measure.

The function $F_ J=\sup _{\alpha \in J}U^\alpha $ is Borel, being a countable supremum of upper semicontinuous functions, and by theorem 5.10.2 it agrees with its regularization $U$ outside a Lebesgue null set. Both $F_ J$ and $U$ are invariant under imaginary translations, so this exceptional set is $\{ (x,\zeta ):(x,\operatorname{Re}\zeta )\in N\} $ for a Borel set $N\subset X\times (0,1)$, which is null by Fubini's theorem. Applying Fubini's theorem to $N$, we find that for almost every $x$ we have $U_ t(x)=\sup _{\alpha \in J}u_ t^\alpha (x)\leq v_ t(x)$, hence $U_ t(x)=v_ t(x)$, for almost every $t\in (0,1)$. If moreover $w_0(x)$ is finite, both sides are finite convex functions of $t$, hence continuous (lemma 17.7.2(1)), so they agree for every $t\in (0,1)$. Thus, for each $t$, $U_ t=v_ t$ almost everywhere, and both are $\theta $-psh, so $U_ t=v_ t$ everywhere.

For the endpoints, note first that, since $U_ t=v_ t\leq (1-t)v_0+tv_1\leq \max \{ v_0,v_1\} $, lemma 17.7.2(1),(3) show that $\lim _{t\to 0}U_ t(x)$ exists at every $x$, as an element of $[-\infty ,\infty )$. For $\alpha \in J$ we have $U_ t\geq u_ t^\alpha $, so by proposition 17.7.4 $\lim _{t\to 0}U_ t\geq \lim _{t\to 0}u_ t^\alpha =u_0^\alpha $ outside a pluripolar set. Outside a countable union of pluripolar sets this holds for all $\alpha \in J$, so $\lim _{t\to 0}U_ t\geq \sup _{\alpha \in J}u_0^\alpha =v_0$ quasi-everywhere, by corollary 5.47.11(1); countable unions of pluripolar sets are pluripolar by theorem 5.47.8. The bound $U_ t=v_ t\leq (1-t)v_0+tv_1$ gives the reverse inequality. Hence $U_ t\to v_0$ almost everywhere as $t\to 0$. Since $w_0\leq U_ t\leq \max \{ v_0,v_1\} $, dominated convergence gives $U_ t\to v_0$ in $L^1(X)$. The endpoint $t=1$ is treated in the same way, after reversing the parameter. $\square$