Proposition 15.35.18.0535F55900CB Suppose $\Lambda _ h(f)\ne \varnothing $, and put $a_0=\inf \Lambda _ h(f)$. For every $a>a_0$, there is $L_ a\in \mathrm{PSH}(X,h)$ such that
\[ U_\lambda \longrightarrow L_ a \quad \text{in }L^1\text{ and in capacity}\quad (\lambda \uparrow a). \]
It satisfies
\[ L_ a\leq U_ a,\qquad \mathrm{MA}^{\mathrm{ext}}_ h(L_ a)=\mathrm e^{aL_ a}f\, \mathrm dV. \]
The equation measures converge in total variation to this finite measure. The left limit is independent of the auxiliary exponent and of the cofinal sequence used to construct it.
Proof.
Choose $\tau \in \Lambda _ h(f)$ with $0<\tau <a$; the upper-interval property gives such a choice. Put $K=K_\tau $ from lemma 15.35.15 and
\[ V_\lambda =U_\lambda +K\log (\lambda /\tau ) \qquad (\lambda \in \Lambda _ h(f),\ \lambda \geq \tau ). \]
This family is increasing. For a sequence $\lambda _ j\uparrow a$ with $\tau \leq \lambda _ j<a$, its members lie between the integrable potentials $V_\tau $ and $V_ a$. The upper-envelope construction in section 5.40 gives
\[ V^-_ a=\left(\sup _{\tau \leq \lambda <a}V_\lambda \right)^* =\left(\lim _ jV_{\lambda _ j}\right)^*\in \mathrm{PSH}(X,h). \]
Indeed cofinality and the order give the same pointwise supremum for every such sequence. Regularization changes it only on a pluripolar set. Dominated convergence between the integrable bounds gives $L^1$ convergence to $V^-_ a$, and lemma 5.29.19 gives convergence in capacity. Since the added constants converge, set
\[ L_ a=V^-_ a-K\log (a/\tau ). \]
Then $U_{\lambda _ j}\to L_ a$ in both topologies and $L_ a\leq U_ a$; the latter inequality survives regularization because $V_ a$ is upper semicontinuous.
The potentials $U_{\lambda _ j}-K$ and their limit $L_ a-K$ are nonpositive. Apply theorem 15.34.2 with exponents tending to $a>0$ and densities $\mathrm e^{\lambda _ jK}f\to \mathrm e^{aK}f$ in $L^1$. It proves the exact equation and total-variation convergence.
Every sequence approaching $a$ from the left has a cofinal increasing subsequence. If potential convergence in either topology, or total-variation convergence of the equation measures, failed, a sequence staying outside a fixed neighbourhood of the proposed limit would therefore have an increasing subsequence, contrary to the preceding argument. This proves the full left limit and measure convergence. Repeating the construction with another $\tau $ gives the same $L^1$ limit; canonical quasi-plurisubharmonic representatives that agree almost everywhere agree everywhere. Thus the auxiliary choice does not change $L_ a$.
$\square$
Pome
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