Theorem 5.25.2FC59B7D406A3

Proof. Let $(u_ j)$ be a sequence in $\mathcal K_\theta $. By upper semicontinuity, there are points $x_ j$ with $u_ j(x_ j)=0$; passing to a subsequence, we may assume that $x_ j\to x$. Cover $X$ by finitely many connected coordinate balls, and on a slightly larger ball around each of them choose a fixed smooth function $r_ i$ such that every $u_ j+r_ i$ is plurisubharmonic; such corrections exist because $\theta $ is smooth. Since $u_ j\leq 0$, these local functions are uniformly bounded above.

On a ball containing $x$ in its interior, locally uniform convergence to $-\infty $ is impossible, because the values $(u_ j+r_ i)(x_ j)=r_ i(x_ j)$ stay bounded below. Local compactness therefore gives a subsequence converging in $L^1$ on this ball. If a second ball meets the first, local compactness on the second ball cannot yield convergence to $-\infty $ on compact subsets of the intersection, since the first limit is finite almost everywhere there. Thus a further subsequence converges in $L^1$ on the second ball. Since $X$ is connected, the intersection graph of the finite cover by connected balls is connected. Proceeding along this graph and passing to subsequences finitely many times, we obtain local $L^1$ convergence on every ball, and a partition of unity then gives convergence in $L^1(X)$ to a function $u$.

The inequalities $\theta +\mathrm{dd}^{\mathrm{c}}u_ j\geq 0$ pass to the limit in the sense of distributions. By proposition 5.22.2, $u$ has a $\theta $-plurisubharmonic representative, still denoted by $u$. Since $u_ j\leq 0$, we have $u\leq 0$ almost everywhere, hence everywhere by lemma 5.24.3.

By lemma 5.25.1, $0=\lim _ j u_ j(x_ j)\leq u(x)$. Thus $u(x)=0$, and the limit belongs to $\mathcal K_\theta $. This proves sequential compactness, which is compactness in the metric space $L^1(X)$. The norm bound follows, since a compact subset of $L^1(X)$ is bounded. $\square$