Lemma 13.2.2D59675D9F6D8

Proof. Let $l\geq k$. By proposition 5.47.23, the level-$l$ and level-$k$ Bedford–Taylor products have the same restriction to $O_ k$; since $O_ k\subset O_ l$, this is the identity $\mathbf1_{O_ k}S_ l=S_ k$. Consequently $S_ l-S_ k=\mathbf1_{O_ l\setminus O_ k}\mathrm{dd}^{\mathrm{c}}u_{1,l}\wedge \cdots \wedge \mathrm{dd}^{\mathrm{c}}u_{p,l}$, and this current is positive by lemma 13.2.1.

Suppose that (13.2.3) holds. The trace measures $S_ k\wedge \beta ^{n-p}$ increase with $k$ and, by (13.2.3), converge to a locally finite measure. Hence the trace mass of the positive current $S_ l-S_ k$ on a compact set $K$ tends to zero as $k,l\to \infty $. By theorem 5.3.7, the total variation on $K$ of each coefficient measure of $S_ l-S_ k$ is at most a fixed multiple of this trace mass. Therefore each coefficient measure of $S_ k$ is a Cauchy sequence in total variation on compact sets, and the limits are the coefficients of a current $S$, which is positive as a weak limit of positive currents. Conversely, suppose that $S_ k\to S$ weakly for a positive current $S$. For every smooth compactly supported $\chi \geq 0$, the numbers $\int \chi S_ k\wedge \beta ^{n-p}$ increase with $k$ and converge to $\int \chi S\wedge \beta ^{n-p}$, which therefore bounds them; choosing $\chi =1$ on $K$ gives (13.2.3).

Assume now that (13.2.3) holds. Letting $l\to \infty $ in $\mathbf1_{O_ k}S_ l=S_ k$ gives $\mathbf1_{O_ k}S=S_ k$. By the convergence in total variation, the trace measure of $S$ is the increasing limit of the trace measures of the $S_ k$, which is $\mu $. By proposition 5.47.23, every pluripolar subset of $U$ has outer $\mu $-measure zero. The total variation of each coefficient measure of $S$ is at most $C_{n,p}\mu $ (theorem 5.3.7), so pluripolar sets have outer measure zero for these total variations as well, and theorem 5.47.20(2), applied to $\mu $ and to these total variations, shows that plurifine open subsets of $U$ are measurable for their completions. In particular $S$ vanishes on $U\setminus \bigcup _ kO_ k$, the union of the pole sets of the $u_ j$, which is a pluripolar Borel set.

Finally, let $E_ k$ be as in the statement. Since $E_ k\subset O_ k$, we have $\mathbf1_{E_ k}S_ k=\mathbf1_{E_ k}S$. By theorem 5.3.7, the total variation of each coefficient measure of $S$ is locally finite and absolutely continuous with respect to the trace measure. Hence $\mathbf1_{E_ k}\to 1$ almost everywhere for each of these variations, and dominated convergence proves the assertion. If the $E_ k$ increase and $U\setminus \bigcup _ kE_ k$ is pluripolar, then $\mathbf1_{E_ k}\to 1$ outside this pluripolar Borel set, which $S$ does not charge. $\square$