Proposition 5.47.23983478512954

Proof. Each set $\{ u_ l>-m\} $ is a subbasic plurifine open set, so $E_ m$ is plurifine open; it is also Borel. On $E_ m$ we have $u_ l>-m$, and hence $u_ l^{(k)}=u_ l=u_ l^{(m)}$ for all $l$ and all $k\ge m$. The truncations are locally bounded plurisubharmonic functions, so theorem 5.47.22 gives $\mathbf1_{E_ m}S_ k=\mathbf1_{E_ m}S_ m$. Taking trace measures, and using $E_ m\subset E_ k$ and the positivity of $\mu _{S_ k}$, we obtain $\mu _ m=\mathbf1_{E_ m}\mu _{S_ k}\le \mathbf1_{E_ k}\mu _{S_ k}=\mu _ k$. By corollary 5.47.11(3), every pluripolar subset of $\Omega $ has outer measure zero for $\mu _{S_ m}$, and hence for $\mu _ m$.

For a Borel set $A\subset \Omega $ put $\mu (A)=\lim _ m\mu _ m(A)\in [0,\infty ]$. If $A$ is the disjoint union of Borel sets $A_ i$, then

by monotone convergence for series of nonnegative terms that increase with $m$. Thus $\mu $ is a Borel measure. Finally, let $N\subset \Omega $ be pluripolar. Since $N$ has outer measure zero for each $\mu _ m$, there are Borel sets $G_{m,i}\supset N$ with $\mu _ m(G_{m,i})<1/i$ for all $m,i\ge 1$. The Borel set $G=\bigcap _{m,i}G_{m,i}$ contains $N$ and satisfies $\mu _ m(G)=0$ for every $m$, hence $\mu (G)=0$. Thus $N$ has outer measure zero for $\mu $. $\square$