Lemma 16.28.432A500AE8E8B3

Proof. In the pluripolar case, choose $b\in \mathcal L_1$ with $b=-\infty $ on $E$ by lemma 16.28.12, and a finite upper bound $C$ for $b$ on $B$. The functions

are competitors for $h_{E,B}$, and they tend to zero almost everywhere as $\epsilon \downarrow 0$. Hence $h_{E,B}^*\geq 0$ almost everywhere, and then everywhere by lemma 5.24.3. Since $h_{E,B}^*\leq 0$, it vanishes.

Now suppose that $E$ is non-pluripolar. Proposition 16.28.10 gives a finite upper bound $M\geq 1$ for $V_ E^*$ on $B$. For each competitor $v$ for $V_ E$, the function $\max (-1,v/M-1)$ is a competitor for $h_{E,B}$. Taking suprema and regularizing gives

As $V_ E^*\geq 0$, the equality $h_{E,B}^*(a)=-1$ forces $V_ E^*(a)=0$.

For the converse, choose $r<R_0<R$ and $k>0$ with $k\log (R_0/r)>1$. The plurisubharmonic function

belongs to $\mathcal L_ k$ and equals $-\infty $ at $a$. For a competitor $u$ for $h_{E,B}$, define $W=\max (u+1,L)$ on $B$ and $W=L$ outside $B$. On the collar $R_0<\lVert z-a\rVert <R$ we have $L>1\geq u+1$, so the two definitions agree there, and $W$ is plurisubharmonic by lemma 5.24.2. On $E$ one has $u=-1$ and $L\leq 0$, so $W=0$; on the ball $B(a,r)$ one has $L<0\leq u+1$, so $W=u+1$. Hence $W/k$ is a competitor for $V_ E$, with $k$ independent of $u$. Taking the supremum over $u$ and regularizing at $a$ gives

Thus $V_ E^*(a)=0$ forces $h_{E,B}^*(a)=-1$. $\square$