Theorem 4.2.180056FED1AF8C

Proof. We sweep the partial sums of a decomposition of $\mu $ into pieces with bounded Green potentials, show that the swept measures increase to a measure $\nu $ satisfying (4.2.17), and then identify $G_ B\nu $ with the regularized reduction. By the last assertion of proposition 4.2.11, $\mu $ is the sum of its restrictions to countably many pairwise disjoint compact sets, and the partial sums $\mu _ n$ of this series, which satisfy $\mu _ n\uparrow \mu $, have bounded Green potentials. Hence theorem 4.2.16 applies to the measures $\mu _ n$ and to the differences $\mu _ m-\mu _ n$ with $m>n$, whose Green potentials are bounded by $G_ B\mu _ m$.

By additivity, $\mu _ m^ A=\mu _ n^ A+(\mu _ m-\mu _ n)^ A\geq \mu _ n^ A$. Hence the measures $\mu _ n^ A$ increase setwise, and $\nu (E)=\lim _ n\mu _ n^ A(E)$ defines a measure $\nu $ on $A$: if a Borel set $E$ is the disjoint union of Borel sets $E_ k$, the monotone convergence theorem for series gives $\nu (E)=\lim _ n\sum _ k\mu _ n^ A(E_ k)=\sum _ k\nu (E_ k)$. Moreover $\nu (A)=\lim _ n\mu _ n^ A(A)\leq \mu (B)$. For every nonnegative Borel function $f$ we have

This is clear for simple functions, and in general it follows by approximating $f$ from below by an increasing sequence of simple functions $f_ k$ and interchanging the suprema over $k$ and $n$ of $\int f_ k\, \mathrm d\mu _ n^ A$.

Testing (4.2.17) for $\mu _ n$ against $\tau =\mu _ n^ A$ and using the Cauchy–Schwarz inequality, we get

so that $I_ B(\mu _ n^ A)\leq I_ B(\mu _ n)\leq I_ B(\mu )$. We claim that $I_ B(\mu _ n^ A)\uparrow I_ B(\nu )$. The sequence increases and is bounded by $I_ B(\nu )$, because $\mu _ n^ A\leq \mu _{n+1}^ A\leq \nu $ and $G_ B\geq 0$. Conversely, applying (4.2.19) to $f=G_ B(x,\cdot )$, we get $G_ B\mu _ m^ A\uparrow G_ B\nu $ pointwise as $m\to \infty $. For $m\leq n$ we have $I_ B(\mu _ n^ A)\geq \int G_ B\mu _ m^ A\, \mathrm d\mu _ n^ A$, and by (4.2.19) the right side tends to $\int G_ B\mu _ m^ A\, \mathrm d\nu $ as $n\to \infty $; letting $m\to \infty $ and using monotone convergence for the fixed measure $\nu $, we obtain $\lim _ nI_ B(\mu _ n^ A)\geq \int G_ B\nu \, \mathrm d\nu =I_ B(\nu )$. This proves the claim, and hence $I_ B(\nu )\leq I_ B(\mu )<\infty $.

Now let $\tau $ be a positive measure on $A$ with finite Green energy. We have $I_ B(\mu _ n,\tau )=I_ B(\mu _ n^ A,\tau )$ for every $n$. As $n\to \infty $, the left side $\int G_ B\tau \, \mathrm d\mu _ n$ increases to $\int G_ B\tau \, \mathrm d\mu =I_ B(\mu ,\tau )$ by monotone convergence, since $\mu _ n$ is the restriction of $\mu $ to an increasing sequence of sets whose union carries $\mu $, and the right side $\int G_ B\tau \, \mathrm d\mu _ n^ A$ increases to $I_ B(\nu ,\tau )$ by (4.2.19). Hence $I_ B(\mu ,\tau )=I_ B(\nu ,\tau )$, which is (4.2.17) for $\nu $.

By monotone convergence, $G_ B\mu _ n\uparrow G_ B\mu $ pointwise on $B$, and we have seen that $G_ B\mu _ n^ A\uparrow G_ B\nu $. Since $G_ B\mu _ n^ A\leq G_ B\mu _ n$, it follows that $G_ B\nu \leq G_ B\mu $ everywhere. Moreover, $G_ B\mu _ n^ A=G_ B\mu _ n$ on $A$ outside a polar set $E_ n$, so $G_ B\nu =G_ B\mu $ on $A\setminus E$, where $E=\bigcup _ nE_ n$ is polar by proposition 4.1.11.

We now identify $\widehat R_{G_ B\mu }^ A$ with $G_ B\nu $. Since $B$ is bounded and $E$ is polar, proposition 4.2.13, applied with $U=B$ and with $E$ in place of $A$, provides a nonnegative superharmonic function $p$ on $B$, locally integrable and hence finite almost everywhere, with $p=+\infty $ on $E$. For every $\varepsilon >0$, the function $G_ B\nu +\varepsilon p$ is nonnegative and superharmonic, and it is at least $G_ B\mu $ at every point of $A$: on $A\setminus E$ because $G_ B\nu =G_ B\mu $ there, and on $E$ because $p=+\infty $. Hence $R_{G_ B\mu }^ A\leq G_ B\nu +\varepsilon p$. Letting $\varepsilon \downarrow 0$, we get $R_{G_ B\mu }^ A\leq G_ B\nu $ wherever $p<\infty $, that is, almost everywhere. Thus $\widehat R_{G_ B\mu }^ A\leq G_ B\nu $ almost everywhere, and hence everywhere by the ball-average formula preceding proposition 4.2.13, since both functions are superharmonic. Conversely, $G_ B\mu _ n\leq G_ B\mu $, so proposition 4.2.13 gives $\widehat R_{G_ B\mu }^ A\geq \widehat R_{G_ B\mu _ n}^ A=G_ B\mu _ n^ A$ for every $n$, and therefore $\widehat R_{G_ B\mu }^ A\geq G_ B\nu $. Consequently $\widehat R_{G_ B\mu }^ A=G_ B\nu $, and we put $\mu ^ A=\nu $; this measure is unique because a measure is determined by its Green potential.

It remains to collect the other conclusions. The inequality $G_ B\mu ^ A\leq G_ B\mu $, the equality quasi-everywhere on $A$, the mass bound and the orthogonality relation have been proved above. If $\nu _1$ and $\nu _2$ are positive measures on $A$ with finite Green energy satisfying (4.2.17) in place of $\mu ^ A$, then $I_ B(\nu _1,\tau )=I_ B(\mu ,\tau )=I_ B(\nu _2,\tau )$ for $\tau =\nu _1$ and $\tau =\nu _2$, so $\nu _1=\nu _2$ by the uniqueness criterion at the beginning of this section; thus (4.2.17) characterizes $\mu ^ A$. If $\mu _1$ and $\mu _2$ have finite Green energy, so has $\mu _1+\mu _2$, because the mutual energy $I_ B(\mu _1,\mu _2)$ is finite; and $\mu _1^ A+\mu _2^ A$ is a positive measure on $A$ with finite Green energy which satisfies (4.2.17) for the source $\mu _1+\mu _2$. By the characterization, $(\mu _1+\mu _2)^ A=\mu _1^ A+\mu _2^ A$, and in the same way $(c\mu _1)^ A=c\mu _1^ A$ for $c\geq 0$. Taking $\tau =\mu ^ A$ in (4.2.17) gives $I_ B(\mu ,\mu ^ A)=I_ B(\mu ^ A)$, and then

which is nonnegative by proposition 4.2.11. Finally, if $\mu $ is supported on $A$, then $\mu $ itself satisfies (4.2.17), so $\mu ^ A=\mu $ by the characterization. $\square$