Theorem 4.2.16F28E6D45483F

Proof. The function $G_ B\mu $ is nonnegative, superharmonic by part (2) of proposition 4.1.16, and bounded on $A$. Applying theorem 4.2.14 to $s=G_ B\mu $, we see that $\widehat R_{G_ B\mu }^ A$ is the Green potential of a positive measure $\mu ^ A$ on $A$ with finite Green energy, and that $G_ B\mu ^ A=G_ B\mu $ quasi-everywhere on $A$. Since $G_ B\mu $ is itself admissible, $G_ B\mu ^ A=\widehat R_{G_ B\mu }^ A\leq G_ B\mu $. The measure $\mu ^ A$ is unique because $-\Delta G_ B\nu =\nu $ for every measure $\nu $, by part (2) of proposition 4.1.16, so that a measure is determined by its Green potential. Since $-G_ B\mu \leq -G_ B\mu ^ A$ on $B$, the comparison of boundary fluxes in part (2) of proposition 4.1.16, applied with $\lambda =\mu $ and $\lambda '=\mu ^ A$, gives the mass bound $\mu ^ A(A)\leq \mu (B)$.

Let $\tau $ be a positive measure on $A$ with finite Green energy. It does not charge polar sets, so we may integrate the equality $G_ B\mu =G_ B\mu ^ A$, valid quasi-everywhere on $A$, against $\tau $. Both integrals are finite, since $0\leq G_ B\mu ^ A\leq G_ B\mu $ and $G_ B\mu $ is bounded on $A$. By Tonelli's theorem and the symmetry of $G_ B$, the two integrals are $I_ B(\mu ,\tau )$ and $I_ B(\mu ^ A,\tau )$, which proves (4.2.17).

Suppose that $\nu _1$ and $\nu _2$ are positive measures on $A$ with finite Green energy which both satisfy (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. If $\mu _1$ and $\mu _2$ satisfy the boundedness hypothesis, so does $\mu _1+\mu _2$, and $\mu _1^ A+\mu _2^ A$ is a positive measure on $A$ with finite Green energy satisfying (4.2.17) for the source $\mu _1+\mu _2$. By the characterization, $(\mu _1+\mu _2)^ A=\mu _1^ A+\mu _2^ A$. The same argument gives $(c\mu _1)^ A=c\mu _1^ A$. $\square$