4.2.15 Sweeping measures and harmonic measureDB98A10B30F4

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 $(c\mu _1)^ A=c\mu _1^ A$. $\square$

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. $\square$

Proof. We first check that $\partial W$ is the disjoint union of $\partial A$ and $\partial B$. Since $W=B\cap (\mathbb R^ d\setminus A)$ is the intersection of two open sets, $\partial W\subset \partial B\cup \partial A$, and $\partial A\subset A\subset B$ does not meet $\partial B$. Conversely, no point of $\partial B\cup \partial A$ lies in $W$. A point of $\partial B$ is a limit of points of $B\setminus A=W$, because the compact set $A\subset B$ has positive distance from $\partial B$. Every small ball around a point of $\partial A$ lies in $B$ and contains points outside $A$, that is, points of $W$. Consequently $\partial A$ and $\partial B$ are disjoint compact sets whose union is $\partial W$, and $f_ x$ is continuous on $\partial W$, because $G_ B(x,\cdot )$ is continuous on $B\setminus \{ x\} \supset A$.

Let $H=H_ Wf_ x$ be the solution of the Dirichlet problem on $W$ with boundary values $f_ x$; it is continuous on $\overline W$ since $W$ is regular, and $H\geq 0$ because $f_ x\geq 0$ and $H_ W$ preserves order. The function $G_ B(x,\cdot )=G_ B\delta _ x$ is superharmonic on $B$ by part (2) of proposition 4.1.16, it is continuous on $B\setminus \{ x\} $, and it tends to zero at $\partial B$ (theorem 3.1.34); hence its boundary limits on $\partial W$ are the values of $f_ x$. Therefore the maximum principle (proposition 3.4.7), applied to the subharmonic function $H-G_ B(x,\cdot )$ on the bounded open set $W$, gives $H\leq G_ B(x,\cdot )$ on $W$. Define $r=H$ on $W$ and $r=G_ B(x,\cdot )$ on $A$. Then $r$ is continuous on $B$, because $H$ and $G_ B(x,\cdot )$ have the same values on $\partial A$, and $0\leq r\leq G_ B(x,\cdot )$.

To see that $r$ is superharmonic, we argue as in Step 2 of the proof of theorem 4.2.14. Let $E$ be a ball with $\overline E\subset B$, and let $q$ be harmonic on $E$ and continuous on $\overline E$ with $q\leq r$ on $\partial E$. Then $q\leq G_ B(x,\cdot )$ on $\partial E$, hence on $E$ by superharmonicity of $G_ B(x,\cdot )$; in particular $r\geq q$ on $E\cap A$. On $E\cap W$ the harmonic function $H-q$ has nonnegative boundary limits: at points of $\partial E$ because $H$ extends continuously to $r\geq q$ there, and at points of $E\cap \partial A$ because $H$ tends to $G_ B(x,\cdot )\geq q$ there. The minimum principle gives $H\geq q$ on $E\cap W$. Hence $r\geq q$ on $E$, and $r$ is superharmonic by theorem 3.4.5.

The function $r$ is admissible for $R_{G_ B(x,\cdot )}^ A$: it is nonnegative and superharmonic on $B$ and equals $G_ B(x,\cdot )$ on $A$. Let $h$ be any admissible function. Then $h\geq r$ on $A$. On $W$, the superharmonic function $h$ satisfies $\varliminf _{y\to \zeta }h(y)\geq h(\zeta )\geq f_ x(\zeta )$ at points $\zeta \in \partial A$, by lower semicontinuity, and $\varliminf _{y\to \zeta }h(y)\geq 0=f_ x(\zeta )$ at points $\zeta \in \partial B$, since $h\geq 0$. Applying proposition 3.4.7 to the subharmonic function $H-h$ on $W$, we get $h\geq H$ there. Hence $h\geq r$ on $B$, and $R_{G_ B(x,\cdot )}^ A=r$ pointwise. Since $r$ is continuous, $\widehat R_{G_ B(x,\cdot )}^ A=r$, that is, $r_ x=r$.

On $W$, both $G_ B(x,\cdot )$ and $G_ W(x,\cdot )$ are locally integrable and satisfy $-\Delta =\delta _ x$ there (theorem 3.1.34). Their difference is therefore distributionally harmonic on $W$; by Weyl's lemma (theorem 3.1.11) it has a harmonic representative, which coincides with the difference off $x$. This harmonic function tends to $f_ x$ at $\partial W$, because $G_ W(x,\cdot )$ tends to zero there. By uniqueness of the solution of the Dirichlet problem (proposition 3.1.10), it equals $H$. The harmonic-measure formula is proposition 3.1.33 applied to $H=H_ Wf_ x$.

Finally, the measure $\delta _ x^ A=-\Delta r_ x$ vanishes on $W$, where $r_ x=H$ is harmonic, and on $\operatorname{Int}A$, where $r_ x=G_ B(x,\cdot )$ is harmonic because $x\notin A$. It is therefore supported on $\partial A$, and its mass is at most $\delta _ x(B)=1$ by theorem 4.2.16.

For the last assertion, write $B=B(c,R)$ and let $A=\overline{B(c,\rho )}$ with $0<\rho <R$, so that $W=\{ y:\rho <|y-c|<R\} $ is a bounded open annulus. It is connected: every point of $W$ is joined by a radial segment in $W$ to the sphere $\partial B(c,(\rho +R)/2)$, which is path-connected because $d\geq 2$. By theorem 3.1.29, it suffices to find, for every $\zeta \in \partial W$, a closed ball that meets $\overline W$ only at $\zeta $. If $|\zeta -c|=R$, the closed ball $\overline{B(2\zeta -c,R)}$ meets $\overline{B(c,R)}\supset \overline W$ only at $\zeta $, as noted at the beginning of subsection 3.1.32. If $|\zeta -c|=\rho $, let $m=(c+\zeta )/2$. Every $y\in \overline{B(m,\rho /2)}$ satisfies $|y-c|\leq |y-m|+|m-c|\leq \rho $, with equality only if $y-m=(\zeta -c)/2$, that is, $y=\zeta $. Since $|y-c|\geq \rho $ on $\overline W$, the closed ball $\overline{B(m,\rho /2)}$ meets $\overline W$ only at $\zeta $. Hence every boundary point of $W$ is regular, and $W$ is a regular domain. $\square$