PomeTheorem 3.3.9.CAE5D89698DF Let $B=B(0,R)\subset \mathbb R^ d$, $d\ge 2$, and let $\sigma $ be surface measure on $S=\partial B$. Let $\mu $ be a finite signed measure on $S$. For $\zeta \in S$ put $C_ s=\{ y\in S:|y-\zeta |\le s\} $. If $\mu (C_ s)/\sigma (C_ s)\to a\in \mathbb R$ as $s\downarrow 0$, then $P_ B\mu (t\zeta )\to a$ as $t\uparrow 1$.
Theorem 3.3.9CAE5D89698DF
Proof. Subtract $a\sigma $, whose Poisson integral is the constant $a$. Thus $F(s)=\mu (C_ s)=o(A(s))$, where $A(s)=\sigma (C_ s)$. In particular $\mu (\{ \zeta \} )=0$. Fix $0<\delta <2R$ so that $|F(s)|\le \varepsilon A(s)$ for $0<s\le \delta $. The ball Poisson kernel at $t\zeta $ is a function of $s=|y-\zeta |$:
\[ k_ t(s)=\frac{R^2(1-t^2)}{s_{d-1}R\, [R^2(1-t)^2+ts^2]^{d/2}}. \]
Here $s_{d-1}$ is the area of the unit sphere. This function is decreasing and smooth for $s\ge 0$, $0<t<1$. Integration of $k_ t(s)=k_ t(\delta )+\int _ s^\delta (-k_ t'(v))\, dv$ against the pushforward of $\mu |_{C_\delta }$, using finite signed Fubini, gives
\[ \int _{C_\delta }k_ t(|y-\zeta |)\, d\mu =k_ t(\delta )F(\delta )+\int _0^\delta (-k_ t'(s))F(s)\, ds. \]
For each fixed $t$ all integrands are bounded and $\mu $ is finite, so this identity also holds when $F$ has jumps. Its absolute value is at most $\varepsilon \int _{C_\delta }k_ t\, d\sigma \le \varepsilon $, by positivity and normalization of the Poisson kernel. On $S\setminus C_\delta $, $\sup k_ t\to 0$, so the remaining integral tends to zero, bounded by $|\mu |(S)\sup _{s\ge \delta }k_ t(s)$. Let $\varepsilon \downarrow 0$. $\square$
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