PomeTheorem 3.3.14.EEE7DD56F49D For a finite signed measure $\mu =f\sigma +\mu _ s$ on $S=\partial B(0,R)\subset \mathbb R^ d$, $d\ge 2$, the ball Poisson integral has nontangential limit $f(\zeta )$ at surface-almost every $\zeta \in S$. Precisely, for every fixed $A\ge 1$, it converges as $x\to \zeta $ in $B$ with $|x-\zeta |\le A(R-|x|)$. The same single full-measure set works for all finite $A$.
3.3.13 Nontangential boundary limits of Poisson integrals3412F805AF75
Proof. We first prove that a positive finite measure $\nu $ with zero spherical cap density at $\zeta $ has nontangential Poisson limit zero there. Put $t=R-|x|$ and $s=|y-\zeta |$. Since $|x-y|\ge t$ and $s\le |x-y|+At\le (1+A)|x-y|$,
\[ P_ B(x,y)\le C_{A,d}\frac{t}{(t+s)^ d}. \]
The numerator in the exact ball kernel is at most $2Rt$; the displayed bound follows with an explicit finite constant depending only on $A,d$ and the unit-sphere area. Given $\varepsilon >0$, choose $\delta >0$ so that $\nu (C_ s)\le \varepsilon \sigma (C_ s)\le C_ R\varepsilon s^{d-1}$ for $0<s\le \delta $. For $2t<\delta $, integrate first on $C_ t$, then on the annuli $2^{j-1}t<s\le 2^ jt\le \delta $. The contributions are bounded by $C\varepsilon $ and $C\varepsilon 2^{-j}$, respectively. Choose the last radius $r=2^ jt$ so that $\delta /2<r\le \delta $; outside $C_ r$ the integral is at most $C t\delta ^{-d}\nu (S)$, which tends to zero. Thus the limsup is at most $C\varepsilon $, and then zero.
By spherical measure differentiation, outside a null set the positive measure $|\mu _ s|$ has zero cap density and $|f-f(\zeta )|\sigma $ has zero cap density. Since $P_ B\sigma =1$,
\[ |P_ B\mu (x)-f(\zeta )|\le P_ B\bigl (|f-f(\zeta )|\sigma +|\mu _ s|\bigr )(x), \]
which tends to zero by the first part. The exceptional set used only these density conditions, not $A$; hence it works for every finite aperture. $\square$
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