Singular circle measures have Poisson integral tending nontangentially to zero almost everywhere

Fix $\epsilon >0$ and choose the compact null set $K$ in [F1]. Split $\mu ={\mu }_K+\rho$, its restrictions to $K$ and its complement; then $\rho (\mathbb{T})<\epsilon$. At a circle point $\zeta \mathrm{\notin }K$, compactness gives $d=\mathrm{dist}(\zeta ,K)>0$ if $K$ is nonempty. For $\mathrm{\mid }z-\zeta \mathrm{\mid }<d\mathrm{/}2$, [F3] gives $P[{\mu }_K](z)\le 4d^{-2}(1-\mathrm{\mid }z{\mathrm{\mid }}^2)\mu (\mathbb{T})⟶0.$ If $K$ is empty, this integral is already zero. This limit holds along every approach within the disc, not only a radius.

Fix $A>1$ and $t>0$, and let $B_{A,t}$ be the circle points where the nontangential limsup of $P[\mu ]$ in ${\mathrm{\Gamma }}_A$ is greater than $t$. For $\zeta \mathrm{\notin }K$, step 1.1 and positivity imply that this limsup is the limsup for $P[\rho ]$. Thus [F2] gives $B_{A,t}\subseteq K\cup \{M_{\mathbb{T}}\rho >t\mathrm{/}(A+1{)}^2\}\mathrm{.}$ The right side is Borel and has Haar measure at most $3(A+1{)}^2\epsilon \mathrm{/}t$. Since $\epsilon$ is arbitrary, $B_{A,t}$ has Haar outer measure zero. This argument does not assume measurability of the cone limsup.

For each integer $j\ge 2$ and $k\ge 1$, apply step 2.1 with $A=j$ and $t=1\mathrm{/}k$. A set of outer measure zero is contained in a Borel null set: choose Borel supersets of mass less than $2^{-l}$ and intersect them, using countable choice. By countable choice choose these null supersets for the countable pairs $(j,k)$ and take their union, a Borel null set. Outside that union, positivity gives limsup zero in every cone of integer aperture $j\ge 2$, hence the limit zero there. Every cone of aperture $A>1$ is contained in one with integer aperture $j>A$, so the same full-measure set works for all $A$. For $\mu =0$ the integral is identically zero throughout.