Fatou limits for Poisson extensions of L1 boundary data

Continuous data converge in every cone. Let $g\in C(\mathbb{T},\mathbb{C})$, $\zeta \in \mathbb{T}$ and $A>1$, and let $\epsilon >0$; choose $\delta >0$ with $\mathrm{\mid }g(\eta )-g(\zeta )\mathrm{\mid }<\epsilon$ for $d(\zeta ,\eta )<\delta$. For $z\in {\mathrm{\Gamma }}_A(\zeta )$ with $r=\mathrm{\mid }z\mathrm{\mid }$ and $u:=\mathrm{Re}(\bar{\zeta }z)\le r$ one has $\mathrm{\mid }r\zeta -z{\mathrm{\mid }}^2=2r(r-u)\le 1+r^2-2u=\mathrm{\mid }\zeta -z{\mathrm{\mid }}^2$ because $(1-r)(1+r-2u)\ge 0$, so for every $\eta \in \mathbb{T}$ the triangle inequality gives $\mathrm{\mid }\eta -r\zeta \mathrm{\mid }\le \mathrm{\mid }\eta -z\mathrm{\mid }+\mathrm{\mid }\zeta -z\mathrm{\mid }\le (1+A)\mathrm{\mid }\eta -z\mathrm{\mid }$, using $\mathrm{\mid }\eta -z\mathrm{\mid }\ge 1-r$ and the cone condition. Hence, by [L5] and the unit mass of the kernel, $\mathrm{\mid }P[g](z)-g(\zeta )\mathrm{\mid }\le {\int }_{\mathbb{T}}P(z,\eta )\mathrm{\mid }g(\eta )-g(\zeta )\mathrm{\mid }dm(\eta )\le \epsilon +2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}(A+1{)}^2{\sup }_{2\pi \delta \le \mathrm{\mid }\theta \mathrm{\mid }\le \pi }{P}_r(\theta ),$ and the last supremum tends to $0$ as $r\uparrow 1$; since $z\to \zeta$ inside ${\mathrm{\Gamma }}_A(\zeta )$ forces $r\to 1$, this is less than $2\epsilon$ for $z$ close enough to $\zeta$. Thus $P[g](z)\to g(\zeta )$ along ${\mathrm{\Gamma }}_A(\zeta )$, and in particular the cone limsup $L_m(g)(\zeta ):={\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta ,z\in {\mathrm{\Gamma }}_m(\zeta )}\mathrm{\mid }P[g](z)-g(\zeta )\mathrm{\mid }$ is $0$ for every integer $m\ge 2$.

The cone limsup is Borel measurable. Fix an integer $m\ge 2$ and let $D_0:={\mathbb{Q}}^2\cap \mathbb{D}$, countable and dense in $\mathbb{D}$ by [L7]. For $j\ge 1$ put $S_j(\zeta ):=\sup \{\mathrm{\mid }P[f](z)-f(\zeta )\mathrm{\mid }:z\in D_0,z\in {\mathrm{\Gamma }}_m(\zeta ),\mathrm{\mid }z-\zeta \mathrm{\mid }<1\mathrm{/}j\}$. For fixed $z\in D_0$ the summand is the product of the constant $\mathrm{\mid }P[f](z)-f(\zeta )\mathrm{\mid }$ restricted to the Borel set $\{\zeta \in \mathbb{T}:\mathrm{\mid }z-\zeta \mathrm{\mid }<m(1-\mathrm{\mid }z\mathrm{\mid }),\mathrm{\mid }z-\zeta \mathrm{\mid }<1\mathrm{/}j\}$; a countable supremum of Borel measurable functions is Borel measurable, so every $S_j$ is Borel measurable and so is $L_m:={\inf }_j{S}_j$. Moreover, since $P[f]$ is continuous on $\mathbb{D}$ by [L4] and $D_0$ is dense, the supremum over the points of $D_0$ in the open set $U_j(\zeta ):={\mathrm{\Gamma }}_m(\zeta )\cap \{\mathrm{\mid }z-\zeta \mathrm{\mid }<1\mathrm{/}j\}$ equals the supremum over all of $U_j(\zeta )$: every point of $U_j(\zeta )$ is a limit of points of $D_0\cap U_j(\zeta )$. Therefore $L_m(\zeta )={\inf }_j{S}_j(\zeta )={\mathrm{lim\hspace{0.17em}sup}}_{z\to \zeta ,z\in {\mathrm{\Gamma }}_m(\zeta )}\mathrm{\mid }P[f](z)-f(\zeta )\mathrm{\mid }$ is exactly the cone limsup, and it is Borel measurable.

Pointwise error bound. Let $g\in C(\mathbb{T},\mathbb{C})$ and let $m\ge 2$ be an integer. By step 1.1, $L_m(g)(\zeta )=0$ for every $\zeta$, and limsup subadditivity gives $L_m(f)(\zeta )\le L_m(f-g)(\zeta )+L_m(g)(\zeta )\le N_m(P[f-g])(\zeta )+\mathrm{\mid }f-g\mathrm{\mid }(\zeta )\le (m+1{)}^2{M}_{\mathbb{T}}(f-g)(\zeta )+\mathrm{\mid }f-g\mathrm{\mid }(\zeta ),$ where the middle inequality uses ${\mathrm{lim\hspace{0.17em}sup}}_z\mathrm{\mid }P[f-g](z)-(f-g)(\zeta )\mathrm{\mid }\le {\mathrm{lim\hspace{0.17em}sup}}_z\mathrm{\mid }P[f-g](z)\mathrm{\mid }+\mathrm{\mid }f-g\mathrm{\mid }(\zeta )$ and the last one is [L3] with aperture $m$ and measure $\nu =(f-g)m$, together with $P[f-g]=P[(f-g)m]$ and $M_{\mathbb{T}}(f-g)=M_{\mathbb{T}}((f-g)m)$ from [L1].

Small measure of the bad sets. Fix an integer $m\ge 2$ and $t>0$. If a point $\zeta$ satisfies $(m+1{)}^2{M}_{\mathbb{T}}(f-g)(\zeta )\le t$ and $\mathrm{\mid }f-g\mathrm{\mid }(\zeta )\le t$, then step 2.1 gives $L_m(f)(\zeta )\le 2t$; hence $\{L_m>2t\}\subseteq \{(m+1{)}^2{M}_{\mathbb{T}}(f-g)>t\}\cup \{\mathrm{\mid }f-g\mathrm{\mid }>t\}\mathrm{.}$ By [L2] and [L8], applied to the $L^1$ function $f-g$, the first set has measure at most $3(m+1{)}^2\mathrm{\parallel }f-g{\mathrm{\parallel }}_1\mathrm{/}t$ and the second at most $\mathrm{\parallel }f-g{\mathrm{\parallel }}_1\mathrm{/}t$, so $m(\{L_m>2t\})\le (3(m+1{)}^2+1)\mathrm{\parallel }f-g{\mathrm{\parallel }}_1\mathrm{/}t$ for every continuous $g$. Given $\epsilon >0$, [L6] supplies a continuous $g$ with $\mathrm{\parallel }f-g{\mathrm{\parallel }}_1<\epsilon$; hence $m(\{L_m>2t\})\le (3(m+1{)}^2+1)\epsilon \mathrm{/}t$ for every $\epsilon >0$, and consequently $m(\{L_m>2t\})=0$.

The exceptional set is null. For each integer $m\ge 2$, the set $\{L_m>0\}={\bigcup }_{k\ge 1}\{L_m>1\mathrm{/}(2k)\}$ is a countable union of Borel sets of $m$-measure zero by steps 1.2 and 3.1, hence is $m$-null by [L9]; the union $B:={\bigcup }_{m\ge 2}\{L_m>0\}$ over the countably many integers $m\ge 2$ is then $m$-null as well.

Conclusion. Let $\zeta \mathrm{\notin }B$, so that the complement of $B$ has full measure. Then $L_m(\zeta )=0$ for every integer $m\ge 2$: for every $\epsilon >0$ there is $\delta >0$ with $\mathrm{\mid }P[f](z)-f(\zeta )\mathrm{\mid }<\epsilon$ for all $z\in {\mathrm{\Gamma }}_m(\zeta )$ with $\mathrm{\mid }z-\zeta \mathrm{\mid }<\delta$. Given $A>1$, choose an integer $m\ge A$; since ${\mathrm{\Gamma }}_A(\zeta )\subseteq {\mathrm{\Gamma }}_m(\zeta )$ by [L1], the same $\delta$ witnesses $P[f](z)\to f(\zeta )$ as $z\to \zeta$ within ${\mathrm{\Gamma }}_A(\zeta )$. Thus the nontangential limit exists and equals $f(\zeta )$ for every $\zeta$ outside the null set $B$. If $f^{\mathrm{\prime }}=f$ almost everywhere is another representative, then $L_m^{\mathrm{\prime }}\le L_m+\mathrm{\mid }f-f^{\mathrm{\prime }}\mathrm{\mid }$ for the corresponding cone limsups, so the bad set for $f^{\mathrm{\prime }}$ is contained in $B\cup \{\mathrm{\mid }f-f^{\mathrm{\prime }}\mathrm{\mid }>0\}$, and the latter is a countable union of null sets by [L8] and [L9]; hence the assertion is independent of the representative.