Bounded harmonic functions have L-infinity Fatou boundary data

Class membership and choice bookkeeping. The hypothesis says that $u$ is complex harmonic with $M={\sup }_{z\in \mathbb{D}}\mathrm{\mid }u(z)\mathrm{\mid }<+\mathrm{\infty }$, so [L1] gives $u\in h^{\mathrm{\infty }}(\mathbb{D})$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}=M$. By [L4] the Axiom of Choice supplies countable choice, so the choice hypotheses of [L2] (the Axiom of Choice) and of [L3] (countable choice) are met.

The boundary data. Apply [L2] with $p=\mathrm{\infty }$, which is allowed because $1<\mathrm{\infty }\le \mathrm{\infty }$: there is a unique $f\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{C})$ with $u=P[f]$, and $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Together with step 1.1 this gives $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=M={\sup }_{z\in \mathbb{D}}\mathrm{\mid }u(z)\mathrm{\mid }$, so $f$ is the promised boundary datum and the norm identity holds.

The boundary datum is integrable. Apply the Hölder inequality of [L6] with the conjugate pair $(p,p^{\mathrm{\prime }})=(\mathrm{\infty },1)$, to $f\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{C})$ and to the constant function $1\in L^1(\mathbb{T},m;\mathbb{C})$: one has $\mathrm{\parallel }f{\mathrm{\parallel }}_1={\int }_{\mathbb{T}}\mathrm{\mid }f\cdot 1\mathrm{\mid }dm\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }1{\mathrm{\parallel }}_1=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, where the last equality is [L5]. Since $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=M<+\mathrm{\infty }$ by step 2.1, this shows $f\in L^1(\mathbb{T},m;\mathbb{C})$.

Nontangential convergence almost everywhere. By step 3.1 the function $f$ lies in $L^1(\mathbb{T},m;\mathbb{C})$ and by step 1.1 countable choice is available, so [L3] applies: there is a set $N\subseteq \mathbb{T}$ with $m(N)=0$ such that for every $\zeta \in \mathbb{T}\setminus N$ and every $A>1$ one has $P[f](z)\to f(\zeta )$ as $z\to \zeta$ within ${\mathrm{\Gamma }}_A(\zeta )$. Since $u=P[f]$ by step 2.1, the same convergence holds with $u$ in place of $P[f]$; the exceptional set does not depend on $A$, and the almost-everywhere representative used is the class $f$ of step 2.1.

Assembly. Steps 2.1, 3.1 and 4.1 produce a unique $f\in L^{\mathrm{\infty }}(\mathbb{T},m;\mathbb{C})$ with $u=P[f]$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, and show that $u(z)\to f(\zeta )$ as $z\to \zeta$ within every fixed nontangential region ${\mathrm{\Gamma }}_A(\zeta )$, $A>1$, for $m$-almost every $\zeta \in \mathbb{T}$; by step 1.1 the norm $\mathrm{\parallel }u{\mathrm{\parallel }}_{h^{\mathrm{\infty }}}$ equals ${\sup }_{z\in \mathbb{D}}\mathrm{\mid }u(z)\mathrm{\mid }$, so all clauses of the Statement hold. The Axiom of Choice is used exactly in step 1.1: it supplies the hypothesis of the representation theorem [L2] and, through dependent and countable choice, the hypothesis of the Fatou theorem [L3]. ∎