= Solution
A <nontangential limit> at $\xi\in\partial\mathbb D$ restricts approach to a fixed <Stolz region> $\Gamma_A(\xi)=\{z\in\mathbb D:|z-\xi|<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The <Fatou theorem for bounded holomorphic functions> says that every $f$ in <H-infinity space> has a finite boundary value $f^*(\xi)$ for almost every $\xi$, with convergence through every fixed <Stolz region> and $|f^*|\le\|f\|_\infty$ <almost everywhere>.
Here is a proof outline. Write $f=u+iv$ with bounded real <harmonic functions> $u,v$. Take a sequence $R_j\uparrow1$. The bounded functions $u(R_j\xi)$ have a subsequence converging in the <weak-star topology> of $L^\infty(\partial\mathbb D)$ by the <Banach-Alaoglu theorem> and <weak-star metrizability of the dual ball>; its $L^1$ predual is <separable>. Denote the limit by $U$. For $|z|<R_j$, the <Poisson integral on the unit disk> gives
$$
u(z)=\int_{\partial\mathbb D}P_{z/R_j}(\xi)u(R_j\xi)\,dm(\xi).
$$
The <Poisson kernels> converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the <weak-star topology> permits passage to the limit:
$$
u(z)=\int_{\partial\mathbb D}\frac{1-|z|^2}{|\xi-z|^2}U(\xi)\,dm(\xi).
$$
Repeat for $v$, obtaining $V$. Thus $f$ is the <Poisson integral> of $F=U+iV\in L^\infty$.
The <Lebesgue differentiation theorem> makes almost every boundary point a <Lebesgue point> of $U$ and $V$. The lemma that a <Poisson integral converges nontangentially at Lebesgue points> then gives $f(z)\to F(\xi_0)$ in every fixed <Stolz region>. To see why, put $\delta=1-|z|$ and let $d(\xi,\xi_0)$ be the shorter angular distance. In $\Gamma_A(\xi_0)$ the kernel satisfies
$$
P_z(\xi)\le C_A\frac{\delta}{\delta^2+d(\xi,\xi_0)^2}.
$$
Split the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a <Lebesgue point>, averages of $|F-F(\xi_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A2^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta)$, so its contribution vanishes. This proves the claimed <nontangential limit>. The same <Lebesgue points> work for every integer aperture $A\ge2$, hence for all finite apertures.
These boundary values recover the interior by the <Poisson integral>, so they determine $f$ uniquely. They satisfy $\|f^*\|_{L^\infty}=\|f\|_\infty$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the <Poisson kernel>. The <dominated convergence theorem> also gives $f(r\xi)\to f^*(\xi)$ in every finite $L^p$ norm. Its negative <Fourier coefficients> vanish, because the <Taylor series> of $f$ has only nonnegative powers. This explains the boundary realization of analytic <Hardy spaces>.
The exceptional set cannot generally be removed. For example, $f(z)=\exp(i\operatorname{Log}(1-z))$ is bounded and <holomorphic> on the <unit disc>, using the <holomorphic logarithm> on the right half-plane: $|f(z)|=e^{-\arg(1-z)}$ lies between $e^{-\pi/2}$ and $e^{\pi/2}$. At the boundary point one its radial values $e^{i\log(1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.
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