Extending a holomorphic function to a radial limit function for almost every angle I've read in several places about the "well known theorem" which states that a holomorphic function on the (open) unit disk $D=\{z\in\mathbb{C}:\ |z|< 1\}$ can be extended to its boundary on almost every angle $\theta$ such that
$$\lim_{r\to 1^-} \int_0^{2\pi} \left|f\left(re^{i\theta}\right)-f\left(e^{i\theta}\right)\right|^2d\theta = 0.$$
Where can I find a proof for that "well known fact"? Can you show an example for which it can be extended for almost every $\theta$, but not for every $\theta$?
 A: This is known as Fatou's theorem. It's not quite true for every holomorphic function though. In the version you are quoting, you have to assume that
$$
\sup_{0 < r < 1} \int_0^{2\pi} |f(re^{i\theta})|^2\,d\theta < \infty.
$$
You should be able to find a proof in most advanced textbooks on complex analysis. Off the top of my head, Rudin's "Real and complex analysis" certainly has it. The usual proof is via Poisson's integral theorem and something called Hardy-Littlewood's maximal function.
For your question about "almost every $\theta$", the extension is defined via radial boundary limits, i.e.
$$
f^*(e^{i\theta}) = \lim_{r\to 1^-} f(re^{i\theta})
$$
and this limit may fail to exist at some (at most measure zero) points on $\partial D$. A simple example would be if $f$ is unbounded on $D$, for example
$$
f(z) = \log(1-z)
$$
with the principal branch of $\log$. Then $f^*(e^{i0})$ doesn't exist. More exotic examples can be constructed by (carefully) choosing a discontinous function on $\partial D$, extending it to a harmonic function $u$, and taking $f = u+iv$ where $v$ is the harmonic conjugate of $u$.
