If $f$ is a holomorphic (on open unit disk D) and bounded satisfying a limit, then f = 0. Let $D \subset \mathbb{C}$ be the open unit disk at origin and $I \subset [-\pi,\pi]$ be an open interval. If $f$ is a holomorphic in D and bounded function satisfying $\lim_{r\to1^-}f(re^{i\theta}) = 0$ for a.e $\theta \in I$ , then f = 0.
 A: I will expand the comment in an answer since it is not difficult as it doesn't use the full power of the existence of radial limits etc: since both the hypothesis and the conclusion are not affected by multiplying with a constant, we can assume $f$ holomorphic, not identically zero and $|f| \leq 1$ on the open unit disc; also we can assume $f(0) \neq 0$ since otherwise by Schwarz lemma, $\frac{f(z)}{z^m}$ satisfies same properties as $f$ (bounded by 1, converging radially to zero on a set $I$ of non-zero Lebesgue measure on the unit circle - e.g the interval of the original problem), where $m$ is the order of the zero of $f$ at the origin.
Numbering the (possible empty or finite set of) zeros of $f$ as $0 < |z_1| \leq |z_2|...$, Jensen'Thm says that for $0<r<1$, $\frac{1}{2\pi}$$\int_{0}^{2\pi} \log|f(re^{i\theta})|d\theta$ = $|\log(f(0))| +  \sum_{|z_n| < r}{\log{\frac{r}{|z_n|}}}$ and the sum is obviously increasing with r since the terms are positive;  $g(re^{i\theta}) = -\log|f(re^{i\theta})|$ is non-negative ($|f| \leq 1$!) and by the above its integral in $\theta$ is decreasing and non-negative as $r$ tends to $1$, hence it is uniformly upper bounded by some $M \geq 0$ so we can apply Fatou's lemma and get (noting that since $|f| \leq 1$ in the disc any radial liminf/limsup is also bounded by 1 in absolute value so its logarithm of the absolute value is negative, so taking a minus, integrating and restricting the integral to a subset only decreases the value of the integral, while by hypothesis the set $I$ where the radial limit of $f$ is zero, has non-zero Lebesgue measure ):
${\infty}$ = $\frac{1}{2\pi}$$\int_{I} -\log(0)d\theta \leq \frac{1}{2\pi}$$\int_{0}^{2\pi} \liminf\limits_{r \rightarrow 1}(-{\log|f(re^{i\theta})|}d\theta)\leq \liminf\limits_{r \rightarrow 1}(-\frac{1}{2\pi}$$\int_{0}^{2\pi} \log|f(re^{i\theta})|d\theta) \leq M$ which is a contradiction
