Factorization of a bounded holomorphic function Let $f$ be a bounded holomorphic function on the unit disk $\mathbb{D} \subset \mathbb{C}$. 
Then we can write $f = BSF$, where $B$ is a Blaschke product, $S$ is a singular function, $F$ is an outer function. 
Assume that there exists a boundary point $\alpha \in \partial\mathbb{D}$ and an open disc $N$ centered at $\alpha$  such that $|f(\lambda)| \geq \delta >0$ for all $\lambda \in \mathbb{D} \cap N$. 
Here is my question: 
If the function $S$ is of the form 
$$
S(z) = \exp \left( -\int \frac{e^{i\theta} + z}{e^{i \theta} - z} \, d\mu(\theta) \right),
$$
where some positive singular measure $\mu$,
then how can we show that the closed support of $\mu$ is contained in $\partial \mathbb{D} \setminus N$?
Any help will be appreciated! 

It is shown that it suffices to show that for any compact subset $K \subset \partial\mathbb{D} \cap N$, $\mu(K) = 0$. 
If $\mu(K) >0$, then $K$ is uncountable. Does this imply that $f$ has a zero in $\partial\mathbb{D} \cap N$? (So this is a contradiction and $\mu(K) = 0$)
 A: The function $f$ is not really defined on $\partial \mathbb{D}$ and certainly isn't holomorphic there, so thinking about it having a zero there isn't productive. 
I would argue as follows. Suppose $\mu$ does not vanish identically in $N$. Then for any number $M$ there is a boundary point $e^{i\varphi}\in \partial\mathbb{D}\cap N$ and $h>0$ such that 
$$
\frac{\mu(\varphi-h, \varphi+h)}{h} > M \tag1
$$
Indeed, if such quotients were bounded, then $\mu$ would be absolutely continuous with respect to the Lebesgue measure. We can also force $h$ to be as small as we wish, so that $z$, defined below, is in $N$. 
Focus on $z = (1-h)e^{i\varphi}$ and show that the Poisson integral
$$
\int \operatorname{Re}\left(\frac{e^{i\theta} + z}{e^{i \theta} - z}\right) \, d\mu(\theta) \tag2
$$
is very large when the quotient in $(1)$ is large. Indeed, the integrand is nonnegative, and the arc $(\varphi-h, \varphi+h)$ contributes a lot of positive  because of (1). An important auxiliary estimate is that the Poisson kernel with respect to $z$ is bounded from below on this arc by a universal positive constant.
Now that you know that (2) is large, it follows that $|S(z)|$ is small, contradicting the assumption that $|S|$ is bounded away from zero. 
