We can easily see that if a sequence of inner functions converges then certainly the limit is still inner (using the fact that a countable union of measure zero sets still has measure zero). This implies that the set of inner functions is closed in $H^2$.
Now, is the set of the singular inner functions closed?
We know that a singular inner function has the form:
$$S(z) = K \text{exp} \left ( -\frac{1}{2\pi} \int_0^{2\pi} \frac{e^{i \theta} + z}{e^{i \theta} - z} \, d\mu(\theta) \right )$$ where $\mu$ is a positive Borel measure singular with respect to the Lebesgue measure and $K$ is a constant of modulus $1$.
Thus if we take $S_n \to S$ in $H^2$ then $S_n \to S$ uniformly on compact sets. Clearly $S$ is inner because $S_n$ is inner, now we need to show it is singular. My question now is, does anyone have a hint how I show this? I know that if $(\mu_n)$ is a sequence of positive finite Borel measures, then by Riesz and Alaoglu we have that a subsequence converge to a finite positive Borel measure $\mu$. Can I use this?