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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?

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You could use Hurwitz's theorem. For some $R$ with $0\lt R\lt1$, $S$ has no zeros on $\partial B(0,R)$, and $R$ can be taken arbitrarily close to 1. By Hurwitz's theorem, $S$ has the same number of zeros in $B(0,R)$ as $S_n$ for sufficiently large $n$. Each $S_n$ is never zero in the open disk, so this shows that $S$ is never zero on $B(0,R)$. Thus $S$ is never zero on the disk. This means that $S$ has no Blaschke factor, hence it is singular.

I do not know whether $(\mu_n)$ converges weak-$*$ to the singular measure of $S$.

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The limit of singular inner functions could be a constant. Whether or not this is a singular inner function will depend on whether you consider that a singular inner function. (The singular measure is the zero measure).

Nonconstant singular inner functions have properties much different from constant functions, such as they must have modulus that tends to zero at some points on the unit circle.