# On the set of zero radial limits of bounded analytic functions

Hi,

Let $f$ be a non-identically zero bounded analytic function in the open unit disk $\mathbb{D}$. It is well-known that $f$ has radial limits almost everywhere on the unit circle $\mathbb{T}$. Let $Z_f$ be the set of points in $\mathbb{T}$ where $f$ has zero radial limit :

$$Z_f:= \{e^{i\theta} \in \mathbb{T} : \lim_{r \rightarrow 1}f(re^{i\theta})=0\}.$$

Then it is also well-known that $Z_f$ has measure zero.

My question is the following :

For which sets $E \subseteq \mathbb{T}$ of measure zero does there exist a bounded analytic function $f$ in $\mathbb{D}$ such that $E=Z_f$?

Remark : It is easy to see that every $Z_f$ is a $F_{\sigma \delta}$, so the question could be wether or not every $F_{\sigma \delta}$ of measure zero is $Z_f$ for some $f$.

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I'm having some problem with LaTex in the definition of $Z_f$... Does someone know how to fix that? – Malik Younsi Mar 10 '13 at 12:17
From "How to write math" (look on the right $\rightarrow$): if you're having problems with the preview (or the post looks wrong), put backticks around any math that contains underscores or asterisks. E.g. write $f'_n=g_{n+1}$. – Giuseppe Tortorella Mar 10 '13 at 12:25
@Giuseppe : Thank you! – Malik Younsi Mar 12 '13 at 20:20

Lohwater and Piranian proved that for every set $F_\sigma$ of measure zero, there exists a bounded analytic function which has radial limits exactly on the complement of this set. (Ann. Acad. Sci. fenn., 239 (1957) 1-17.)