Measure $\mu$ is 0. I am reading the paper of Balazard ,Saias and Yor. Let, $$f(z)=(s-1)\zeta(s) $$ where $s=\frac{1}{1-z}$ and $\zeta(s)$ denotes the Riemann zeta function. Denote by $$\exp\left[\int_{-\pi}^{\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}d\mu(\theta)\right]$$ the singular interior factor of $f.$
Denote by $$f^*(e^{i\theta})=\lim_{r\to 1^-}f(re^{i\theta})$$
Then by Generalised Jensen's formula $$\frac{1}{2\pi}\int_{-\pi}^{\pi} \log|f^*(e^{i\theta})|d\theta=\log|f(0)|+\sum_{|\alpha|<1,f(\alpha)=0}\log\frac{1}{|\alpha|}+\int_{-\pi}^{\pi}d\mu(\theta)$$

The measure $\mu$ associated to the singular interior factor of $f$ is 0. For this it suffices to reuse the argument developed by Bercovici and Foias for the interior factor of the functions $(\theta-\theta^s)\zeta(s)\frac{(s+\frac{1}{2})}{s}$

Question How does the measure $\mu$ associated to the singular interior factor of $f$ is $0$?.
 A: This result involves a bunch of theorems in Hardy Space theory which can be found in the classic references by Duren or Garnett (chapter II especially)
I will quote the relevant theorems and then explain how they imply the result:

*

*If $f \in H^p(\mathbb D), p>0$, then $f$ has non-tangential finite limit on the unit circle ae $d\theta$; if we call that $f^*(e^{i\theta})$, one has $\log |f^*| \in L^1(d\theta), |f^*| \in L^p(d\theta)$


*There is a (unique up to constants of modulus one) decomposition $f=BSF$ where $F$ is outer (non-zero and determined by $\log |F^*|=\log |f^*|$), with $|F^*|=|f^*|$ ae, $B$ is a Blaschke product with the zeroes of $f$ inside the unit disc and $S$ is a singular inner function which is determined by a positive singular measure (wr $d\theta$) $\mu$; both $|B^*|=|S^*|=1$ a.e. $d\theta, |B(z)|, |S(z)| <1, |z|<1$ (unless either is trivial so identically $1$)


*If $f$ is analytically continuable across some arc $I$ of the unit circle, then $B,S,F$ are so too (and of course vice versa) which concretely means that $B$ has no zeroes accumulating there and the measure $\mu$ has support disjoint from $I$


*If the singular factor is non-trivial (ie measure $\mu \ne 0$, so $S$ is not $1$), then there is a point $e^{i\theta}$ for which $f(z) \to 0, z \to e^{i\theta}$ non-tangentially; this is very easy to see when $\mu$ is a point mass at some $e^{i\theta}$ so $S(z)=e^{-c\frac{e^{i\theta}+z}{e^{i\theta}-z}}, c>0$ hence $S$ decays exponentially at $e^{i\theta}$ while $F$ cannot increase too much since $F^*$ is in some $L^p$
Now for $f(z)=(s-1)\zeta(s), s=\frac{1}{1-z}$, one easily shows that it is in $H^p$ for all $p<1/2$ so it has a decomposition $f=BSF$ as above; we know that $(s-1)\zeta$ is analytically continuable across the finite part of $\Re s =1/2$ which corresponds with $f$ being so across $|z|=1, z \ne 1$; by Theorem 3 above, it follows that the singular measure $\mu$ which determines $S$ is concentrated at $1$ or trivial. If we were in the first case we would get $\lim_{r \to 1, r<1}f(r)=0$ (Theorem 4 above)
But now $x=\frac{1}{1-r}>0, x\to \infty, r \to 1$ and  $f(r)=(x-1)\zeta(x)$ by defintion; but this would mean $\lim_{x \to \infty, x>0}(x-1)\zeta(x)=0$ and that is of course not true as the limit is infinity; hence $S=1$ and the measure $\mu$ is zero. Incidentally $B=1$ is equivalent of course with RH so $f$ is outer iff RH is true
