Given a positive singular measure $\mu$ on $[-\pi,\pi]$, we define a singular inner function by
$$S(z)=\exp\left(-\int_{-\pi}^{\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}\,d\mu(\theta)\right).$$
It is stated in many different sources that the radial limits $\lim_{r\rightarrow1^{-}}S(re^{2\pi it})$ equal 0 for $\mu$-a.e. $t\in\overline{\text{supp}}(\mu)$. None of them give a proof, and most of them say it is "easy to see." I personally have not been able to figure it out. Can anyone either give me a proof or direct me to one?
If $\mu$ is a positive singular measure on $[0,2\pi)$ then $$\lim_{h\to0}\frac{\mu((x-h,x+h))}{2h}=+\infty$$for $\mu$-almost every $t$. That much I know is in Rudin Real and Complex Analysis, for example.
It follows that if $\mu$ is a singular measure on the circle then $$\lim_{r\to1}P[\mu](re^{it})=+\infty$$for $\mu$-almost every $t$, where $P[\mu]$ is the Poisson integral (as in Rudin). This is just because the Poisson kernel $P(r,t)$ satisfies $$P(1-\delta,t)\ge \frac c{\delta}\chi_{(-\delta,\delta)}.$$
If you note that $|e^{x+iy}|=e^x$ you're done if you just show that the real part of the integral in the exponent is $P[\mu]$; look at the formula.