Let $\mu$ be a real measure on the circle $\mathbf{T}$. Then the function $f(z)=\int_\mathbf{T} \mathrm{Im}\left(\frac{\zeta+z}{\zeta-z}\right) d\mu(\zeta)$ is harmonic on the unit disc and its radial limits exist almost everywhere.
Does the Hilbert transform of $\mu$ equal the radial limit function of $f$ on $\mathbf{T}$ almost everywhere and why?
The Hilbert transform of $\mu$ is $(H\mu)(z)=p.v.\frac{1}{\pi}\int_\mathbf{T}\frac{d\mu(\zeta)}{z-\zeta}$ for $z\in\mathbf{T}$. (Please check, I'm not sure)
I have come so far to see, that the Hilbert transform and the radial limits of $f$ behave somewhat similarly as can be seen from noting that $\mathrm{Im}\left(\frac{e^{it}+re^{is}}{e^{it}-re^{is}}\right)\rightarrow \cot(s/2)$ as $r\rightarrow 1$ and $\cot(s/2)$ behaves similarly as $2/s$ when $|s|$ is small.
(I would also appreciate if somebody knew the answer just for the case that $\mu$ has continuous density w.r.t. the Lebesgue measure on $\mathbf{T}$ or similar special cases.)