Let $\mu$ be a finite complex Borel measure on the circle/unit interval, and let $\mathscr{C}\left\{ d\mu\right\}$ denote the Cauchy Transform: $$\mathscr{C}\left\{ d\mu\right\} \left(z\right)\overset{\textrm{def}}{=}\int_{0}^{1}\frac{d\mu\left(t\right)}{1-e^{-2\pi it}z},\textrm{ }\forall\left|z\right|<1$$ Using the Dominated Convergence Theorem (DCT), one can show that: $$\lim_{r\uparrow1}\left(1-r\right)\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)=\mu\left(\left\{ t\right\} \right)$$ for all $t\in\mathbb{R}/\mathbb{Z}$ (where we identify $\mathbb{R}/\mathbb{Z}$ with $\left[0,1\right))$. In particular, the DCT allows us to write: $$\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)=\int_{0}^{1}\left(\lim_{r\uparrow1}\frac{1-r}{1-re^{2\pi i\left(t-\tau\right)}}\right)d\mu\left(\tau\right)=\int_{0}^{1}\mathbf{1}_{\left\{ t\right\} }\left(\tau\right)d\mu\left(\tau\right)=\mu\left(\left\{ t\right\} \right)$$ where $\mathbf{1}_{\left\{ t\right\} }$ is the indicator function for the set $\left\{ t\right\}$.
I'm investigating what happens when the above is modified by using fractional powers. As a particular example, fix $\omega\in\left(0,1\right)$, and let: $$f\left(z\right)=\frac{1}{\left(1-z\right)^{\omega}}$$ Then, $f\left(e^{2\pi it}\right)$ is in $L^{1}\left(\mathbb{R}/\mathbb{Z}\right)$, and, for the absoutely continuous measure $dF\left(t\right)\overset{\textrm{def}}{=}f\left(e^{2\pi it}\right)dt$, we have that: $$\mathscr{C}\left\{ dF\right\} \left(z\right)=f\left(z\right),\textrm{ }\forall\left|z\right|<1$$ and so: $$\lim_{r\uparrow1}\left(1-r\right)^{\omega}\mathscr{C}\left\{ dF\right\} \left(re^{2\pi it}\right)=\mathbf{1}_{\left\{ 0\right\} }\left(t\right)$$ I would like to know if the right-hand side of this limit can be expressed in terms of the measure $dF\left(t\right)$. More generally, I would like to know how (if at all possible) to express:$$\lim_{r\uparrow1}\left(1-r\right)^{\omega}\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)$$ in terms of the measure $\mu$; that is, to know how to generalize the right-hand side of the formula: $$\lim_{r\uparrow1}\left(1-r\right)^{1}\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)=\mu\left(\left\{ t\right\} \right)$$ to deal with the case where the exponent $1$ is $\omega\in\left(0,1\right)$.
My attempt to do this one on my own has led me to an impass, as follows. Multiplying and dividing by $\left(1-e^{2\pi i\left(t-x\right)}z\right)^{-\omega}$, observe that: $$\left(1-r\right)^{\omega}\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)=\int_{0}^{1}\left(\frac{1-r}{1-re^{2\pi i\left(t-x\right)}}\right)^{\omega}\frac{d\mu\left(x\right)}{\left(1-re^{2\pi i\left(t-x\right)}\right)^{1-\omega}}$$ i.e. (viewing this as a convolution against the measure $d\mu$):$$\left(1-r\right)^{\omega}\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)=\int_{0}^{1}\left(\frac{1-r}{1-re^{2\pi ix}}\right)^{\omega}\frac{d\mu\left(t-x\right)}{\left(1-re^{2\pi ix}\right)^{1-\omega}}$$ Here: $$\sup_{x\in\mathbb{R}/\mathbb{Z}}\sup_{r\in\left(0,1\right)}\left|\left(\frac{1-r}{1-re^{2\pi ix}}\right)^{\omega}\right|\leq1$$ with: $$\lim_{r\uparrow1}\left(\frac{1-r}{1-re^{2\pi ix}}\right)^{\omega}=\mathbf{1}_{\left\{ 0\right\} }\left(x\right)$$ On the other hand, since $\omega\in\left(0,1\right)$, note that: $$\sup_{r\in\left(0,1\right)}\left|\frac{1}{\left(1-re^{2\pi ix}\right)^{1-\omega}}\right|\leq\frac{1}{x^{1-\omega}\left(1-x\right)^{1-\omega}}$$ holds for almost every $x\in\left[0,1\right)$, and that the upper bound is integrable: $$\int_{0}^{1}\frac{dx}{x^{1-\omega}\left(1-x\right)^{1-\omega}}=\int_{0}^{1}x^{\omega-1}\left(1-x\right)^{\omega-1}dx=\textrm{B}\left(\omega,\omega\right)=\frac{\left(\Gamma\left(\omega\right)\right)^{2}}{\Gamma\left(2\omega\right)}$$ where $\textrm{B}$ and $\Gamma$ are the Beta and Gamma functions, respectively. Hence, by the Dominated Convergence Theorem, the $r$-limit should be able to brought into the integral, but it's not clear how to make sense of the almost-everywhere limit of the integrand: $$\lim_{r\uparrow1}\left(1-r\right)^{\omega}\mathscr{C}\left\{ d\mu\right\} \left(re^{2\pi it}\right)=\int_{0}^{1}\underbrace{\left(\lim_{r\uparrow1}\frac{\left(1-r\right)^{\omega}}{1-re^{2\pi i\left(x-t\right)}}\right)}_{?}d\mu\left(t\right)$$