Let $\sigma$ be complex measure on $[-\pi, \pi]$ which is singular with respect to the Lebesgue measure and let $P_r (t)=\frac{1-r^2}{1+r^2-2r\cos t }$ for $[-\pi,\pi]$. Let $\mathfrak{A} (x)= \sigma ((-\pi, x))$ for $x \in (-\pi, \pi)$. I want to justify that integration by parts holds, that is,
\begin{align*} \int_{-\pi}^{\pi} P_r (t) d\sigma(t) = P_r(\pi)\mathfrak A(\pi) - P_r(-\pi)\mathfrak A(-\pi)-\frac{1}{2\pi}\int_{-\pi}^{\pi} \partial_t P_r(t) \mathfrak{A}(t)dt \end{align*}
I can easily show this using Fubini's theorem when the functions are absolutely continuous: \begin{align*} \int_{-\pi}^{\pi} f (t) g'(t) dt = f(\pi)g(\pi) - f(-\pi)g(-\pi)-\frac{1}{2\pi}\int_{-\pi}^{\pi} f'(t) g(t)dt \end{align*}
How do I proceed to show this for the Poisson kernel and the singular measure?
I do not agree with the formula given, for example with the factor $1/(2\pi)$ in the right-hand side.
We have to use Fubini theorem instead of integration by parts. I call $\mu$ the measure $|\sigma|$, so $\mu$ is a finite positive measure and $\sigma = h \mu$ for some complex function $h$ with constant modulus $1$. Then $$\int_{-\pi}^\pi P_r'(t) \mathfrak{A}(x) dt = \int_{-\pi}^\pi \Big[ P_r'(t) \int_{]-\pi,\pi[} 1_{s<t}~h(s)~d\mu(s) \Big]dt.$$ Since \begin{eqnarray*} \int_{-\pi}^\pi \int_{]-\pi,\pi[} \big| P_r'(t) 1_{s<t}~h(s)\big| ~d\mu(s) \Big]dt &\le& \int_{-\pi}^\pi \int_{]-\pi,\pi[} \frac{1-r^2}{(1-r)^2}~d\mu(s) dt \\ &=& \frac{1+r}{1-r} \mu]-\pi,\pi[ < +\infty, \end{eqnarray*} one can apply Fubini theorem. $$\int_{-\pi}^\pi P_r'(t) \mathfrak{A}(x) dt = \int_{]-\pi,\pi[} \Big[ \int_{-\pi}^\pi 1_{s<t}~P_r'(t)~dt\Big] ~h(s)~d\mu(s).$$ $$\int_{-\pi}^\pi P_r'(t) \mathfrak{A}(x) dt = \int_{]-\pi,\pi[} \Big[ (P_r(\pi)-P_r(s)\Big] ~d\sigma(s).$$ $$\int_{-\pi}^\pi P_r'(t) \mathfrak{A}(x) dt = P_r(\pi)\mathfrak{A}(\pi) - \int_{]-\pi,\pi[} P_r(s)~d\sigma(s).$$