Here's what I need to show:
Let $\mu$ be any finite signed Borel measure on $[-\pi,\pi]$. Let $g: [-\pi,\pi] \to \mathbb R$ be defined by $g(\theta ) = \int_{0}^{\theta} d\mu (t)$ if $\theta >0$ and $g(\theta ) = -\int_{\theta}^{0} d\mu (t)$ if $\theta <0$. Prove that $d\mu = g' d\lambda$ where $\lambda$ is Lebesgue measure on $[-\pi,\pi]$.
I am not entirely sure to how to proceed but here's my attempt:
We need to prove that $\mu (A)=\int_{A} g' d\lambda$ for every Borel set $A$. I tried to write $g'(x)= \lim_{n\to 0} \frac{g(x+1/n) - g(x)}{1/n}$. I tried to use Fubini theorem to complete the proof but I am not getting what I desire. Hints would be appreciated!
NOTE: This result may or may not be true! I made this claim in order to justify the proof of Fatou's theorem in Koosis - Introduction to $H_p$ spaces.
I want to show that $\frac{1}{2\pi}\int_{-\pi}^{\pi} P_r (t) d\mu(t) = P_r(\pi)g(\pi)-P_r(-\pi)g(-\pi)-\frac{1}{2\pi}\int_{-\pi}^{\pi}\partial_t P_r (t) g(t) dt$, where $P_r$ is the Poisson kernel, which is essentially integration by parts. I know this result holds for absolutely continuous functions but I was generalising this statement. Any hints to prove this will be appreciated!