If we know that $$\int_{\mathbb{R}_+}\frac{d|\mu|(t)}{t+1}<\infty,$$ how can I prove that $$\int_{\mathbb{R}_+}\frac{z}{1+tz}d\mu(t)<\infty$$ where $z \in \mathbb{C}$, $\mu$ is a complex Radon Measure? In this case, $|\mu|$ stands for the total variation of measure $\mu$.
I had the follow hint: $$\frac{1}{1+tz}-\frac{1}{1+t}=\frac{t(1-z)}{1+tz}\frac{1}{1+t}.$$
Thank you everyone!
This is not true.
Consider $z = -1$ and $d\mu = - \chi_{(0,2)}(x) \, dx$, where $\chi_M$ is the characteristic function/indicator function of the set $M$.
We then have $d |\mu| = \chi_{(0,2)}(x) \, dx$ and hence
$$ \int_{\Bbb{R}_+} \frac{1}{1+t} \, d|\mu|(t) = \int_0^2 \frac{1}{1+t} \, dt< \infty, $$
but
$$ \int_{\Bbb{R}_+} \frac{z}{1+tz} \, d\mu(t) = \int_0^2 \frac{1}{1-t} \, dt = \infty, $$ because the singularity at $t = 1$ is not integrable w.r.t. the Lebesgue measure.