(This may look very silly to you but I don't understand the reason that was given to me, nor do I have the knowledge to find out by myself).
The domain for $z$ is the open unit disc $D=\{z\in\mathbb{C} | \vert z \vert < 1\}$.
Here $\sigma$ is a real-valued function of bounded variation on $[0,2\pi]$. $f(z):=\int_0^{2\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}d\sigma (\theta).$
In any compact $K\subset D$, the modulus of the integrand is bounded by $(1+\lambda)/(1-\lambda)$, so $|f(z)|$ is bounded by $\int_0^{2\pi}\frac{1+\lambda}{1-\lambda}|d\sigma (\theta)| = \frac{1+\lambda}{1-\lambda}V_{[0,2\pi]}(\sigma) $ ($V_{[0,2\pi]}(\sigma)$ is the total variation of $\sigma$).
Why does this show that $f(z)$ is holomorphic in $D$? I was told "by Weirstrass' theorem" but that theorem deals with sequences of holomorphic functions.
Thank you.