Extending the path of integration to the boundary of the domain of definition of $g:\Omega \to \mathbb{C}$ Working in Shakarchi and Stein's Complex Analysis, they show an example (p.231) of a conformal map from the $\mathbb{H}  = \{z: Im(z)>0 \}$ to $\{z: 0 < arg(z) < \alpha z\}$ using the transformation $f(z) = z^{\alpha}$ where $0 < \alpha < 2$.  At one point they write 
$$
 z^{\alpha}=\int^z_0 f'(\zeta) d\zeta 
$$
So I'm having trouble working out justifying this equality, since $f'$ is not defined on the boundary of $\mathbb{H}$. I am assuming what is meant by $\int^z_0 f'(\zeta) d\zeta$ is $lim_{z_0 \to 0} \int^z_{z_0} f'(\zeta) d\zeta$, and we want to prove this limit is well-defined.  The book says it follows by continuity and Cauchy's Theorem.

There is no problem at $0$ due to continuity because (example) if $f:[a,b]\to \Bbb X$ is not defined at $c\in [a,b]$ then for every $ε>0$ ,$\int_a^b f(x)dx=\int_a^{c-ε} f(x)dx +\int_{c+ε}^{b} f(x)dx$ .
