Proving a function is holomorphic in $\mathbb{C}\setminus[0,1]$ 
Let $h(t):[0,1]\to\mathbb C$ be a continuous function. Prove that $f:\mathbb C\setminus[0,1]\to\mathbb C $ defined by $f(z)=\int_0^1\frac{h(t)}{z-t}$  is holomorphic.

My attempt:
$$\lim_{z\to0}\frac{f(z+z_{0})-f(z_{0})}{z}=\lim_{z\to0}\frac{\int_0^1\frac{h(t)}{z+z_0-t}-\int_0^1\frac{h(t)}{z_0-t}}{}=\\\lim_{z\to0}\frac{{}{\int_0^1}(\frac{(z_{0}-t)h(t)}{(z+z_{0}-t)(z_{0}-t)}-\frac{(z+z_{0}-t)h(t)}{(z_{0}-t)(z+z_{0}-t)})dt}{z}=\lim_{z\to0}\frac{{}{\int^1_0}\frac{-zh(t)}{(z+z_{0}-t)(z_{0}-t)}dt}{z}=\\-\lim_{z\to0}\int_{0}^{1}\frac{h(t)}{(z+z_{0}-t)(z_{0}-t)}dt=-\int_{0}^{1}\frac{h(t)}{(z_{0}-t)^{2}}dt$$
Where I'm not sure about the last equality.
Is my solution correct? Any help would be appreciated.
 A: Let $g$ be a function of two variables, say $t$ in $[0, 1]$ and $z$ in some disk about a point $z_{0}$ in the complex plane. We say $g(t, z) \to g(t, z_{0})$ uniformly as $z \to z_{0}$ if
$$
\sup_{t \in [0, 1]} |g(t, z) - g(t, z_{0})| \to 0
$$
as $z \to z_{0}$. If this condition holds, we can bring the limit in $z$ inside the integral:
$$
\lim_{z \to z_{0}} \int_{0}^{1} g(t, z)\, dt = \int_{0}^{1} g(t, z_{0})\, dt.
$$
The triangle inequality for integrals implies
\begin{align*}
  \biggl|\int_{0}^{1} g(t, z)\, dt - \int_{0}^{1} g(t, z_{0})\, dt\biggr|
  &= \biggl|\int_{0}^{1} [g(t, z) - g(t, z_{0})]\, dt\biggr| \\
  &\leq \int_{0}^{1} |g(t, z) - g(t, z_{0})|\, dt \\
  &\leq \sup_{t \in [0, 1]} |g(t, z) - g(t, z_{0})|,
\end{align*}
and the upper bound approaches $0$ as $z \to z_{0}$ by hypothesis.
In your situation we can fix $z_{0}$ and take
$$
g(t, z) = \frac{h(t)}{(z - t)(z_{0} - t)}.
$$
If, say, the distance from $z_{0}$ to the interval $[0, 1]$ is $r > 0$, and if we assume $|z - z_{0}| < r/2$, then $g(t, z) \to g(t, z_{0})$ uniformly in the above sense as $z \to z_{0}$. (This is not exactly your notation, but I hope expresses the ideas more clearly.)
