Show $F(z)=\int_{0}^{1}{g(t)\over t-z}dt$ is holomorphic in $\Bbb{C}\setminus[0,1]$. Limit problem. Show $\int_{0}^{1}{g(t)\over t-z}dt$, $g(t):[0,1]\to \Bbb{R}$ a continuous function, is holomorphic in $\Bbb{C}\setminus[0,1]$. Trying to simplify ${F(z+h)-F(z)\over h}$ I arrived at $\int_{0}^{1}{g(t)\over (t-z)(t-(z+h))}dt$, and I don't know how I better compute its limit as $h\to 0$. Can I put the limit inside? How? Following Calculus, I only know I can do this for $n$ functions (countable ones) uniformly converging to $F$ but this is not the case. In addition, I don't know how to relate to $[0,1]$ points in this process so at to truly be sure how they can or can't impact the convergence of the limit above. I would appreciate your help. 
Edit: I added more tags mentioning Real Analysis because I don't want to ignore the probability that one whose expertise is Real Analysis and not Complex Analysis will be able to give a Real-Analysis-based explanation 
 A: HINT:
Please double check your result of ${F(z+h)-F(z)\over h}$
$${F(z+h)-F(z)\over h}=\frac{1}{h}(\int_{0}^{1}{g(t)\over t-(z+h)}dt-\int_{0}^{1}{g(t)\over t-z}dt)$$
$${F(z+h)-F(z)\over h}=\frac{1}{h}(\int_{0}^{1}{[(t-z)-(t-z-h)] g(t)\over (t-(z+h))(t-z)}dt)$$
$${F(z+h)-F(z)\over h}=\frac{1}{h}(\int_{0}^{1}{h g(t)\over (t-(z+h))(t-z)}dt)$$
$$\lim\limits_{ h\to 0  }{ {F(z+h)-F(z)\over h} }= \lim\limits_{ h\to 0  }{  \frac{1}{h}(\int_{0}^{1}{h g(t)\over (t-(z+h))(t-z)}dt)}$$
$$\lim\limits_{ h\to 0  }{ {F(z+h)-F(z)\over h} }= \lim\limits_{ h\to 0  }{  \frac{h}{h}(\int_{0}^{1}{ g(t)\over (t-(z+h))(t-z)}dt)}$$
$$\lim\limits_{ h\to 0  }{ {F(z+h)-F(z)\over h} }= \lim\limits_{ h\to 0  }{  (\int_{0}^{1}{ g(t)\over (t-(z+h))(t-z)}dt)}=\int_{0}^{1}{ g(t)\over (t-z)^2}dt$$
A: HINT:
$$\begin{align}
\left|\dfrac{F(z+h)-F(z)}{h}-\int_0^1 \frac{g(t)}{(t-z)^2}\,dt\right|&=\left|\frac1h\int_0^1 g(t)\left(\dfrac{1}{t-(z+h)}-\dfrac{1}{t-z}-\frac{h}{(t-z)^2}\right)\,dt\right|\\\\
&=\left|\int_0^1 g(t)\left(\dfrac{1}{(t-z)(t-z-h)}-\frac{1}{(t-z)^2}\right)\,dt\right|\\\\
&=\left|\int_0^1 \dfrac{hg(t)}{(t-z)^2(t-z-h)}\,dt\right|\\\\
&\le\int_0^1 \dfrac{|h|\,|g(t)|}{(t-z)^2|t-z-h|}\,dt\\\\
\end{align}$$

SPOILER ALERT Scroll over the highlighted area to reveal the solution

We note that $|t-z-h|=\sqrt{(t-h)^2-2\text{Re}(z)(t-h)+|z|^2}\ge \left|\text{Im}(z)\right|$.  Therefore, we have $$\int_0^1\left|\dfrac{hg(t)}{(t-z)^2(t-z-h)}\right|\,dt\le \dfrac{h}{\left|\text{Im}(z)\right|}\int_0^1\dfrac{|g(t)|}{(t-z)^2}\,dt \tag 1$$Since the integrand in $(1)$ is bounded and independent of $h$, then it is easy to see that $$\lim_{h\to 0}\int_0^1\left|\dfrac{hg(t)}{(t-z)^2(t-z-h)}\right|\,dt=0$$And this completes the proof that $F(z)$ is differentiable.

