Just a small proof verification. I'm a bit weirded out by the end of my proof in particular.
Suppose $h: [0,1] \to \mathbb{C}$ is continuous. Consider the function $$H(z) = \int_0^1 \frac{h(t)}{t-z} dt$$ where $z \notin [0,1]$. Then $H(z)$ is analytic on $\mathbb{C} / [0,1]$
Proof: It suffices to check that $H$ is complex differentiable for $z \in \mathbb{C} / [0,1]$ so let's just hit the definition of complex differentiable.
$$\lim_{r \to 0} \frac{H(z+r)-H(z)}{r} = \lim_{r \to 0} \int_0^1 \bigg[\frac{h(t)}{t-(z+r)} - \frac{h(t)}{t-z}\bigg] \frac{1}{r} \ dt $$
$$ =\lim_{r \to 0} \int_0^1\frac{rh(t)}{(t-z)(t-(z+r))} \frac{1}{r} dt$$
$$ =\lim_{r \to 0} \int_0^1\frac{h(t)}{(t-z)(t-(z+r))} dt $$
This suggests that the limit is probably $$\int_0^1 \frac{h(t)}{(t-z)^2} dt$$ which is what one would probably expect (i.e you can just differentiate under the integral), but I need to show that the integral and limit do indeed commute. I think I can this by showing that $$G_r(t) = \frac{h(t)}{(t-z_0)(t-(z_0+r))} \to \frac{h(t)}{(t-z_0)^2}$$ uniformly as $|r| \to 0$ for arbitrary $z_0$ fixed. So let's hit up that uniform convergence definition. That is fix $z_0 \in \mathbb{C} / [0,1]$ $$|G_r(t) -G_0(t)| = \bigg|\frac{h(t)}{(t-z_0)(t-(z_0+r))} - \frac{h(t)}{(t-z_0)^2 } \bigg| $$ $$ = \bigg| \frac{h(t)}{(t-z_0)^2(t-(z_0+r))} \bigg| = \bigg|\frac{h(t)}{(t-z_0)^2} \bigg| \bigg|\frac{r}{(t-(z_0+r)} \bigg|$$ By continuity on a compact interval, the L.H.S is bounded above by $$M \bigg|\frac{r}{t-(z_0+r))}\bigg|$$ Choosing $r$ sufficiently small so that $(z_0+r) \notin [0,1]$ I believe that should bound $\bigg|\frac{1}{(t-(z_0 +r))} \bigg|$ above my some constant $C$ and I think we're done.