Reading Exercise 8 of Chapter 16, I imagine Rudin interrogating the reader.
Let $E\subset\mathbb R$ be a compact set of positive measure, let $\Omega=\mathbb C\setminus E$, and define $f(z)=\int_E \frac{dt}{t-z}$. Now answer me!
a) Is $f$ constant?
b) Can $f$ be extended to an entire function?
c) Does $zf(z)$ have a limit at $\infty$, and if so, what is it?
d) Is $\sqrt{f}$ holomorphic in $\Omega$?
e) Is $\operatorname{Re}f$ bounded in $\Omega$? (If yes, give a bound)
f) Is $\operatorname{Im}f$ bounded in $\Omega$? (If yes, give a bound)
g) What is $\int_\gamma f(z)\,dz$ if $\gamma$ is a positively oriented loop around $E$?
h) Does there exist a nonconstant bounded holomorphic function on $\Omega$?
Part h) appears to come out of the blue, especially since $f$ is not bounded: we found that in part (e). But it is part (f) that's relevant here: $\operatorname{Im}f$ is indeed bounded in $\Omega$ (Hint: write it as a real integral, notice that the integrand has constant sign, extend the region of integration to $\mathbb R$, and evaluate directly). Therefore, $f$ maps $\Omega$ to a horizontal strip. It's a standard exercise to map this strip onto a disk by some conformal map $g$, thus obtaining a bounded function $g\circ f$.
Firstly, you can prove it with real methods, splitting the function and integrals into real and imaginary parts, and using the characterisation of holomorphic functions by the Cauchy-Riemann equations.
We have, however, nicer methods at our disposal in complex analysis. One very nice tool is
Theorem (Morera):
Let $U \subset \mathbb{C}$ open, and $f \colon U \to \mathbb{C}$ a continuous function. If, for all closed triangles $\Delta \subset U$, you have
$$\int_{\partial \Delta} f(z)\, dz = 0,$$
then $f$ is holomorphic on $U$.
(Cauchy's integral theorem provides the converse.)
Armed with that, we let $\Delta$ an arbitrary closed triangle in $U$ and compute
$$\begin{align}
\int_{\partial \Delta} F(z)\,dz &= \int_{\partial \Delta} \int_\gamma f(\omega,\,z) \, d\omega\,dz\\
&= \int_\gamma \int_{\partial \Delta} f(\omega,\, z)\,dz\, d\omega & \text{(Fubini)}\\
&= \int_\gamma 0\, d\omega & \text{(Cauchy)}\\
&= 0,
\end{align}$$
hence conclude by Morera's theorem that $F$ is holomorphic in $U$.
Since the partial derivative $\frac{\partial f}{\partial z}$ is continuous by the premise, the function
$$G(z) = \int_\gamma \frac{\partial f}{\partial z}(\omega,\, z)\,d\omega,$$
is continuous on $U$ (even holomorphic, by Morera), and we can compute
$$\begin{align}
\int_\alpha G(z)\,dz &= \int_\alpha \int_\gamma \frac{\partial f}{\partial z}(\omega,\, z)\,d\omega\,dz\\
&= \int_\gamma \int_\alpha \frac{\partial f}{\partial z}(\omega,\, z)\,dz\,d\omega & \text{(Fubini)}\\
&= \int_\gamma f(\omega,\, \alpha(1)) - f(\omega,\,\alpha(0))\,d\omega\\
&= F(\alpha(1)) - F(\alpha(0))
\end{align}$$
for each path of integration $\alpha \colon [0,\,1] \to U$. That implies that $F'(z) = G(z)$.
(To make it explicit, fix $z_0 \in U$ and choose $\alpha$ a straight line segment connecting $z_0$ and $z \in U$ - for $r := \lvert z-z_0\rvert$ small enough, the segment is contained in $U$. Then
$$\begin{align}
\left\lvert\frac{F(z) - F(z_0)}{z-z_0} - G(z_0)\right\rvert &= \left\lvert\frac{1}{z-z_0}\int_\alpha G(\zeta) - G(z_0)\, d\zeta\right\rvert\\
&\leqslant \frac{1}{\lvert z-z_0\rvert}\int_0^1 \lvert G(z_0 + t(z-z_0)) - G(z_0)\rvert\cdot \lvert z-z_0\rvert\, dt\\
&\leqslant \max_{\lvert w-z_0\rvert \leqslant r} \lvert G(w) - G(z_0)\rvert,
\end{align}$$
and the continuity of $G$ in $z_0$ shows the desired convergence.)
Best Answer
Let $p\in \Omega$, and let $\varepsilon>0$ be small enough so that $B(p,\varepsilon) \subset \Omega$. $\DeclareMathOperator{\d}{d} \newcommand{\dx}{\d\!}$ Let us show that $h$ is holomorphic on $B(p,\varepsilon)$. Since $B(p,\varepsilon)$ is simply connected, by Morera's it suffices to show that for every piecewise differentiable closed curve $\gamma$ in $B(p,\varepsilon)$, $\int_\gamma h(z)\dx z = 0$.
Let $\gamma\colon [a,b] \to B(p,\varepsilon)$ be such a closed curve, and up to a reparametrization, assume that $|\gamma'(s)|=1$ for all $s$. Since $|f(\gamma(s),t)\gamma'(s)| = |f(\gamma(s),t)| \geqslant 0$ on $B(p,\varepsilon)\times X$, we have \begin{align} \int_{[a,b]} \int_X |f(\gamma(s),t)\gamma'(s)|\dx\mu(t)\dx s &= \int_X\int_{[a,b]} |f(\gamma(s),t)|\dx s\dx\mu(t) \\ &\leqslant (b-a)\int_X \sup_{z\in \gamma([a,b])} |f(z,t)|\dx\mu(t) \\ & <+ \infty \end{align} by the third assumption. It follows by Fubini's Theorem that \begin{align} \int_{[a,b]} \int_X f(\gamma(s),t)\gamma'(s)\dx\mu(t)\dx s & = \int_X \int_{[a,b]} f(\gamma(s),t)\gamma'(s)\dx s\dx\mu(t) \\ &= \int_X \int_{\gamma}f(z,t)\dx z \dx\mu(t) \\ &= \int_X 0 \dx\mu(t) \\ &= 0, \end{align} the last equality being true since $f(\cdot,t)$ is holomorphic on $B(p,\varepsilon)$ which is simply connected. Hence, $\int_\gamma h(z)\dx z = 0$. This being true for all $\gamma$, $h$ is holomorphic on $B(p,\varepsilon)$. In particular, $h$ is holomorphic at $p$. This being true for all $p \in \Omega$, $h$ is holomorphic on $\Omega$.