non constant bounded holomorphic function on some open set this is an exercise I came across in Rudin's "Real and complex analysis" Chapter 16.
Suppose $\Omega$ is the complement set of $E$ in $\mathbb{C}$, where $E$ is a compact set with positive Lebesgue measure in the real line. 
Does there exist a non-constant bounded holomorphic function on $\Omega$? 
Especially, do this for $\Omega=[-1,1]$.

Some observations:
Suppose there exists such function $f$, then WLOG, we may assume $f$ has no zeros points in $\Omega$ by adding a large enough positive constant, then, $\Omega$ is simply-connected implies $\int_{\gamma}fdz=0$, for any closed curve $\gamma\subset \Omega$, how to deduce any contradiction?

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 $g\circ f$. 
