Continuous extension of analytic functions Is it possible to prove the following statement or is there a counter-example:
Let $H=\{y>0\}$ be the upper half plane in the complex plane. If $f$ is an analytic function on $H$ and its real part is continuously extendable on the closure of $H$, which is $\overline H$, then $f$ is continuously extendable on $\overline H$.
 A: Here's the idea:
Let $\mathbb H$ denote the upper half plane.
Let $U=H\setminus i\cdot[0,1]$ and $f\colon U\to i\mathbb H$ given by the Riemann mapping theorem.
It should be not too difficult to write down $f$ explicitly. Via a suitable automorphism of $\mathbb H$, we may assume that that $f$ is "symmetric" and then one sees that $f^{-1}$ extends to the boundary.
This implies that $\Re f(z)\to 0$ if $z\to a\in\mathbb R$, but $\lim_{z\to0}f(z)$ does not exist.

Here's something more explicit:
$g(z)=iz+\frac iz$ maps $S^1\to[-2i,2i]$ and $i\mathbb R\cup\{\infty\}\to \mathbb R\cup\{\infty\}$.
For the regions it constitutes among others a biholomorphic map 
$$-i\mathbb H\setminus \overline{ \mathbb D}\longrightarrow \mathbb H\setminus (0,2i]$$
that extends continuously to the boundary - but for the inverse the continuation to the boundary fails.
We find that the inverse 
$$ \mathbb H\setminus (0,2i]\longrightarrow -i\mathbb H\setminus \overline{ \mathbb D}$$
is given by 
$$ g^{-1}(z) = \frac{-iz+\sqrt{-z^2-4}}{2} $$
where we take the square root with positive real part (which exists because $z$ is neither real nor in $(0,2i]$).
If $a\in\mathbb R$ then $$\Re g^{-1}(z)= \frac{\Im z+\Re\sqrt{-z^2-4}}{2}\to 0 $$
as $z\to a$.
However, as $t\to 0^+$ we have
$$ g^{-1}((\pm1+i)t)=\frac{t\mp t+\sqrt{\pm2it^2-4}}{2} \to\pm i,$$
whence $f:=g^{-1}$ is a counter-example as asked for.

Remark: Taking the problem statement literally (I implicitly assumed above that $U$ should be a region and not just an open set), a simpler counterexample can be given:
Let $U=(-1,0)\times(0,i)\cup (0,1)\times(0,i)$, $I=(-1,1)$, $f(z)=0$ if $\Re z<0$ and $f(z)=i$ if $\Re(z)>0$.  
EDIT: After reformulation of the question, this answer no longer applies.
(It is an example of 

$U\subseteq \mathbb H$, $\mathbb R\subseteq \partial U$, $f\colon U\to \mathbb C$ analytic, $g\colon U\cup \mathbb R\to \mathbb R$ continuous, $g|_U=\Re f$ and there is no continuous extension of $f$ to $U\cup\mathbb R$.

whereas the edited problem statement asks

$f\colon \mathbb H\to \mathbb C$ analytic, $g\colon \mathbb H\cup \mathbb R\to \mathbb R$ continuous, $g|_{\mathbb H}=\Re f$. Is there always a continuous extension of $f$ to $\mathbb H\cup\mathbb R$?

)
A: Define
$$ g(z)= \int_0^{2\pi}\frac{1}{1-e^{-it}z}\,\mathrm dt$$
on the unit disk. It is obtained by computing the Cauchy integral to achieve $g(z)=i$ for $z\in S^1$ with positive imaginary part and $g(z)=0$ for $z\in S^1$ with negative imaginary part.
Tranfering this via a bihomomorphic map $\mathbb H\to \mathbb D$ to the upper half plane, i.e. letting
$$ f(z)=g\left(\frac{iz + 1}{z+i}\right)$$ 
then should do the trick.
A: A correct answer has not been given yet. Unfortunately I cannot take the bounty back. :(
