Construct a harmonic function that appears to be discontinuous on the unit circle. Construct a harmonic function $u$ in $D(0,1)$ that satisfies 
$$
lim_{r \to 1^-}u(re^{i\theta})  = \begin{cases} 
      1 & 0 < \theta < \pi \\
      0 & \pi < \theta < 2\pi 
   \end{cases}
\]
$$
My go-to technique for these types of functions is to construct a continuous funciton on the boundary of the unit circle then say that the answer is the solution to the Dirichlet Boundary Value Problem.  That's not possible here, but I believe that I may need to create a uniform sequence of continuous functions on the boundary of the disk.  I'm stuck.  Can anyone help?
 A: Generally, for a piecewise continuous function on the boundary of the unit circle,
the function defined by the Poisson kernel is still harmonic in the
unit disk, and has the given boundary values at all points where the
boundary function is continuous, see e.g.
Harmonic function in a unit disk with jump boundary data.
Therefore the solution to your boundary value problem can be computed
as
$$
 u(r e^{i \theta}) = \frac {1}{2\pi} \int_0^\pi P_r(\theta - t) \, dt \, ,
$$
compare Poisson kernel.
However, in your special case, a geometric observation helps to get
an explicit solution.
The function 
$$
 v(z) = \operatorname{Arg}(1-z) - \operatorname{Arg}(z+1)
$$
(where both arguments are chosen in the range $(-\pi/2, \pi/2)$ )
is harmonic in the strip $\{ z \mid -1 < \operatorname{Re}z < 1 \}$.
From Thales' theorem
one can see that
$$
 v(z) = \begin{cases}
 -\pi/2 && \text{ if } |z| = 1, \operatorname{Im} z > 0 \\
 \pi/2 && \text{ if } |z| = 1, \operatorname{Im} z < 0 \\
\end{cases}
$$
so that
$$
 u(z) = 1 - \frac 2\pi v(z)
$$
has the desired properties.
A: My first thought went to the square wave:
$$ u(z) =\frac{1}{2}\left(1+ \frac{4}{\pi}\,\text{Im}\left(\text{arctanh}\, z\right)\right) $$
clearly solves the problem.
