Analytic Continuation of a function defined in integral

Question

Let $C$ be the boundary of a unit circle. The function $f:\mathbb{C} \backslash C \to \mathbb{C}$ is defined by $f(z)=1-\frac{1}{2\pi i} \int_C \frac{1}{\zeta-z} d\zeta$.

I found that $f(z)=0$ when $z$ is inside the unit circle and $f(z)=1$ when $z$ is outside the unit circle. How can I use this information to determine whether $f$ has an analytic continuation $g$ that is analytic in the whole complex plane?

Answer 1

Suppose that there is an entire function $g$ such that $g$ is an analytic continuation of $f$. Let $z_0 \in \mathbb C$ such that $|z_0|=1.$ Then we have, since $g$ is continuous at $z_0$,

$$ g(z_0)= \lim_{z \to z_0, |z|<1}f(z)=0$$

and

$$ g(z_0)= \lim_{z \to z_0, |z|>1}f(z)=1,$$

a contradiction.