Continuity on the set $K_{1/3}\times K_{1/3}$. Let $K_{1/3}$ be the usual Cantor set, $Q=K_{1/3}\times K_{1/3}$ and $z_{n,k}$ the centers of the $4^n$ squares of the n-th step in the construction of $Q$. In some lecture notes I read it is stated without a proof that the sequence 
$$g_n(z)=4^{-n}\sum\limits_{k=1}^{4^n} (z-z_{n,k})^{-1}$$ converges to a function $g$ which is continuous on $S^2$ and holomorhic off $Q$. The only thing that challenges me is the continuity of $g$ on the points of $Q$. Any ideas?
 A: One way to look at it, leaving a lot of details to you: 
There is a discrete probability measure $\mu_n$ such that $$g_n=K*\mu_n,$$where $K(z)=1/z)$. These measures have a weak limit $\mu$, which is the "uniform measure" on $Q$. So at least away from $Q$, the limit is $$g(z)=\mu*K(z).$$
Now this measure has the property that $$\mu(D(z,3^{-n}))\le c4^{-n}.$$Which for a warmup you can use to show that $\mu*|K|$ is bounded. In fact if you write $K=\sum K_n$ where $K_n$ is continuous and roughly speaking supported near $|z|=3^{-n}$ then $\mu*K_n$ is continuous and you can use that estimate on $\mu$ to show $\sum\mu*K_n$ converges uniformly.

Edit To show $\mu*|K|$ is bounded: First, let $K'(z)=K(z)$ for $|z|<1$, $0$ for $|z|\ge1$. It's enough to show that $\mu*|K'|$ is bounded. Let $$A_n=\{z:3^{-n-1}\le|z|<3^{-n}\}.$$Then $$|K'|\le c\sum_{n=0}^\infty3^{n}\chi_{A_n},$$so $$\int |K'|\,d\mu\le c\sum 3^{n}\mu(A_n)
\le c\sum 3^{n}\mu(D(0,3^{-n}))\le c\sum 3^n4^{-n}<\infty.$$That gives a bound on $\mu*|K'|(0)$; the same bound applies at every point, with the same argument.
To get continuity, consider instead $\sum \phi_n$, where $\phi_n$ is continuous and supported in $\{3^{-n-2}<|z|<3^{-n}$, and $\sum\phi_n=K'$ near the origin. Each $\mu*\phi_n$ is continuous, and arguing as above shows that $\sum\mu*\phi_n$ converges uniformly, in fact $||\mu*\phi_n||_\infty\le c(3/4)^n$.
