How pathological can a convex function be?

You can get non-differentiability at a countable infinite set by considering $$\sum_{n=1}^\infty a_n|x-c_n|$$ where $a_n>0$ and $(a_n)$ goes to zero rapidly.

Others have provided examples in which the functions have countable sets of points at which $f$ is nondifferentiable. However, the answer to the question on whether this set of nondifferentiable points has measure zero is affirmative. This is even true for convex functions $f:\mathbb{R}^n\to \mathbb{R}$ and is just a consequence of Rademacher's theorem plus the fact that continuous convex functions in normed spaces are locally Lipschitz continuous.

Here are the details:

Let $f:\mathbb{R}^n\to \mathbb{R}$ be convex, and let $\{x_n\}$ be a enumeration of the elements of $\mathbb{R}^n$ with rational coordinates. It is easy to see that this set is dense in $\mathbb{R}^n.$ For each $x \in \mathbb{R}^n,$ define

$$\epsilon(x)= \sup\{\epsilon>0: f \textrm{ is locally Lipschitz on the ball }B(x,\epsilon) \}.$$ Since convex functions are locally Lipschitz continuous, we have that $\epsilon(x)$ is well defined and positive. Now it is easy to prove that $$\mathbb{R}^n= \bigcup_{n\in \mathbb{N}} B(x_n, \epsilon(x_n)).$$ Let $ND$ be the set of points at which $f$ is nondifferentiable. Applying Rademacher's theorem on each $B(x_n, \epsilon(x_n))$ we find that the set $ND\cap B(x_n, \epsilon(x_n))$ has measure zero. Therefore

$$\mu(ND)\leq \sum_{n=1}^\infty \mu(ND\cap B(x_n, \epsilon(x_n)))=0,$$ as desired.

Hope this helps

The set of points with non-differentiability can be dense in the support e.g for $x\in[-1,1]$ take

$$x\mapsto\sum_{i=0}^\infty\sum_{j=0}^{2^i-1}4^{-i}|2^{1-i}j+2^{-i}-1-x|$$