Can the measure of zeroes of a harmonic function be positive?

Question

Let $u$ be a non-constant harmonic function of two variables defined, say, in the unit disk (or on the half plane for example). It is known that $u$ can vanish on some lines, as it discussed in here. But can $u$ vanish on a set of positive measure ?

Answer 1

Here's a way to fill in the gap in @Yiorgos' solution. The problem is local, so we may as well work on a simply connected domain.

Let $v$ be a harmonic conjugate of $u$, so that $f = u+iv$ is holomorphic. It follows from Cauchy-Riemann's equations that $f' = u'_x -iu'_y$. In particular, the gradient of $u$ vanishes exactly at the points where $f' = 0$. But $f'$ is also holomorphic, so the zero set of $f'$ is a discrete set (unless $f' \equiv 0$, but this only happens when $u$ is constant).

Hence, removing the discrete (thus measure zero) set of points where $f'=0$, the rest of the zero set is a countable union of graphs of $C^1$-functions.

(To see this last claim, take a countable exhaustion $K_1 \subset K_2 \subset \cdots$ of the unit disc minus $f^{-1}(0)$ by compact sets. For each point in $K_j$, use the implicit function theorem to see that the zero set of $u$ locally is the graph of a $C^1$ function. Use compactness to select a finite number of such graphs to cover the entire zero set in $K_j$, and take the union over all $j$.)