Are the singular points of harmonic function on the disk always of measure zero? Let $f : \mathbb D^2 \to \mathbb R$ be a smooth function with no singular points, i.e. $df \neq 0$ on $\mathbb D^2$. (Here $\mathbb D^2$ is the closed unit disk in $\mathbb R^2$).
Let $\omega:\mathbb D^2 \to \mathbb R$ be the harmonic function corresponding to the Dirichlet problem imposed by $f$, i.e $\omega|_{\partial \mathbb D^2}=f|_{\partial \mathbb D^2}$.
Is it true that $d\omega \neq 0$ on a set of full measure in $M$?
*By harmonic I mean $\Delta \omega=\omega_{xx}+\omega_{yy}=0$.
 A: After speaking with a colleague, here is a possible solution:
Sine $\omega$ is harmonic, it is real-analytic. In particular its partial derivatives $\omega_x,\omega_y:\mathbb D^2 \to \mathbb R$ are real-analytic. Define $A=\{ p\in \mathbb D^2 \, | \, \omega_x(p)=\omega_y(p)=0\}$. We shall prove $A$ has measure zero. Indeed, $A$ is closed. So, if it had positive measure, then it would contain an accumulation point in $\operatorname{Int} \mathbb D^2$  (see the lemma below). By the identity theorem, this would imply $\omega_x=\omega_y=0$ on $\operatorname{Int} \mathbb D^2$, which would imply $\omega$ is constant on $\mathbb D^2$. However, $f$ cannot be constant* on $\partial \mathbb D^2$ which is a contradiction to $f|_{\partial \mathbb D^2}=\omega|_{\partial \mathbb D^2}$.



*

*$f$ is not constant on $\partial \mathbb D^2$: Indeed, we assumed that $df \neq 0$ everywhere, so $\min f,\max f$ are both obtained on $\partial \mathbb D^2$, and are distinct, since $f$ is not constant on the whole disk. 



Lemma: A closed subset $A \subseteq \mathbb D^2$ of positive measure contains an accumulation point in $\operatorname{Int} \mathbb D^2$.
Proof:
Assume by contradiction that $A$ does not contain an accumulation point in $\operatorname{Int} \mathbb D^2$. This implies (by compactness), that for any natural $n$, $A \cap B_{1-1/n}$ is finite, where $B_r$ is the closed disk of radius $r$. Thus, $A \cap \operatorname{Int} \mathbb D^2=\cup_n A \cap B_{1-1/n}$ is countable, hence of measure zero. Thus $A$ has measure zero, contradiction.
