A function $f:[0,1]\rightarrow\mathbb{R}$ is called singular continuous, if it is nonconstant, nondecreasing, continuous and $f^\prime(t)=0$ for almost every $t\in(0,1)$.
Let $f$ be a singular continuous function and $T$ the set where $f$ is not differentiable.
Question: Is $T$ closed?
Example:
A classical example of such a function is the so-called devil's staircase, obtained as a limit of increasing step functions constructed in a similar way as the classical Cantor set. In that case, the set in question is the Cantor set, which is closed.
Note/Motivation:
I am aware of the duality between nondecreasing functions and positive measures. The corresponding analogue of a singular continuous function is a measure singular with respect to the Lebesgue measure. The set $T$ corresponds in that case to the support of the measure, which is by definition closed.
That makes me think $T$ would have to be closed too, but I'm looking for a direct proof without use of measure theory (or a counterexample).
The Cantor function $f$ can be modified to become differentiable at one point $x_0$ of the Cantor set, thus answering your question in the negative. The modification is $F=f\circ \psi$ where $\psi:[0,1]\to[0,1]$ is smooth and strictly increasing, with nonzero derivative except at one point $t_0\in (0,1)$, with $\psi(t_0)=x_0$. (A suitable cubic polynomial will do the job.) Outside of $t_0$, the function $\psi$ is a diffeomorphism and thus does not change the differentiability in either way.
Recall that $f$ is $C^\alpha$-continuous with $\alpha=\log_2 3>\frac12$; see, e.g., Prove that Cantor function is Hölder continuous. Since $\psi'(t_0)=0$, we have $\psi(t)-\psi(t_0) = O((t-t_0)^2)$ as $t\to t_0$. Therefore, $F(t)-F(t_0)=o(t-t_0)$, which makes $F$ differentiable at $t_0$.