Cantor function, strictly increasing?

Question

On Wikipedia: https://en.wikipedia.org/wiki/Cantor_function it says that the cantor function is non-decreasing. But I am wodering, if we restrict our domain to the cantor-set, does then the function become strictly increasing?

Answer 1

No. When you create the Cantor "devil staircase", call it $V(t)$, you fix the value of the function on all the open intervals that are removed in the construction of the Cantor set. Since $V(t)$ is continuous, the value at the end-points of each removed interval (which are included in the Cantor set) must be the same. Therefore, $V(t)$ is not strictly increasing.

Edit: $V(t)$ is not strictly increasing even if it is restricted to the cantor set, since the endpoints of the removed intervals belong to the Cantor set (although the set contains much more than those, of course), and $V(t)$ attains the same value at the two endpoints of each such interval.

However, it is possible to construct a strictly increasing function whose derivative is zero a.e.

Edit: if you're curious on how this function can be built, look here.