Strictly monotone real function: stationary point, non-differentiable point If we have a real function $f$ that is strictly increasing (or strictly decreasing), what can we say about measure and cardinality of stationary points/points with no derivative.
In particular:
-Is the set of stationary point countable? Measure zero?
-Is the set of point where derivative does not exist countable? Measure zero?
-If it have derivative everywhere (but not necessary continuous derivative) what can we say about its set of stationary point?
What I have so far:
-Derivative does not exist due to discontinuity is accounted for, only countable of them.
-Inverse always exist and is also monotone, and stationary point of $f$ corresponde to point with no derivative of $f^{-1}$, so if we solve the 2nd one we might be done with first one, or perhaps the other way round, depend on what the answer is.
-I am thinking of tweaking Cantor's staircase to make it strictly increasing, while still have stationary point almost everywhere. But cannot figure out how. The main problem is that I do not know anything about the derivative of a function even if I know those of functions that converge uniformly to it.
Thank you for your help.
 A: *

*Is the set of stationary point countable? Measure zero?


It can have positive measure. Let $K\subset [0,1]$ be a fat Cantor set, and define $$f(x)=\int_0^x (\operatorname{dist}(t,K))^2\,dt \tag{1}$$ 
 This function is strictly increasing, since $K$ is nowhere dense. It is differentiable with continuous derivative which vanishes at every point of $K$. So this answers your third question too. 
The inverse $f^{-1}$ is not differentiable on $f(K)$; however  a result stated below implies that   $f(K)$ has measure zero. 


*

*Is the set of point where derivative does not exist countable? Measure zero?


It need not be countable (take Cantor function $+x$), but it has measure zero, by a theorem of Lebesgue. 

Aside: with a bit of machinery, (1) can be improved. It is known that for every closed set (in particular for $K$) there exists a $C^\infty $ smooth function  $g$ that vanishes precisely on that set. An antiderivative of $g^2$ is $C^\infty$ smooth, and has zero derivative everywhere on $K$. 
