# Differentiable A.E. with Positive Derivative Implies Increasing

Claim: Let $$f : [0,1] \to [0,1]$$ be continuous and differentiable almost everywhere on $$[0,1]$$. If the derivative of $$f$$ is positive wherever it exists, then $$f$$ is strictly increasing.

Here's my fallacious proof:

By way of contradiction, suppose there exist $$x < y$$ in $$[0,1]$$ such that $$f(x) \ge f(y)$$. I think I can say by continuity (intermediate value theorem?) that $$f(x) \ge f(z)$$ whenever $$z \in [x,y]$$. Now, if for every $$z \in (x,y]$$ we had that $$f$$ wasn't differentiable on $$(x,z)$$, then this would contradict the fact that $$f$$ is differentiable almost everywhere. Hence, there must exist a $$z \in (x,y]$$ such that $$f$$ is differentiable on $$(x,z)$$. By the mean value theorem, there is a $$c \in (x,z)$$ such that $$f'(c) = \frac{f(z)-f(x)}{z-x} \le 0$$, which is a contradiction. Hence, $$f$$ must be strictly increasing.

As Ryan Unger pointed out in the chatroom, I haven't given a terribly convincing reason why $$f$$ should be differentiable on any open interval in $$[0,1]$$, let alone $$(x,z)$$. So, my question is twofold. First, is the above claim true; is there any way to salvage my proof?

My next question is, does there exist a continuous function which is differentiable almost everywhere but the set of points of differentiability contains no intervals? I was thinking maybe the fat cantor set could help...?

EDIT: I should point out that $$f$$ doesn't have to have the domain and codomain that I gave it; I only specified those because I'm thinking about Thompson's group $$F$$.

• See Goldowsky-Tonelli theorem. – Gabriel Romon Sep 8 '19 at 14:38
• @GabrielRomon Any statement of Goldowsky-Tonelli's theorem I've found says the theorem holds for countable sets. Can it be generalized to sets of measure $0$? – user193319 Sep 8 '19 at 15:08

What about $$f(x) = 1-\phi(x)$$, where $$\phi$$ is the Cantor function? $$\phi$$ maps $$[0,1]$$ onto $$[0,1]$$, is continuous and non-decreasing, $$\phi'(x)$$ exists and is equal to $$0$$ in the complement of the Cantor set (hence a.e.).
• Take $f(x)=x-\phi(x)$ instead an you get an actual counterexample, with a derivative that is positive whenever it exists. – hmakholm left over Monica Sep 8 '19 at 16:26