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 $F$.
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.).