Suppose you have a function $f:[0,1]\to[0,1]$ which is strictly increasing and $f'(x)=0$ Lebesgue almost everywhere $x\in[0,1]$. Is it possible to say something particular about the lateral derivatives at points where $f$ is not differentiable? More specific: could we have a point $x$ where $f$ is not differentiable, but $D_{-}f(x), D_{+}f(x)\in(0,\infty)$?
(An example of such function could be the Cantor function)
It is possible to have some points like that. I'll construct an example with $D_-f(1/2) = 3/2$ and $D_+f(1/2)=1/2$.
Let $g$ be the Cantor function. We'll take $f(1/2) = 3/4$. I want to construct $f$ so that for $0 \le x \le 1/2$, $(3/2) x - (1/2-x)^2 < f(x) < (3/2) x + (1/2-x)^2$. This will imply that $D_-f(1/2) = 3/2$. Let $$ \eqalign{a_0 &= 0\cr a_{k+1} &= \frac{5 - \sqrt{21 - 24 a_k}}{4}\cr}$$ which makes $$\frac{3}{2} a_{k+1} - \left(\frac{1}{2} - a_{k+1}\right)^2 = \frac{3}{2} a_k$$ and $$ \frac{3}{2} a_{k+1} \le \frac{3}{2} a_k + \left(\frac{1}{2} - a_k\right)^2 $$ Note that $a_k$ is an increasing sequence with limit $1/2$. On the interval $[a_k, a_{k+1}]$ define $$ f(x) = \frac{3}{2} a_k + \frac{3}{2} (a_{k+1} - a_k) g\left(\frac{x - a_k}{a_{k+1}-a_k}\right) $$ and note that since $0 \le g \le 1$, the desired inequalities are true.
Similarly we can construct $f$ on $[1/2, 1]$ so that $D_+f(1/2) = 1/2$.
