I learned [this question](https://math.stackexchange.com/questions/235689/pointwise-existence-of-radon-nikodym-derivative-sufficient-for-absolute-continui) from math.stackexchange, which is equivalent to ask that if $f:[0,1]\to \mathbb{R}$ is a continuous function with bounded variation, does 
$$g(x):=\lim\_{\epsilon\to 0}\frac{f(x+\epsilon)-f(x-\epsilon)}{2\epsilon}$$
exist for every $x\in[0,1]$ imply that $f$ is absolutely continuous?


Moreover, if we do not know whether $f$ is of bounded variation or not, what can we say about the differentiability of $f$? For example, if $g\equiv 0$, will $f$ be a constant?


Any help is appreciated. 


**Edit**: I think the original question cited from math.stackexchange has been solved by the asker himself there, which is based on the Vitali covering theorem for Radon measures on $\mathbb{R}^n$. For my own question, where $f$ is not assumed to be of bounded variation a priori, Jack Huizenga's answer is good enough for me.



Does a weaker condition than vanishing derivative imply a function being constant?    
The ordinary proof that 0 derivative implies constant rests on the mean value theorem. Your function $g(x)$ is the "symmetric derivative" of $f$, so we should look for a "quasi-mean value theorem" for symmetric derivatives. A quick google search came up with the paper 


<http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-27/No-3/27%283%298-5%28279-301%29.pdf>


Here it is shown (Theorem 2) that if $f$ is continuous on $[a,b]$ and symmetric differentiable on $(a,b)$ with symmetric derivative $f^s$, then there are points $\xi,\eta\in (a,b)$ with 


$$f^s(\eta) \leq \frac{f(b)-f(a)}{b-a} \leq f^s(\xi).$$


Clearly then if $f^s = 0$ we must have that $f$ is constant.


