Differentiability at a point $x$ with $f$ differentiable in $\mathbb R\backslash\{x\}$ I have a real function that satisfies:


*

*$f:\mathbb R\rightarrow\mathbb R$ is differentiable at $x$ for $x\neq x_0$.

*There is a full-measure set $T$ such that for any sequence $t_n\in T$ with $t_n\stackrel{n\rightarrow\infty}{\longrightarrow} x_0$ we have $\lim_{n\rightarrow\infty} f^\prime (t_n) = c$.

*$f$ is non-decreasing 


I'm trying to show that $\mathbf{f}$ must also be differentiable at $\mathbf{x_0}$. 
My idea is that if there is some sequence $x_n$ with $x_n\stackrel{n\rightarrow\infty}{\longrightarrow} x_0$ and  $\lim_{n\rightarrow\infty}\frac{f(x_n)-f(x_0)} {x_n-x_0} \neq c$ then I can approximate the $x_n$ and $x_0$ with points in $T$ but I have to be careful with the choice of $\varepsilon$ and $\delta$ in approximation because we have two limits here (the derivative itself and the limit of the derivatives). I'm not sure if this would work because it seems that I would need to change the order of limits. Besides, in real analysis, there are some many pathological cases that perhaps there is even a counter-example. Any suggestions on how to attack this problem?
 A: If you let $f(x)=0$ for $x\le 0$ and $f(x)=1$ for $x>0$ you get a counterexample. The result you want requires that you assume that $f$ is continuous at $x_0$.
Assuming that, there's probably a more elementary proof than the following simple but not quite so elementary proof:
Say $x_0=0$ for convenience. Subtract a constant so $f(-1)=0$.
Now since $f$ is continuous at $0$ and differentiable elsewhere it is continuous. So there exists a finite measure $\mu$ on $[-1,1]$ such that $$f(x)=\mu([-1,x])\quad(x\in[-1,1]).$$
Now $\mu=\mu_a+\mu_s$ where $\mu_a$ is absolutely continuous and $\mu_s$ is singular (wrt Lebesgue measure). We must have $\mu_s([-1,1]\setminus\{0\})=0,$ since otherwise there would exist $x\ne0$ such that $f$ was not differentiable at $x$. And $\mu_s(\{0\})=0$ since $f$ is continuous at $0$.
So $\mu_s=0$, which says that $f$ is absolutely continuous. In particular, $$\frac{f(x)-f(0)}{x}=\frac1x\int_0^xf'(t)\,dt,$$and now the assumption about $f'$ tending to $c$ on a set of full measure shows that $f'(0)=c$.
