Prove that if the derivative $f'(x)$ of a function exists on the measurable set $E$, then $f'(x)$ is measurable on $E$. Prove that if the derivative $f'(x)$ of a function exists on the measurable set $E$, then $f'(x)$ is measurable on $E$.
We are told to only consider 1 dimensional spaces,that f is  a  measurable function  in one variable.that is, f is  a  measurable function  in one variable. 
Following is my solution, I am not sure whether I am on the right track.Can someone have a look? Many thanks, I can explain further if I should. 
Thanks in advance!

 A: Sorry for the late answer
It seems to me there are two flaws in your proof.
Actually, for $x \in E$, $h_n(x)$ needs not tend towards $f'(x)$. Consider for instance$$f : x \longmapsto\begin{cases}
1&\text{if $x \le 0$ or $x \notin \mathbb{Q}$}\\
1+x^2&\text{otherwise}
\end{cases}$$
Then you can check that $f$ is differentiable on $]-\infty,0]$ with derivative $0$, but $f$ is discontinuous on $]0,+\infty[$. Thus $g : x \mapsto\begin{cases}
1&\text{if $x \le 0$}\\
0&\text{if $x>0$}
\end{cases} \ \ $ so $h_n(0) \underset{n \to + \infty}{\longrightarrow} + \infty$, and so $\big(h_n(x)\big)_{n \ge 1}$ does not converge to $f'(0) = 0$. Note that, using a Cantor set, you can adapt this so that $h_n$ does not converge to $f'(x)$ on a subset of $E$ with positive measure.
Moreover, your last claim at the end of the proof (with some set being countable, or having zero measure) also seems false to me.

More constructively, I think that you can follow the standard reasoning to prove your result.
For $n \in \mathbb{N}^*$, denote $g_n : x \mapsto n \left ( f\big( x + \frac{1}{n} \big) - f(x) \right)$. We consider that $f'$ is defined on $E$. Then $(g_n)$ converges pointwise to $f'$ on $E$. 
We take an closed set $C \subset \mathbb{R}$, and we want to prove that $f'^{-1} (C)$ is measurable.
For every $x \in f'^{-1}(C)$, for $k \in \mathbb{N}^*$, $d \big( g_k(x), C \big) \le d \big( g_k(x), f'(x) \big) \underset{k \to + \infty}{\longrightarrow} 0$. Now for $n \in \mathbb{N}^*$, we denote the open set $C_n = \left \{ y \in \mathbb{R},\ d(y, C) < \frac{1}{n} \right \}$. Using the previous inequality, we get that for all $n \in \mathbb{N}^*$, there exists $m>n$ such that $g_m(x) \in C_n$, so $x \in g_m^{-1} (C_n)$. Moreover, $x \in E$. Hence 
$$ f'^{-1}(C) \subset \bigcap \limits_{n \in \mathbb{N}^*} \bigcup \limits_{m \ge n} E \cap g_m^{-1} (C_n)$$
Now for the other inclusion, with $x$ in the RHS set, you have a sequence of integers $(m_n)$ such that for all $n > 1$, $m_n \ge n\ $ and $\ d \big ( g_m(x), C \big) \le \frac{1}{n}$, so using pointwise convergence, as $x \in E$, $d \big( f'(x), C \big) = 0$, and $C$ is closed, so $f'(x) \in C$, and thus $x \in f'^{-1}(C)$.
Hence $f'^{-1}(C) = \bigcap \limits_{n \in \mathbb{N}^*} \bigcup \limits_{m \ge n} E \cap g_m^{-1} (C_n)$ and every set $g_m^{-1}(C_n)$ is measurable because $C_n$ is Borel and $g_m$ is measurable (because $f$ is measurable). As $E$ is also measurable, we get an intersection of unions of measurable sets, so $f'^{-1}(C)$ is measurable.
Finally, we get that $f'$ is measurable on $E$.
