measurability with zero measure 
Let $f : [0,1] \rightarrow \Bbb R$ is arbitrary function , and $E \subset \{ x \in [0,1] | f'(x)$  exists$\}$.

How to prove this statement:

If $E$ is measurable with zero measure then $f(E)$ is measurable with zero measure.

If $E$ is measurable with zero measure then $m(E)=0$ that $m :$ {measurable} $\rightarrow \Bbb R$, $m^* : P(\Bbb R) \rightarrow \Bbb R$,$A:= \{ x \in [0,1] | f'(x)$ is  be exist$\}$. so $m(E)=m^*(E) <= m^*(A)$

hot_queen's answer looks fine to me. This is not an independent answer, it's a minor variation and a comment on hot_queen's answer; I'm posting it as an answer for better readability (and who knows, someone might be fooled into upvoting it).
I'm reorganizing the covering lemma part of the argument. As observed earlier, it suffices to discuss the situation where $|f'|\le 1$ on $E$. For each $x\in E$, we can then find an interval $I_x=(x-h_x, x+ h_x)$ so that $|f(I_x)|\le 2 |I_x|$. We can also demand that $I_x\subset U$ for some open set $U\supset E$ with $|U|<\epsilon$.
The intervals $I_x$ ($x\in E$) cover $E$, and we can find a subcollection $I_j$ (necessarily countable) that still covers $E$ and has overlap at most $2$: $\sum \chi_{I_j}\le 2$. This is a version of Besicovich's covering lemma; in one dimension, this is quite easy and immediately plausible. Now
$$
|f(E)|\le \sum |f(I_j)| \le 2 \sum |I_j| \le 4 \left| \bigcup I_j \right| < 4\epsilon ,
$$
as desired.
