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 $m(E)=m^*(E) <= m^*(A)$

Hint. Let $E_n=\{x\in E:|f'(x)|\le n\}$. Prove that each $f(E_n)$ has measure $0$. 
Comment. The above hint is perhaps not very helpful. I realized I do not know how to show that $m(f(E))=0$ even is I assume that, say, $|f'(x)|\le\frac13$ for all $x\in E$. On the other hand if one could solve the problem under this additional assumption, then one would also be able to prove in general that $m(f(E_n))=0$ for each $n$, where $E_n$ are as defined in the hint. Indeed we could define, say, $g(x)=\frac1{3n}f(x)$, and then $|g'(x)|\le\frac13$ for each $x\in E_n$. If we knew that $m(g(E_n))=0$ then clearly we would also have $m(f(E_n))\le3n\cdot m(g(E_n))=0$. The question has not yet been answered. 
Edit. I do not follow my own hint. I know that in addition to, say, $|f'(x)|\le\frac13$ for all $x\in E$, after throwing out at most countably many points from $E$, we may also assume that for each $x\in E$ and each $t>0$ both $(x-t,x)\cap E$ and $(x,x+t)\cap E$ are uncountable. This might be relevant to help cover $E$ with certain open intervals ... but it is getting a bit vague, and I have not done it, and there might be something easier or better. 
