$E \subset \{ x | f \text{ is differentiable at }x\}$. Then $|E|=0 \Rightarrow |f(E)|=0$. 
For $f:[0,1]\to \mathbb{R}$ let $E\subset\left\{x \mid f'(x) \text{exists}\right\}$. Prove that if $|E|=0$, then $|f(E)|=0$.

My attempt:
Let $E_{nk}=\left\{x\in [0,1]|\frac{|f(x+h)-f(x)|}{h}\leq n, |h|< \frac{1}{k} \right\}$. 
I am not sure where to go from here, but  I think that $E\subset \bigcup E_{nk}$, but I am not sure.
Any hints? Thank you.
 A: 
Claim: Let $E_n = \{x \in E : |f'(x)|\le n\}$, then $\lambda(f(E_n)) \le n\lambda(E_n)$.

Notice that $E = \bigcup E_n$, hence the desired result follows from the fact that
$$\lambda(f(E_n)) \le n\lambda(E_n) \le n\lambda(E) = 0.$$ Indeed, to conclude from here, we can use the fact that $f(E) \subset \bigcup f(E_n)$, which is a countable union of null sets.
Proof of Claim: Let $\epsilon > 0$ be given. Notice that the fact that the derivative exists and that it is bounded in $E_n$ means that for $x \in E_n$ $$\lim_{y \to x}\frac{f(y) - f(x)}{y - x} = f'(x),$$ hence there exists $\delta > 0$ such that for every $y \in (x - \delta, x + \delta) \cap [0,1]$ $$|f(y) - f(x)| \le (n + \epsilon)|y - x|. \tag 1$$ 
The idea we get from $(1)$ is that a scalar multiple of the length of an interval can be used to control the length of its image. Essentially we want to exploit the fact that $f$ behaves as a Lipschitz function on $E_n$. Unfortunately this does not hold for every interval, but it works for intervals in $E_n$. 
The next thing to do is to prove that we have "enough" (very tiny) intervals in $E_n$ to make the previous argument work. Indeed, let's define 
$$E_{n,m} := \big\{x \in E_n : \lambda(f(I)) \le (n + \epsilon)\lambda(I), \text{where}\ I\subset [0,1]\ \text{is an interval}, x \in I,\text{and}\ \lambda(I) < \frac 1m \big\}.$$ 
Then $E_{n,m} \subset E_{n,m+1}$ and $E_n = \bigcup_m E_{n,m}$. This last equality is not obvious at all: to prove that $E \subset \bigcup_m E_{n,m}$ we need to show that for every $x \in E_n$ we can find $m$ for which $x \in E_{n,m}$. This can be done reasoning as in the first step of the proof, choosing $m$ accordingly ($2m^{-1} < \delta$ works.)
We can now focus on the $E_{n,m}$. The trick that I have seen used in this situation is the following: by outer regularity consider $U_{n,m}$ open such that $\lambda(E_{n,m}) + \epsilon \ge \lambda(U_{n,m})$. Since $U_{n,m}$ is open we can decompose it into countably many disjoint open intervals of length less than $\frac 1m$. Let's call this family of intervals $\mathcal{I}$. Notice that we can apply $(1)$ to each element in $\mathcal{I}$! This implies that
\begin{align}
\lambda(f(E_{n,m})) \le &\ \lambda\Big(\bigcup_{I \in \mathcal{I}}f(I)\Big) \le \sum_{I \in \mathcal{I}}\lambda(f(I)) \le (n + \epsilon)\sum_{I \in \mathcal{I}}\lambda(I) \\
= &\ (n + \epsilon)\lambda\Big(\bigcup_{I \in \mathcal{I}} I\Big) = (n + \epsilon)\lambda(U_{n,m}) \\ \le &\ (n + \epsilon)(\lambda(E_{n,m}) + \epsilon).
\end{align} 
Notice that here it is important that the intervals are disjoint.
Recall that $E_{n,m} \uparrow E_n$ to get that $\lambda(f(E_{n,m})) \le (n + \epsilon)(\lambda(E_{n}) + \epsilon)$ and conclude letting $m \to \infty$ and then $\epsilon \to 0^+$. $\blacksquare$

As a remark, I should mention that I am not worrying about measurability since everything is a subset of the null set $E$. On the other hand the claim still holds without any assumption on the set $E_n$ defined as above (including measurability). In this case you can use the exact same argument replacing the Lebesgue measure with the outer measure. (you can then reintroduce the measure when you define the (measurable) open sets $U_{n,m}$.
A: Hint: Note that if $f$ is differentiable at $x$, then the limit 
$$ \frac{f(x+ h) - f(x)}{h}$$
exists as $h \to 0$. So whenever $|h|$ is small, the above expression is bounded by a constant. 
