Derivatives and Lebesgue measure. Is it true that, if $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuously differentiable, then for all $E\subseteq\mathbb{R}$ with $\lambda(E)=0$, it holds that $\lambda(f(E))=0$.
My thoughts would be applying the mean value theorem to f and get the measure of both sides. But the proof seems not that simple... 
Is there a simple proof about this?
Unless my intuition is wrong and the proposition doesn't hold.
Thanks in advance...
 A: Hint: It's  enough to do this for  compact $E$ (why?). Say $E\subset[a,b]$. Now $f'$ is bounded on $[a,b]$ so $|f(t)-f(s)|\le c|t-s|$ for $t,s\in[a,b]$.
Cover $E$ by finitely many intervals (contained in $[a,b]$) of total length less than $\epsilon$...
A: The key in the argument is that if $f$ is continuously differentiable it  would lipschtiz continuous  in every interval $[-M,M]$.  
So suppose first that $E$ is bounded then we have $\lambda(E)=0$ and $E \subseteq [\alpha,\beta]$ for some real numbers $\alpha , \ \beta$. 
Then since $f$ is continuously differentiable in $[\alpha, \beta]$ it is going to be bounded with a bound $M >0$ and now its easy to prove by using the mean value theorem that $f$ is going to be lipschitz continuous with constant $M>0$.  
Hence for every $x,y \in [\alpha,\beta]$  the following inequality holds :
$|f(x)-f(y)| \leq M|x-y|$ 
Now let $ \epsilon >0$ since $\lambda (E) =0$ there is an open cover of intervals $((\alpha_n,\beta_n))_{n \in \mathbb{N}}$ such that $ E \subseteq \cup_{n=1}^{\infty}(\alpha_n, \beta_n)$ and $\lambda(E) \leq \sum_{n=1}^{\infty} ( \beta_n - \alpha_n) < \dfrac{\epsilon}{M}$.
Now $f(E) \subseteq \bigcup_{n=1}^{\infty} f(\alpha_n , \beta_n)$ so 
$\lambda(f(E)) \leq \lambda(\bigcup_{n=1}^{\infty}f(\alpha_n , \beta_n)) \leq \sum_{n=1}^{\infty} f(\alpha_n , \beta_n ) \leq $ 
$M \sum_{n=1}^{\infty} ( \beta_n - \alpha_n) < \epsilon.$ 
So we have $\lambda (f(E)) =0. $
Now for general case $E \subseteq \mathbb{R} $ we take intersections $E \cap [-n,n]$.
