A $\mathcal{C}^1$ function $f: \mathbb{R} \rightarrow \mathbb{R} $ preserve Jordan measurability I'm having some difficulty showing that for any Jordan measurable set $E \in \mathbb{R}$ and any $\mathcal{C} ^1$ function $f: \mathbb{R} \rightarrow  \mathbb{R} $. $f(E)$ is also Jordan measurable.
This is what I got so far:
First I claim that if $I \subset \mathbb R $ is compact and $N \subset I $ is of Lebesgue measure zero then for any $\mathcal{C}^1$ function $f: \mathbb{R} \rightarrow  \mathbb{R} $ , $f(N)$ is also a null set.
I'm using the fact that $f$ is $L$-Lipschitz on $I$ thus can't magnify a sets volume by more than factor of $L$... 
Next, $E$ is Jordan measurable set, so it's boundary is a set of measure zero. 
$$E=L \uplus N$$
When $L=\{x \in E |f'(x)\neq 0 \}$ and $N=\{x \in E |f'(x)=0\}$. From Sard's theorem we get that $f(N)$ has Lebesgue measure 0 .
I'm really not sure if I got it right up until here, but assuming this is correct, now I need to show that $L$  is Jordan measurable. 
From the Inverse function theorem, for every $x \in L$  there are open sets: $U_x \subseteq \mathbb {R} $ containing $x$ and $V_x \subseteq \mathbb {R} $ containing $f(x)$ so that $f: U_x \rightarrow  V_x $ is a diffeomorphism. We can write 
$$ L = \bigcup_{x \in L} U_x$$ and $\partial L \subset \bigcup_{x \in L} \partial U_x$  Since $f$ is a diffeomorphism $ U_x \rightarrow  V_x $ we can say that $\bigcup_{x \in L} \partial f(U_x) = \bigcup_{x \in L}  f(\partial U_x) $ 
I don't really know how to proceed from here since I want to say that L can be presented as a countable union of sets $U_x$ but I can't see how I can prove it.
 A: Summing up we have a $\,\mathcal C^1$ function $f:\mathbb R \rightarrow \mathbb R$ and a Jordan measurable set $L \subseteq\mathbb R$  . Assuming that $f'(x) \neq 0$ for every $x \in L$, it is asked to show that $f(L)$ is Jordan measurable.
It is enough to prove that $\;\partial f(L) \subseteq f(\partial L)$ .
We use the fact that $f$ is continuous and open on $L$ that is maps open subsets to open subsets.
Of course $f$ is continuous because differentiable so $f(\overline L)$ is compact, in particular closed.
Then, since $f(L) \subseteq f(\overline L)$, one has also $$\overline {f(L)} \subseteq f(\overline L)$$ In fact $\overline {f(L)}$ is the smallest closed set that includes $f(L)$ .
Now $f$ is open on $L$ (apply the inverse function theorem) so $f(\text {int} L)$ is open.
Then, since $f(\text {int} L) \subseteq f(L)$, one has also $$f(\text {int} L) \subseteq \text {int} f(L)$$ In fact $\,\text {int} f(L)$ is the greatest open set that  is included in $f(L)$ .
Then $$\partial f(L) \,\subseteq\,\overline {f(L)} \setminus \text {int} f(L) \subseteq f(\overline L) \,\setminus\, f(\text {int} L) \,\subseteq\, f(\partial L)$$ so $\partial f(L)$ has measure $0$ .
