# Absolutely continuous function with bounded derivative on an open interval is Lipschitz

I've come across a question which states that one can prove $$f:\mathbb{R} \rightarrow \mathbb{R}$$ is Lipschitz iff $$f$$ is absolutely continuous and there exists $$M \in \mathbb{R}$$ such that $$|f'(x)|≤ M$$ almost everywhere.

I've only ever seen this fact proven for functions on closed and bounded intervals. Is it possible to show this on all of $$\mathbb{R}$$? I'm skeptical as the main fact I'd like to use which represents $$f$$ as an indefinite integral ( since $$f$$ is absolutely continuous) requires the domain to be a closed and bounded interval. However I can't find a counterexample.

$$\Rightarrow:$$ Suppose that $$f$$ is Lipschitz continuous, then there exists $$M>0$$ such that $$|f(x)-f(y)|\leq M|x-y|$$ for all $$x,y\in\mathbb{R}$$. Let $$\varepsilon>0$$ be given. Define $$\delta=\frac{\varepsilon}{2M}>0$$. Let $$\{I_{n}=(a_{n},b_{n})\mid n\in\mathbb{N}\}$$ be a countable family of pairwisely disjoint open intervals such that $$\sum_{n=1}^{\infty}\mu(I_{n})<\delta$$. (Here, $$\mu(A)$$ denotes the Lebesgue measure of a set $$A$$.) $$\begin{eqnarray*} & & \sum_{n=1}^{\infty}|f(b_{n})-f(a_{n})|\\ & \leq & \sum_{n=1}^{\infty}M\mu(I_{n})\\ & < & \varepsilon. \end{eqnarray*}$$ This shows that $$f$$ is absolutely continuous. From the standard theory of any real analysis textbook, for each $$N\in\mathbb{N}$$, $$f'(x)$$ exists a.e. for $$x\in[-N,N]$$. Let $$A_{n}=\{x\in[-N,N]\mid f'(x)\mbox{ does not exist}.\}$$, then $$\{x\in\mathbb{R}\mid f'(x)\mbox{ does not exist}\}=\cup_{N}A_{N}$$ which has measure zero. That is, $$f'(x)$$ exists a.e. Let $$x\in\mathbb{R}$$ for which $$f'(x)$$ exists. For any $$y>x$$, we have $$|\frac{f(y)-f(x)}{y-x}|\leq M$$. Letting $$y\rightarrow x+$$, then we have $$|f'(x)|\leq M$$.
$$\Leftarrow:$$ Suppose that $$f$$ is absolutely continuous and there exists $$M>0$$ such that $$|f'(x)|\leq M$$ a.e.. Let $$x_{1},x_{2}\in\mathbb{R}$$ with $$x_{1}. From the standard theory with $$f$$ restricted on $$[x_{1},x_{2}]$$, we have that $$f(x_{2})-f(x_{1})=\int_{x_{1}}^{x_{2}}f'(x)dx$$. Hence $$\begin{eqnarray*} & & |f(x_{2})-f(x_{1})|\\ & \leq & \int_{x_{1}}^{x_{2}}|f'(x)|dx\\ & \leq & M(x_{2}-x_{1}). \end{eqnarray*}$$ Therefore, $$f$$ is Lipschitz continuous.
If $$f$$ is absolutely continuous then by the fundamental theorem of calculus Lebesgue version) $$f'$$ exists (although here we are assuming this already), $$f'$$ is integrable (in any compact interval $$[x,y]$$ and $$f(y)-f(x)=\int^y_xf'(t)\,dt,\quad x\leq y$$ If $$|f'|\leq M$$ almost surely, then $$|f(y)-f(x)|\leq M|y-x|$$ which means $$f$$ is Lipchitz
The converse is much easier. Suppose $$|f(x)-f(y)| for all $$x\leq y$$, then for any finite number of finite disjoint intervals $$[a_j,b_j]$$ we have $$\sum_j|f(b_j)-f(a_j)|\leq For $$\varepsilon>0$$, let $$\delta=\varepsilon/M$$ so that if $$\sum_j|b_j-a_j|<\delta$$, then $$\sum_j|f(b_j)-f(a_j)|<\varepsilon$$.