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.
 A: $\Rightarrow:$ 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 $x_{1}<x_{2}$. 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.
A: 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 $|f(x)-f(y)|<M|x-y|$ for all $x\leq y$, then for any finite number of finite disjoint intervals $[a_j,b_j]$ we have
$$\sum_j|f(b_j)-f(a_j)|\leq <M\sum_j|b_j-a_j|$$
For $\varepsilon>0$, let $\delta=\varepsilon/M$ so that if $\sum_j|b_j-a_j|<\delta$, then $\sum_j|f(b_j)-f(a_j)|<\varepsilon$.
