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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.

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$\Rightarrow:$ 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.

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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 $\sum_j|f(b_j)-f(a_j)|<\varepsilon$.

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