Lipschitzness of $f$ on a high-probabilty subset from $\|f(X)-f(Y)\|\le \|X-Y\|$ with high probability for iid $(X,Y)$ Consider two iid random vectors $(X,Y)$ both in $R^k$ and $f:R^k\to R^m$. Assume that for some $\epsilon>0$ (that may be assumed small if necessary, say, $\epsilon<0.01$), it holds that
$$P \Big( \|f(X)-f(Y)\| \le \|X-Y\|\Big) \ge 1-\epsilon.$$
The norm is the Euclidean norm.
The goal is to derive from this high-probabilty bound on the product measure some Lipschitzness condition on $f$. I am looking of a reverse, in some sense, of the simple observation that if the restriction of $f$ to some $A$ is 1-Lipschitz and $P(X\in A)\ge1-\epsilon/2$ then the above probability bound is true.
Question:
Assuming $P ( \|f(X)-f(Y)\| \le \|X-Y\|) \ge 1-\epsilon$,
Is it always true that there exists a measurable $A\subset R^k$ such that the restriction of $f$ to $A$ is 1-Lipschitz (or C-Lipschitz for some C) and such that $P(X\in A)\ge 1-g(\epsilon)$ for some $g$ with $g(\epsilon)\to_{\epsilon\to0}0$?
An answer would still be of interest to me if $k=m=1$ and $X,Y\sim$Uniform$[0,1]$.
 A: Here is an example to show that there can not be a uniform $C$ such that the restriction to $A$ is $C$-Lipschitz, whenever $A$ is a set of positive measure. I believe it can be tweaked to get an example such that the restriction to any set of positive measure is not Lipschitz. The rough idea is that $f$ is $C$-Lipschitz on large scales, but not on any small scale, which then shows that $A$ can never have points of density.
Explicitly, fix a large constant $L>0$ and let $f:\mathbb{R} \to \mathbb{R}$ be defined as $f(x) = L |x|$ for $|x|\le \epsilon/L$, extended periodically with period $2\epsilon/L$. (This is a triangle wave with minimum 0, maximum $\epsilon$, and slopes $\pm L$.) Now let $X$ and $Y$ be uniformly distributed independent variables on $[0,1]$. Since $|f(x)-f(y)| \le \epsilon \le |x-y|$ whenever $|x-y| \ge \epsilon$, we have that $P(|f(X)-f(Y)| \le |X-Y|) \ge 1-2\epsilon$.
Now let $C$ be any constant with $0<C<L$, let $x \in [0,1]$ be arbitrary, and let $B = \{ y \in [0,1] : |f(x)-f(y)| \le C |x-y| \}$. Since the slopes near $x$ are larger than $C$ in absolute value, it is immediate that $x$ is an isolated point in $B$, which implies that any measurable set $A \subset [0,1]$ on which $f$ is $C$-Lipschitz has no points of density, so $A$ must have measure zero.
