Let $\mu$ be a probability measure on $\mathbb R^n$ and let $P$ be a compact polytope in $\mathbb R^n$. For any $x \in \mathbb R^n \setminus P$, let $p(x) \in P$ be (unique!) point in $P$ which is closest to $x$ and let $d(x) := \|x-p(x)\|$ be the distance from $x$ to $p$ and $u(x) := (x-p(x))/d(x) \in S\_{n-1}$, where $S\_{n-1}$ is the unit-sphere in $\mathbb R^n$.



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> **Question.** *Is it true that that for every $\epsilon>0$, there exists a measurable set $P^\epsilon$ such that $P \subseteq P^\epsilon$, $\mu(P^\epsilon) \le \mu(P) + \epsilon$, and the mapping $x \mapsto u(x)$ is Lipschiitz-continuous on $\mathbb R^n \setminus P^\epsilon$ ?*
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For example, if $P$ is just a half-space, $u$ is a constant, and thus Lipschitz-continuous on $\mathbb R^n\setminus P$.


**Related.** [Conditions for Lipschitz continuity of boundary normal vector to closed set with "unique closest-point" poperty (e.g closed convex sets)](https://mathoverflow.net/q/412656/78539)



On the Lipschitz continuity of the unit-normal vector field of a polytope
**EDIT:** The OP changed the question while I was writing this answer. In this answer, we take the domain of $u$ to be $\{x\in\mathbb{R}^n:d(x)\geq\epsilon\}$ for some fixed $\epsilon>0$. This is a different use of $\epsilon$ than in OP's current problem formulation.


Let $B\_\epsilon$ denote the open ball centered at the origin of radius $\epsilon$. Below, we show how to factor $u$ as the composition of a $2$-Lipschitz map $g\colon\mathbb{R}^n\to\mathbb{R}^n$ with a $(1/\epsilon)$-Lipschitz map $f\colon(\mathbb{R}^n\setminus B\_\epsilon)\to S^{n-1}$.


To analyze each map, we will use the fact that [projection onto a nonempty closed convex set is contractive](https://math.stackexchange.com/questions/3809431/geometric-proof-that-projections-on-convex-sets-are-contractive).


First, define $g$ by $g(x):=x-p(x)$, where $p$ denotes projection onto $P$. Then
$$
\|g(x)-g(y)\|=\|(x-p(x))-(y-p(y))\|\leq\|x-y\|+\|p(x)-p(y)\|\leq2\|x-y\|.
$$
Next, define $f$ by $f(x):=x/\|x\|$. Since the domain of $f$ is the complement of $B\_\epsilon$, we can instead write $f(x)=q(x)/\epsilon$, where $q$ denotes projection onto the closure of $B\_\epsilon$. Then
$$
\|f(x)-f(y)\|=\|q(x)/\epsilon-q(y)/\epsilon\|=(1/\epsilon)\|q(x)-q(y)\|\leq(1/\epsilon)\|x-y\|.
$$
Combining these estimates, we have that $u=f\circ g$ is $(2/\epsilon)$-Lipschitz.


