On the Lipschitz continuity of the unit-normal vector field of a polytope 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 $\mathbb R^n$.

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 $\mathbb R^n \setminus P^\epsilon$ ?

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)
 A: 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.
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.
