This post shows that if $K$ is a compact subset of $\mathbb R^n$, then the set of points in $\mathbb R^n$ which are at a distance $1$ from $K$ have Lebesgue measure 0.
Does the same conclusion not follow if $K$ is replaced by any subset of $\mathbb R^n$?
Here is my (possibly flawed) argument.
Let use take $n=2$ for simplicity. Let $S$ be an arbitrary subset of $\mathbb R^2$ and $A = \{x\in \mathbb R^2:\ d(x, S) = 1\}$. We want to show that $A$ has Lebesgue measure $0$. Let $I^2 = \{x\in \mathbb R^2:\ 0\leq x_1, x_2\leq 1\}$. For any $v\in \mathbb Z^n$ let $Q_v$ be the unit square defined as $Q_v = \{v+x:\ x\in I^2\}$. Thus $Q_0 = I^2$.
Let $S_v = Q_v\cap S$ for any $v\in \mathbb Z^2$ and let $A_v$ be the set of points in $\mathbb R^2$ which are at a distance $1$ from $S_v$. Note that each $A_v$ has Lebesgue measure $0$ since $d(x, S_v) = d(x, \bar S_v)$ for any $x\in \mathbb R^2$, and $\bar S_v$ is compact. So we will be done if we can show that $A \subseteq \bigcup_{v\in \mathbb Z^2} A_v$. Indeed, let $x\in A$ be arbitrary and assume that $x\notin A_v$ for any $v$. Clearly, if $v$ is far away from $x$ then $d(x, S_v) > 2$. Thus, it follows that there is $\epsilon>0$ such that $d(x, S_v) > 1 + \epsilon$ for all $v$. But then $x$ cannot lie in $A$, a contradiction, and we are done.
I may be consistently overlooking a simple mistake.
