To show that the set point distant by 1 of a compact set has Lebesgue measure $0$ Could any one tell me how to solve this one?

Let $K$ be a compact subset of $\mathbb{R}^n$, and $$A:=\{x\in\mathbb{R}^n:d(x,K)=1\}.$$
Show that $A$ has Lebesgue measure $0$.

Thank you!
Suppose $x_0\in A$. Let $B := \{ d(x,K) &lt; 1\}$. 

Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) &lt; 1\}$ is a subset of $B$. 

This implies that for all $\epsilon &lt; 1/2$, we have that
$$ \mu(B_\epsilon(x_0) \cap B) \geq \frac1{2^n} \mu(B_\epsilon(x_0)) $$
where $\mu$ is the Lebesgue measure. (The factor $1/2^n$ is very loose: Since a sphere tangent to $x_0$ is contained in $B$, locally an orthant centered at $x_0$ is contained in $B$.)

Hence we have that for every $x_0\in A$, 
$$ \limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} &lt; 1 $$

By the Lebesgue differentiation theorem, the set of $x_0\in A$, for $A$ measurable, such that the above condition holds must be measure zero. Hence $A$ has measure zero.
$ f_K : \mathbb{R}^n \rightarrow \mathbb{R} $ given as $ f_K(x) = d(x,K) $ is clearly Lipschitz as $ |f_K(x)-f_K(y)| \leq |x-y|,\ A = f_K^{-1}(\{1\})$. A Lipschitz function $f$ take measure 0 sets to measure 0 sets as if $\mu$ is lebesgue measure for $\mathbb{R}^n$  then $ \mu(f(A))\leq (Lip(f))^n\mu(A) $(you can prove this is either using outer measure arguments or very trivially using equality of lebesgue and hausdorff measure).