To show that the set point distant by 1 of a compact set has Lebesgue measure $0$
Suppose $x_0\in A$. Let $B := \{ d(x,K) < 1\}$.
Observe that since $K$ is compact, there exists $y_0\in K$ such that $d(x_0,y_0) = 1$. By definition $B_1(y_0) = \{x: d(x,y_0) < 1\}$ is a subset of $B$.
This implies that for all $\epsilon < 1/2$, 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 $x_0\in A$, $$ \limsup_{\epsilon \to 0} \frac{\mu(B_\epsilon(x_0) \cap A)}{\mu(B_\epsilon(x_0))} \leq \frac{2^n-1}{2^n} < 1 $$
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.
I was going to make this a comment but it occurred to me there might be sufficient interest that perhaps I should not bury it in a comment.
At the beginning of the paper below Erdős gives a short proof (that he attributes to Tibor Radó) making use of the Lebesgue density theorem that $E_r$ has Lebesgue measure zero, where $E$ is a closed set in ${\mathbb R}^n$ and $$E_r \, = \; \{ x \in {\mathbb R}^n : \; d(x,E)=r \} $$
Paul Erdős, Some remarks on the measurability of certain sets, Bulletin of the American Mathematical Society 51 #10 (October 1945), 728-731.
Later in this paper (item 6), Erdős proves the stronger result that, for $K$ compact, the Hausdorff $(n-1)$-measure of $K_r$ is finite, and hence $E_r$ has $\sigma$-finite Hausdorff $(n-1)$-measure when $E$ is closed.
More precise results can be found in a 1985 paper by Oleksiv/Pesin Zbl 573.28010 [English translation: Mathematical Notes 37 (1985), 237-242], and I'm sure there are quite a few related results in the literature. For instance, each of the sets $E_r$ is $[1]$-very porous in the sense defined in this conference talk of mine, and among other things I gave a short argument there that each such set has $\sigma$-finite packing $(n-1)$-measure.
The analysis of results related to these in infinite dimensional normed spaces has also generated a fair amount of interest. One possible entry point into this is Ludek Zajicek's 1983 paper Differentiability of the distance function and points of multi-valuedness of the metric projection in Banach space.
(Added 12 Weeks Later) Someone recently gave me a vote on this answer and, in looking at what I wrote, it occurred to me that a much better reference for the comment "one possible entry point into this" that I made in the last paragraph is the following book:
Joram Lindenstrauss, David Preiss, and Jaroslav Tišer, Fréchet Differentiability of Lipschitz Functions and Porous Sets in Banach Spaces, Annals of Mathematics Studies #179, Princeton University Press, 2012, x + 425 pages. Zbl 1241.26001 (a review)
Princeton University Press web page for the book
Amazon.com web page for the book
Disclaimer: The following approach has a flaw in the second step, because the set $A$ could be a nowhere dense set with positive measure like e.g. a fat Cantor set as Davide Giraudo pointed out in his comment. I don't see any way to work around this gap, but leave the answer here in the hope that it might still prove useful in some way.
$A$ is compact. Since $A$ is clearly bounded, it only needs to be shown that it's closed. Let $x_i$ be a convergent sequence in $A$ with limit point $x$. Then $x$ is also in $A$, because for each $x_i$ you find a $y_i$ in $K$ with $d(x_i,y_i)=1$. The $y_i$ have a convergent subsequence with some limit point $y$. Because $d$ is continuous it follows that $d(x,y)=1$. Further, one can show that $d(x,y)\geq1$ for all $y\in K$ (Otherwise, say $d(x,y)<1-\epsilon$ for some $y\in K$, then $d(x_i,y)\leq d(x,x_i)+d(x,y)<\epsilon/2+1-\epsilon<1$ for some $x_i$). Thus $x\in A$.
If $A$ is closed and Lebesgue measurable with $\lambda(A)>0$ it must contain at least one non-empty open set of $\mathbb R^n$, namely the interior of $A$. This in turn would contain a closed ball, call it $\bar B_\delta(x)$ for some $x\in A$ and $\delta>0$.
$K$ cannot contain any point of the open ball $B_{1+\delta}(x)$, because otherwise there would be a point in $\bar B_\delta(x)$ with distance to $K$ smaller than 1. But, then $x$ would have distance larger then $1$ to $K$, i.e. $x\not\in A$, contrary to the assumption that $\bar B_\delta(x)\subset A$.
A possible remedy: If we define $B=\bigcup_{y\in K}\bar B_1(y)$, then $A=\partial B$, since the boundary of $B$ contains exactly the points which have distance $1$ to $K$. Since $B$ is a relatively "smooth" set (in the sense that it cannot have infinitely small and dense holes like a Cantor set), this may carry over to its boundary and salvage step 2. by ruling out "weird sets" like the fat Cantor.