intersection of boundaries have Lebesgue measure 0 Suppose $A$ and $B$ are two sets in $R^n$ such that $\overline{A}\cap B \cup \overline{B}\cap A$ is empty
then $\partial A \cap \partial B$ has $n$-dimensional Lebesgue measure $0$ (where $\partial X=\overline X -\overset{\circ}{X} $ is the boundary of $X$).
I don't know how to prove this fact. In general, the boundary doesn't have to have measure 0, so one has to use the assumptions but how?
Any help appreciated.
 A: For a counterexample with $n=1$, take $A$ and $B$ to be disjoint  open subsets of $[0,1]$ defined as follows. 
Start with $C_0 = [0,1]$.  At each stage $n \ge 1$, $C_{n-1}$ will be a finite union of closed intervals; take two disjoint finite sets of points $E_n$ and $F_n$ in the interior of $C_{n-1}$ such that every point of $C_n$ is within distance $1/n$ of $E_n$ and of $F_n$.  Take $A_n = \bigcup_{x \in E_n} (x-\epsilon, x+\epsilon)$ and $B_n = \bigcup_{x  \in F_n} (x-\epsilon, x+\epsilon)$ where $\epsilon > 0$ is small enough that $A_n$ and $B_n$ are disjoint, their closures are contained in the interior of $C_{n-1}$, and have measure $< 2^{-n}/4$.  Let $C_{n} = C_{n-1} \backslash (A_n \cup B_n)$.
Finally, take $A = \bigcup_n A_n$ and $B = \bigcup_n B_n$.
By construction, $A$ and $B$ are disjoint open sets, and $\partial A \cap \partial B = [0,1] \backslash (A \cup B)$ has measure $> 0$.
A: This is false in general. For instance, you can take a Jordan curve $C$ in ${\mathbb R}^2$ of positive 2-dimensional measure (an "Osgood curve") and let $A, B$ be the  components of ${\mathbb R}^2 -C$.   
