Open set whose boundary is not a null set I've just seen a theorm about a bounded set $A \subset \mathbb{R}^n$.
$$\chi_A \text{ is Riemann integrable} \Longleftrightarrow \partial A \text{ is a null set}$$
Then, I wonder if there's any open set whose boundary is not a null set. Can you give me some example for that?
 A: Let $\{r_k,k\in\Bbb N\}$ an enumeration of the rational numbers of $[0,1]$. Consider the map $t\mapsto \bigcup_{n\geq 0}\left(r_n-t\cdot 2^{-n}, r_n+t\cdot 2^{-n}\right)=O_t$. Then $O_t$ is an open set, and $f\colon t\mapsto \lambda(O_t)$ is increasing, and continuous (in fact $1$-Lipschitz continuous). Hence for $\delta\in (0,1)$ you can find $t$ such that $\lambda(O_t)=\delta$, and $O_t$ is a dense subset of $[0,1]$. Its boundary is $[0,1]\setminus O_t$, which has measure $1-\delta\neq 0$.
More generally, you can find an open subset of $\Bbb R$ whose boundary has measure $m$, where $r$ is an arbitrary fixed positive number (you also may have $m=+\infty$ when you take $\{r_k\}$ an enumeration of the rational (of the whole real line) and $O:=\bigcup_{k\geq 1}(r_k-2^{-k},r_k+2^{—k})$.
A: Consider $A$ a fat Cantor set, for example a set of measure $\frac13$. $A$ is closed, so its complement is open, and also $A$ is nowhere dense therefore its complement is dense. That is to say that $U=[0,1]\setminus A$ is open, so for the Lebesgue measure $m$ we have: $$m(\partial U)=m([0,1]\setminus U)=m(A)=\frac13.$$
