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 $O:=\bigcup_{k\geq 1}(r_k-2^{-k},r_k+2^{—k})$.