Domain whose boundry has non zero volume.

Let $(0,1) \cap \mathbb Q = \{ q_1,q_2, \dots\}$. We define $A = (0,1) \cap\bigcup_k (q_k-4^{-k}, q_k+4^{-k})$. By construction we have $\bar A = [0,1]$ but an easy estimate shows that $\lambda( A) < 1$ where $\lambda$ is the Lebesgue measure.

If you want a connected example, you can take $\left(A \times (0,1)\right) \cup \left((0,1) \times (0, 1/2)\right) \subset \mathbb R^2$. This one is actually contractible.

Let $ν \colon ℕ → [0,1]∩ℚ$ be a enumeration of the rationals there. Then the set $M := \bigcup_{k∈ℕ} B_{r(k)}\big(ν(k)\big)$ with $r(k) = \tfrac{1}{3}·\big(\tfrac{1}{2}\big)^k$ has measure less than $\sum_{k=1}^∞ 2r(k) = \tfrac{1}{3}·1 < 1$ (as $B_{r(k)}$ has measure $2r(k)$), so it’s not all $[0,1]$. But $M$ lies dense in $[0,1]$ since it contains all the rationals. Therefore, the boundary of $M$ has to be $[0,1]\setminus{M}$, having nonzero measure, viz measure $\tfrac{2}{3}$. Also $M$ is open because it’s the union of open balls.