Let $\Omega \subset \mathbb R^n$ be open, connected, and bounded. We assume that the boundary $\partial\Omega$ is locally the graph of a continuous (or differentiable) function, i.e., for all $x_0 \in \partial\Omega$ there are a rotation $Q\in \mathbb R^{n,n}$,an open set $U$ with $x_0\in U$, and a function $h: (QU)'\to \mathbb R$ such that$$\partial \Omega \cap U = Q^{-1} graph(h),$$where $(QU)'= \{ (x_1\dots x_{n-1}): \ x\in QU\}$.
Does this imply that locally the set $\Omega$ is on one side of the boundary?
I.e.,after rotation the set is locally above (or below) the graph of a continuous function. In the notation above this means$$Q(\Omega \cap U) \subset \{y\in \mathbb R^n: \ y_n > h(y_1 \dots y_{n-1}) \}\quad\text{ or } \quadQ(\Omega \cap U) \subset \{y\in \mathbb R^d: \ y_n < h(y_1 \dots y_{n-1}) \}.$$
I suspect the answer to be 'yes', but have difficulty finding a proof.For $n=2$ this should follow from the Jordan curve theorem.
The context is that some authors define bounded $C^{m,\alpha}$ domains in $\mathbb R^n$ asopen, connected, and bounded sets that are locally above the graph of a $C^{m,\alpha}$function. So the question is, whether/why 'domain is locally above the graph' can be replaced by 'boundary is locally equal to the graph'.
Edit: In Adams' book on Sobolev spaces it is mentioned that the answer to this question is yes, without any hints or references.
