Let $g: A\rightarrow B$ be a bijective function which is locally lipschitz, where $g^{-1}$ is continuous and $A,B\subset R^{n}$ open subsets. Prove that if $D\subset A$ rectifiable, then $g(D)$ is rectifiable.
I am thinking about this problem, and don't really see why will we be needing for the inverse to be a continuous function as well. I am not very sure on my reasoning when claiming $g(\partial D)=\partial g(D)$.
I know that $\partial D$ has measure $0$ (since $D$ is rectifiable) and $g$ maps sets of measure $0$ to sets of measure $0$, but other than that, I don't really know what else may I need to use.
I appreciate any help given.