Let $A$ be a semi-algebraic subset of $\mathbb{R}^n$ that is semi-algebraically homeomorphic to $(0,1)^k$ with $k<n$.
I would require a result stating that $\mathcal{L}^n(A)=0$ where $\mathcal{L}^n$ is the Lebesgue measure in $\mathbb{R}^n$. It seems intuitive that this should be true, since $A$ is a k-dimensional manifold in $\mathbb{R}^n$, i.e., a set of lower dimension.
One way to prove this is that lower-dimensional manifolds have lower Hausdorff dimension, and a set of lower Hausdorff dimension has Lebesgue measure zero. A semialgebraic set is a finite union of smooth manifolds of various dimensions - see Bochnak-Coste-Roy Theorem 3.3.11 and 3.3.14. You can find the definition and basic properties of Hausdorff dimension in Folland's "Real Analysis: Modern Techniques and Their Applications" for example.