Let $A\subset\mathbb{R}^n$ open $g:A\to\mathbb{R}$ of class $C^{1}$ and $g'(x)\not=0$ in each $x\in A$ then I want to compute $dV$ in the differentiable manifold $ M = g^{-1}(0)$.
The thing is that we have the result that says: Let $M\subset\mathbb{R}^n$ be a differentiable manifold of dimension $n-1$ with orientation $\mu$ so $$dV(x)((v_1)_p, \ldots, (v_{n-1})_p) = \det\begin{pmatrix}\nu_x & v_1 & \cdots & v_{n-1}\end{pmatrix}.$$
where $\nu_p\in(M_p)^\perp\subset\mathbb{R}^n$ is the unitary normal vector, this is, $\nu_p$ is the vector such that $[\nu_p, (v_1)_p, \ldots, (v_{n-1})_p]$ is the right-handed orientation in $\mathbb{R}^n$ for any basis $\{(v_1)_p, \ldots, (v_{n-1})_p\}$ of the tangent space $M_p$ with orientation $[(v_1)_p, \ldots, (v_{n-1})_p] = \mu_p$
Then it turns out that the problem is only to prove that $M$ is orientable, but how can I prove this with the hypotheses I got? Or How can I solve this problem?
NOTE: The notation and definitions are from the book Calculus in manifolds by Spivak the problem isn't taken from that book.
THANKS FOR YOUR HELP :)
ATTEMPT:
First we give to our differentiable manifold the orientation that has $\mathbb{R}^n$ then we pick the vector $\nu = \frac{\nabla g}{|\nabla g|}$, since this is well define because $\nabla g$ doesn't vanish this give us a unitary normal vector, this new $\nu$ has a direction, and therefore if it is wrong oriented we take $- \nu$ BUT I DONT UNDERSTAND WHY IF $\mathbb{R}^n$ is orientable then so is $M$ Thnks :)
Since $\nabla g$ is non-vanishing in $A$ by hypothesis, the normalized gradient $\nu = \nabla g/\|\nabla g\|$ is a continuous unit normal field on the hypersurface $M$. Since the ambient space is orientable, $M$ itself is orientable.