Existence and Uniqueness of a Volume Element on an Oriented, Pseudo Inner Product Space? The definition of orientation on a vector space I'm using is as an equivalence class of ordered bases, where two bases are related if and only if the determinant of the change of basis is positive. Now, one has the following theorem

Theorem.
Let $(V, g, \mathcal{Or})$ be an oriented $n$-dimensional inner product space over $\Bbb{R}$. Then, there is a unique volume form $\omega$ on $V$, such that for every positively-oriented, ordered, orthonormal basis $\{e_1, \dots, e_n\}$ of $V$, we have
\begin{align}
\omega(e_1, \dots, e_n) &= 1.
\end{align}
In fact for any positively-oriented basis $\{f_1, \dots, f_n\}$, if we let $\{\varphi^1, \dots, \varphi^n\}$ be the dual basis for $V^*$, then
\begin{align}
\omega &= \sqrt{\det \left[ g(f_i, f_j)\right]}\,\,\varphi^1 \wedge \dots \wedge \varphi^n
\end{align}

The sketch of the proof is: pick some positively-oriented, ordered, orthonormal basis $\{e_1, \dots, e_n\}$, and  denote $\{\epsilon^1, \dots, \epsilon^n\}$ to be the dual basis. Next, let $\{f_1, \dots, f_n\}$ be any positvely-oriented basis, with $\{\varphi^1, \dots, \varphi^n\}$ its dual basis, and let $A$ be the change of basis matrix $(f_j = \sum_{i}A_{ij}e_i)$. Then,
\begin{align}
\epsilon^1 \wedge \dots \wedge \epsilon^n &= \det(A) \cdot \varphi^1 \wedge \dots \wedge \varphi^n.
\end{align}
In this case, it's relatively straight-forward to show that $A^tA = [g(f_i,f_j)]$. From this, it is easily deduced that we can define $\omega = \epsilon^1 \wedge \dots \wedge \epsilon^n$, this definition is basis-independent, and that $\omega$ is unique.

My question is whether there is an analogous statement in the pseudo-Riemannian case; i.e when we require $g:V \times V \to \Bbb{R}$ to only be bilinear, symmetric, and non-degenerate (as opposed to positive-definite). I tried to adapt the same proof to this situation. However, here's the trouble I faced: let $\{e_1, \dots, e_n\}$ be positively-oriented and orthonormal with respect to $g$, so that $g(e_i,e_j) = \pm \delta_{ij}$. Next, let $\{f_1, \dots, f_n\}$ be any positively-oriented basis. After some calculations, I found that
\begin{align}
(A^tA)_{ij} &= \sum_{k, \beta, \alpha}g(e_k, e_{\alpha}) g(e_{\alpha}, f_i) g(e_k, e_{\beta}) g(e_{\beta}, f_j).
\end{align}
In the Riemannian case, the $g(e_{(\cdot)}, e_{(\cdot)})$ would just become a Kronecker delta with a $+$ sign, and then (making use of the formula for expressing vectors in terms of an orthonormal basis) it immediately reduces to $g(f_i,f_j)$. However, in general, the minus signs cause trouble, and I'm not sure how to further simplify this.
So, in the general case, is there even a unique volume element determined only from $g$ and the orientation? If so, how does one go about proving it? What other modifications (if any) to the theorem are necessary?
 A: Having had some more time to think about this, I realized that I've unnecessarily confused myself, and that I'd been trying to prove $A^tA = [g(f_i,f_j)]$ in the pseudo-Riemannian case, which is actually incorrect. The fix however, is very simple. For fun, I'll include the modified statement, along with two proofs.

Theorem.
Let $(V, g, \mathcal{Or})$ be an $n$-dimensional, oriented, pseudo inner-product space over $\Bbb{R}$. Then, there is a unique volume form $\omega$ on $V$, such that for every positively-oriented, ordered, $g$-orthonormal basis $\{e_1, \dots, e_n\}$ of $V$, we have
  \begin{align}
\omega(e_1, \dots, e_n) &= 1.
\end{align}

The first proof makes use of the fact that a (pseudo) inner product, $g$, on $V$ induces for each $q \in \Bbb{N}$, a pseudo-inner-product $g_q$ on the subspace $\mathcal{A}^q(V)$ of alternating $q$-tensors over $V$.

Proof $1$.
Let $\{e_1, \dots, e_n\}$ be a positively oriented, $g$-orthonormal basis of $V$, and $\{\epsilon^1, \dots, \epsilon^n\}$ the dual basis. Suppose that the number of $-1$ in the matrix $[g(e_i,e_j)]$ is $\#$. Then,
  \begin{align}
g_n\left( \epsilon^1 \wedge \dots \wedge \epsilon^n, \epsilon^1 \wedge \dots \wedge \epsilon^n\right) &= \det\bigg( g_1\left(\epsilon^i, \epsilon^j\right)\bigg) = \det\bigg( g\left(e_i, e_j\right)\bigg) = (-1)^{\#}
\end{align}
  If $\mu$ is any other volume-form, which satisfies $g_n(\mu, \mu) = (-1)^{\#}$, then since $\mathcal{A}^n(V)$ is one-dimensional, there is a $c\in \Bbb{R}\setminus\{0\}$ such that $\mu = c \cdot \epsilon^1 \wedge \dots \wedge \epsilon^n$. Then
  \begin{align}
(-1)^{\#} &= g_n(\mu, \mu) = c^2 g_n(\epsilon^1 \wedge \dots \wedge \epsilon^n,\epsilon^1 \wedge \dots \wedge \epsilon^n) = c^2 (-1)^{\#}.
\end{align}
  Hence, $c^2 = 1$, so $c = \pm 1$, which means $\mu = \pm \epsilon^1 \wedge \dots \wedge \epsilon^n$.
In particular, if $\{f_1, \dots f_n\}$ is a positively oriented, $g$-orthonormal basis of $V$, and $\{\varphi^1, \dots, \varphi^n\}$ its dual basis, then
  \begin{align}
\varphi^1 \wedge\dots \wedge\varphi^n &= \pm\epsilon^1 \wedge \dots \wedge \epsilon^n;
\end{align}
  but in fact since both bases are positively-oriented, it is easily shown that the volume forms induced by them must be proportional by a positive constant. Hence,
  \begin{align}
\varphi^1 \wedge \dots \wedge\varphi^n = \epsilon^1 \wedge \dots \wedge \epsilon^n. \tag{$*$}
\end{align}
Hence, we can define $\omega := \epsilon^1 \wedge \dots \wedge \epsilon^n$, and the above argument shows that this definition is independent of the choice of basis. Lastly, it is clear that
  \begin{align}
\omega(e_1, \dots, e_n) = (\epsilon^1 \wedge \dots \wedge \epsilon^n)(e_1, \dots, e_n) = 1,
\end{align}
  which completes the proof of both existence and uniqueness of $\omega$.

The second proof establishes more in the process:

Proof 2.
Let $E = \{e_1, \dots, e_n\}$, $F = \{f_1, \dots, f_n\}$ be positively-oriented, ordered, bases for $V$, and we assume that $E$ is $g$-orthonormal. Also, let $E^* = \{\epsilon^1, \dots, \epsilon^n\}$ and $F^* = \{\varphi^1, \dots, \varphi^n\}$ be the respective dual bases for $V^*$. Lastly, let $T:V \to V$ be the isomorphism such that $f_j = T(e_j)$, for all $j \in \{1, \dots, n\}$. Then, we have the following relations between the various matrix-representations:
  \begin{align}
[g]_F &:= [g(f_i, f_j)] \\
&= [g(T(e_i), T(e_j))] \\
&= \left([T]_E \right)^t \cdot [g]_E \cdot [T]_E \tag{basic linear algebra}
\end{align}
  Hence, by taking the determinant of both sides, and using elementary properties of the determinant, we find that
  \begin{align}
\det [g]_F &= (\det T)^2 \cdot \det [g]_E
\end{align}
  Since $E$ is a $g$-orthonormal basis, we have that $\det [g]_E = \pm 1$. It follows that
  \begin{align}
|\det(T)| &= \sqrt{\left|\det [g]_F \right|}
\end{align}
  Also, one of the definitions of $\det T$ is that it is the unique constant such that
  \begin{align}
\epsilon^1 \wedge \dots \wedge \epsilon^n &= (\det T) \cdot \varphi^1 \wedge \dots \wedge \varphi^n
\end{align}
  Since we assumed $E$ and $F$ are positively-oriented, it follows that the volume forms they induce must be proportional by a positive constant. Hence, it follows that
  \begin{align}
\epsilon^1 \wedge \dots \wedge \epsilon^n &= \sqrt{\left|\det [g]_F \right|} \cdot \varphi^1 \wedge \dots \wedge \varphi^n.
\end{align}
  From this formula, it follows that if we further assume $F$ is $g$-orthonormal, then the determinant on the RHS is $1$, so we obtain exactly the same relationship as in $(*)$ of proof 1. Therefore, we can define
  \begin{align}
\omega &:= \epsilon^1 \wedge \dots \wedge \epsilon^n,
\end{align}
  and this definition is basis-independent. The fact that $\dim \mathcal{A}^n(V) = 1$ shows that $\omega$ is unique.


I found proof $1$ more conceptually clear: the pseudo-inner product $g$ induces one on every space of alternating tensors, and by using the fact $\dim \mathcal{A}^n(V) = 1$, it follows that there are exactly two volume forms which are "normalized", to $(-1)^{\#}$, and these volume forms only differ by a sign. Then, the orientation $\mathcal{Or}$ helps us to pick one of these two volume forms.
The second proof is more "down-to-earth", but necessarily more computationally tedious, because it doesn't require us to define the inner product on each space of alternating tensors (that by itself requires a little bit of work to show everything is well-defined etc). It only requires basic linear algebra, and knowing the relationship between the determinant and wedge products to prove that $\omega$ is well-defined and unique. 
One advantage though is that for computational purposes, it gives us an explicit formula for the volume element, in terms of any positively-oriented basis. Such a formula could potentially be useful in certain concrete situations, for example, when performing integration on manifolds, and choosing a chart which is particularly suited to the specific problem at hand. In such a case, orthogonalizing, or normalizing the tangent vectors in the might be more tedious than simply calculating a determinant (for example, the tangent vectors in the spherical-coordinate chart are not normalized).
