A volume element on an $n$-dimensional semi-Riemannian manifold $M$ is a smooth $n$-form $w$ such that $w(e_1,\cdots, e_n) = \pm1$ for every frame on $M$.
How do I prove
A semi-Riemannian manifold $M$ has a (global) volume element if and only if $M$ is orientable.
If $M$ has a volume form $w$, then we can define an orientation by stating that a frame $(e_1,\ldots,e_n)$ is positively (resp. negatively) oriented if $w(e_1,\ldots,e_n)$ is positive (resp. negative).
For the other direction, suppose $M$ is a (closed) orientable $n$-manifold. We can construct a volume form in two different ways:
Approach 1: Let $\mathscr{U}_{\alpha \in \mathcal{A}}$ be a cover of $M$ such that each open set $U_\alpha$ is the domain of an orientation-preserving chart $\phi_\alpha : U_\alpha \to \mathbb{R}^n$. Letting $dy_1 \wedge \cdots \wedge dy_n$ be the volume element on $\mathbb{R}^n$, the pullback $(\phi_\alpha)^*( dy_1\wedge \cdots \wedge dy_n)$ satisfies $$(\phi_\alpha)^*( dy_1\wedge \cdots \wedge dy_n)(e_1,\ldots,e_n)>0$$ for all oriented frames. Now take a partition of unity $\rho_\alpha: M \to [0,1]$ subordinate to the cover $\mathscr{U}$ and define a smooth $n$-form $\tilde w$ on $M$ by $$\tilde w_x = \sum_{\alpha \in \mathcal{A}} \rho_\alpha(x) \cdot (\phi_\alpha)^*_x (dy_1\wedge \cdots \wedge dy_n).$$ It follows that $\tilde w$ satisfies $\tilde w (e_1,\ldots,e_n)>0$ for all oriented frames. Note that we haven't made use of the semi-Riemannian structure yet. Now define $f: M \to \mathbb{R}$ by $f(x)=1/\big(\tilde w_x(\hat e_1,\ldots,\hat e_n)\big)$, where $\hat e_1,\ldots,\hat e_n$ is an (orthogonal) oriented frame for $T_x M$. This is independent of the choice of frame and is smooth. The $n$-form $w$ defined by $w_x=f(x) \cdot \tilde w_x$ now satisfies $w(e_1,\ldots,e_n)= \pm1$ for oriented (orthogonal) frames $(e_1,\ldots,e_n)$.
Approach 2: Let $\phi: U \to \mathbb{R}^n$ be an orientation-preserving chart providing local coordinates $x_1,\ldots,x_n$. Then we can write the metric in local coordinates as $\sum_{i,j} G_{ij} \,dx_i dx_j$, where $G_{ij}$ is the smooth function $$G_{ij}(p)=\left\langle \frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j} \right \rangle_p.$$ Viewing $G$ as an $n\times n$ matrix (that is symmetric and nondegenerate) you can define $$w = \big|\det G \big|^{1/2}\, dx_1 \wedge \cdots \wedge dx_n. $$ We can define such an $n$-form at any point of $M$, and we claim that it is independent of the choice of local coordinates. [See proof below.] To see that $w$ is a volume form, consider an orthogonal frame $(e_1,\ldots,e_n)$. At each point we can find a linear transformation $A$ such that $(e_1,\ldots,e_n)=A(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n})$. Then \begin{align*} w(e_1,\ldots,e_n)&=w \left(A\left(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n}\right)\right)\\ &= (\det A) w\left(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n}\right)\\ &= (\det A) \big| \det G \big |^{1/2} \, dx_1 \wedge \cdots \wedge dx_n \left(\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n}\right)\\ &= (\det A) \big| \det G \big|^{1/2} \\ &= \pm \big|\det A \big| \big| \det G \big|^{1/2} \\ &= \pm \big| \det (A^T G A) \big|^{1/2}.\end{align*} Now we are abusing notation and regarding $A$ as an $n \times n$ matrix with respect to the basis $\frac{\partial}{\partial x_1},\ldots,\frac{\partial}{\partial x_n}.$ Observe that $$(A^T G A)_{ij}= \langle e_i, e_j \rangle.$$ In particular, the diagonal entries are $\pm1$ and the off-diagonal entries are zero. Thus $$w(e_1,\ldots,e_n)=\pm \big| \det( A^T G A)\big |^{1/2}=\pm |\pm 1|^{1/2}= \pm 1.$$
Proof that $w$ is well-defined. To that end, suppose $\psi: V \to \mathbb{R}^n$ is another orientation-preserving chart providing local coordinates $x_1',\ldots,x_n'$. We can similarly define the functions $G_{ij}'$ and $n$-form $$w'= \big| \det G'\big|^{1/2} \, dx_1' \wedge \cdots dx_n'.$$ If $U \cap V$ is nonempty, a direct computation shows that $w$ and $w'$ agree on the overlap: In general, top-dimensional forms in overlapping charts transform according to the rule $$dx_1' \wedge \cdots \wedge dx_n' = (\det J )dx_1 \wedge \cdots \wedge dx_n,$$ where $J$ is the Jacobian matrix associated to the coordinate change, i.e. $J_{ij}=\partial x_i'/\partial x_j.$ Therefore \begin{align*} w'&=\big| \det G'\big|^{1/2} \, dx_1' \wedge \cdots \wedge dx_n' \\ &= \big| \det G'\big|^{1/2} \left( \det J\right) dx_1 \wedge \cdots \wedge dx_n. \end{align*} It now suffices to show that $\big| \det G'\big|^{1/2}( \det J)=\big| \det G \big|^{1/2}$. Observe that \begin{align*} G_{ij}&= \left \langle \frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}\right\rangle \\ &= \left \langle \sum_{k=1}^n \frac{\partial x_k'}{\partial x_i} \frac{\partial }{\partial x_k'},\sum_{\ell=1}^n \frac{\partial x_\ell'}{\partial x_j} \frac{\partial }{\partial x_\ell'} \right \rangle\\ &= \sum_{k=1}^n \sum_{\ell=1}^n\frac{\partial x_k'}{\partial x_i} \frac{\partial x_\ell'}{\partial x_j} \left \langle \frac{\partial }{\partial x_k'}, \frac{\partial }{\partial x_\ell'} \right \rangle\\ &= \sum_{k=1}^n \sum_{\ell=1}^n\frac{\partial x_k'}{\partial x_i} \frac{\partial x_\ell'}{\partial x_j} G_{k\ell}'. \end{align*} We can use this to show that $G= J^T G' J$, from which it follows that $$ \big| \det G \big|^{1/2} = \big| \det(J^T G' J) \big|^{1/2}= \big| \det G' \big|^{1/2}\big| (\det J)^2 \big|^{1/2}= \big| \det G' \big|^{1/2} \det J,$$ where we use the fact that $\det J >0$ because both charts are orientation-preserving. $\square$