Derivation of the volume element (which uses the metric tensor)? I have often seen $\sqrt{-g}$ in integrals, especially actions, where $g=\mathrm{det}(g_{\mu \nu})$. Does anyone know of a derivation that shows that this is indeed the volume element which must be utilized? I have searched online, but all the literature I have encountered assumes you accept it as the volume element.
 A: I) Pragmatically speaking, the most important property of $\sqrt{-g}$ for model building purposes, is not per se the fact that $\sqrt{-g}d^{4}x$ measures the volume element of a 4-dimensional Parallelepiped with infinitesimal edges $dx^0, \ldots, dx^3$. 
II) A more important property is that $\sqrt{-g}d^{4}x$ transforms as a scalar (i.e. is invariant) under general coordinate transformations 
$$x^{\prime \mu} ~=~ f^{\mu}(x^0 , \ldots, x^3).$$ 
In more detail, the factor $\sqrt{-g}$ transforms as a density under general coordinate transformations, cf. also this Phys.SE answer.
III) Example: The Einstein-Hilbert action 
$$S_{EH}[g]~:=~\int R \sqrt{-g}d^{4}x$$ 
is invariant under change of coordinates, because the scalar curvature $R$ and  volume element $\sqrt{-g}d^{4}x$ are both invariant under general coordinate transformations. Thus the action $S_{EH}[g]$ is a geometric quantity that doesn't depend on the choice or coordinates.
IV) The factor $\sqrt{-g}$ is the so-called canonical density on a Lorentzian manifold. If the Lorentzian manifold is endowed with another density $\rho$, one could in principle use that to build actions and get interesting theories.
V) Finally let us mention that on a symplectic manifold $(M,\omega)$, the canonical density is instead given by the Pfaffian 
$$ {\rm pf}(\omega_{ij}) $$
of the symplectic matrix $\omega_{ij}$.
