Volume form is "always positive" The Wikipedia article on volume forms states the following:

In mathematics, a volume form on a differentiable manifold is a nowhere-vanishing top-dimensional form (i.e., a differential form of top degree).

Later it is mentioned that:

Any oriented pseudo-Riemannian (including Riemannian) manifold has a natural volume form. In local coordinates, it can be expressed as $\omega ={\sqrt {|g|}}dx^{1}\wedge \dots \wedge dx^{n}$

Is this the only volume form that we can define on a Riemannian manifold or is it just the most natural (since it is constructed from the Jacobian) For example, could we prove that:
$$\omega = dx^1 \wedge \dots \wedge dx^n$$ may be vanishing ?
 A: If $\omega$ is a volume form on $M$, then $f \omega$ is also a volume form for any smooth positive function $f$ on $M$. So if there is one volume form, then there are infinite many volume forms!
The volume form $\sqrt{g} \ dx_1 \wedge \dots \wedge  dx_n$ is the natural volume form in the context of general relativity because it actually corresponds to the physical volume.
[For example, if $M$ is a spherical surface, so the metric is $ds^2 = d\theta^2 + \sin^2 \theta \ d\phi^2$, then $\sqrt{g} \ d\theta \wedge d\phi = \sin \theta \ d\theta \ d\phi$ is the familiar volume form in spherical coordinates that actually describes the real physical volume.]
Another nice thing about the "natural" volume form is that it is given by the expression $\sqrt{g} \ dx_1 \wedge \dots \wedge dx_n$ in all local coordinate charts. Your suggestion of $dx_1 \wedge \dots dx_n$ will look very different in a different coordinate frame - and then you would have to decide how to define this outside your chosen coordinate patch as well, which may not even be possible.
[Back to the sphere example, picking $d\theta \wedge d\phi$ as the volume form wouldn't actually work because there would be no way of smoothly extending this to the north and south poles (which are not covered by the $(\theta, \phi)$ coordinate system).]
