TITLE: Why do I need densities in order to integrate on a non-orientable manifold?
QUESTION [35 upvotes]: Integration on an orientable differentiable n-manifold is defined using a partition of unity and a global nowhere vanishing n-form called volume form. If the manifold is not orientable, no such form exists and the concept of a density is introduced, with which we can integrate both on orientable and non-orientable manifolds.
My question is: On a non-orientable n-manifold, every n-form vanishes somewhere, but shouldn't I be able to chose an n-form with say a countable number of zeros, which would then constitute a set of measure zero and thus allow me to use n-forms (with zeros) for global integration also on non-orientable n-manifolds?

REPLY [4 votes]: In fact, for the purpose of integrating functions on a non-orientable manifold, you don't directly need densities. Every (connected) manifold $M$ has an orientable double cover (connected if and only $M$ is not orientable), $\pi: \tilde M \to M$. Then, upon fixing an orientation and a volume form $\mathcal V$ on $\tilde M$, you can define the integral of a function $f$ on $M$ by
$$\frac{1}{2}\;\int_{\;\tilde M}\left(f\circ\pi\right) \mathcal V\,.$$
Of course, if $M$ is not orientable, the volume form $\mathcal V$ cannot be chosen so as to be invariant under the involution that swaps the two points in each fiber of $\pi$, otherwise it would be the pullback of a volume form on $M$. 
This has the disadvantage of not being very canonical in general. But, if $M$ is equipped with a Riemannian metric $g$, then there is a canonical volume form on $\tilde M$, namely that induced by $\pi^* g$. So, for instance, the volume of a Riemannian Moebius strip or of a Riemannian Klein bottle is well defined.