If we take to be a manifold equipped with an orientation given by an orientation form . Then is nowhere zero, and for any positively oriented basis of at any point .
The integral of over all of must surely be even greater than the integral over the image of , since we can cover by orientation-preserving singular -cubes, and none of them can ever contribute a negative to the integral.
If we further suppose that is compact, we can cover by finitely many such singular cubes, and the integral on each is well-defined. Using a partition of unity as usual this shows us that the integral over all of exists and, further, must be strictly positive. In particular it’s not zero.
But now suppose that also has an empty boundary. Since is a top form, we know that — it’s closed in the de Rham cohomology. But we know that it cannot also be exact, for if for some -form then Stokes’ theorem would tell us that
since is empty.
And so if is a compact, oriented -manifold without boundary, then there must be some -forms which do not arise from taking the exterior derivatives of -forms. If is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on which are not the divergence of any vector field on .