Volume of a closed and orientable manifold is positive Let $g_{ij} dx^i \otimes dx^j$  be a Riemannian metric on an orientable and closed manifold $M$ (dimension $n$). Let $dvolg=\sqrt{det(g_{ij})} dx^1\wedge \cdots \wedge dx^n$. Show $\int_M dvolg>0$. I wonder where does the "closedness" of $M$ come in the proof. Or do we need the manifold to be "closed" at all?

"Closed" in this context means "compact without boundary." If the manifold is not compact, there is no guarantee that the volume is finite. But if the volume is finite, then the conclusion still holds. 
I suspect that whoever wrote the problem was working in a context where integrals of differential forms had been defined either on compact manifolds with or without boundary, or for compactly supported differential forms. If $M$ is noncompact, then neither of these conditions holds, and you have to interpret the integral as an improper integral (for example, defined as the limit over compact subdomains, or something like that).
