False: densities and top forms coincide on nonorientable manifolds

Density bundle and smooth density fields: For a smooth manifold $M^n$, with boundary allowed, the density bundle is $DM={\coprod }_{p\in M}D(T_{p}M)$. In coordinates $x$, let $\mathrm{\mid }dx\mathrm{\mid }=\mathrm{\mid }d{x}^1\cdots d{x}^n\mathrm{\mid }$ be the density taking value one on the coordinate frame. On overlaps, $\mathrm{\mid }dy\mathrm{\mid }=\mathrm{\mid }\det D_{x}y\mathrm{\mid }\mathrm{\mid }dx\mathrm{\mid }\mathrm{.}$ A smooth density is a section with smooth real coefficient in these frames. Its support is the closure of its nonzero locus. The absolute determinants are positive smooth transition functions and satisfy the cocycle identities by the chain rule. A countable atlas and thm-vector-bundle-construction-from-a-smooth-cocycle therefore give a smooth line bundle. For boundary charts the same gluing proof uses half-space product charts; smoothness of transitions follows from their local extensions, and Hausdorffness and second countability follow as for the supplied cocycle construction. The fibers are lines by prop-one-densities-form-a-one-dimensional-vector-space. When $n=0$ the empty frame trivializes $DM=M\times \mathbb{R}$.

Existence of positive smooth densities: Assuming ${\mathrm{A}\mathrm{C}}_{\omega }$, every smooth manifold, with or without boundary, admits a smooth positive density.

Orientability is equivalent to a nowhere-vanishing top form: Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. A smooth manifold is orientable if and only if it has a nowhere-vanishing smooth top-degree form.