A density integral on the Mobius band

The quotient map is open because the inverse image of an image-open set is the union of its translates. A rectangle with s-width less than one is disjoint from all its nontrivial translates, so maps homeomorphically onto its image; at t=1 or t=-1 use a half-rectangle. For two inequivalent points only finitely many translates of a bounded neighborhood of one can approach a bounded neighborhood of the other; shrink to separate these finitely many translates. Their saturated neighborhoods are disjoint, proving Hausdorffness. Images of rational rectangles form a countable base. The transition maps are restrictions of $T^k$, hence smooth, so these charts define a smooth manifold with boundary. It is compact as the image of $[0,1]\times [-1,1]$.

The seam transition has determinant $-1$, with absolute value one, so the local densities glue and are positive. Equivalently $T^{\ast }\mathrm{\mid }dsdt\mathrm{\mid }=\mathrm{\mid }dsdt\mathrm{\mid }$ by the pullback formula; the same holds for all integer powers.

Use the single finite parametrization from $D=(0,1)\times (-1,1)$ to the quotient. It is a diffeomorphism onto the open complement of seam and boundary, extends continuously from the closed rectangle, and is smooth up to each edge in target coordinates. Its image closure is B and its pulled-back density coefficient is one. The density parametrization formula therefore gives ${\int }_B\delta ={\int }_0^1{\int }_{-1}^{1}1dtds=2$. Seam and boundary are covered by that formula’s null-boundary control.