Flat Mobius strip density measure

The quoted strip construction gives a Hausdorff second-countable smooth nonorientable surface. Its chart transitions are powers of $T$, with derivative $\mathrm{diag}(1,(-1{)}^k)$. Thus $(ds{)}^2+((-1{)}^{k}dt{)}^2=d{s}^2+d{t}^2$ and $\mathrm{\mid }\det D{T}^k\mathrm{\mid }=1$; both the quadratic form and the unit density agree on overlaps and descend smoothly.

In such a chart an orthonormal frame has column matrix $A$ satisfying $A^{\mathsf{T}}A=I$. Taking determinants gives $(\det A{)}^2=1$, so the density on that frame is $\mathrm{\mid }\det A\mathrm{\mid }=1$. Conversely a density with this normalization must have coefficient one on the coordinate frame, which is orthonormal. This verifies the claimed normalization directly without a general Riemannian volume theorem. The positive finite coefficient also gives a Radon measure.

Let $q$ be the quotient map. The set $W=q((0,1)\times (-1,1))$ is a chart: no two points in this open strip are related by a nonzero power of $T$. Its mass is $1\cdot 2=2$. Its complement is the seam $S=q(\{0\}\times (-1,1))$, a Borel set because $W$ is open. In the seam chart $q((-1\mathrm{/}4,1\mathrm{/}4)\times (-1,1))$, $S$ is the coordinate line $s=0$; it has measure zero (or cover it by rectangles of arbitrarily small width). Hence ${\mu }_r(M)={\mu }_r(W)+{\mu }_r(S)=2$. The omitted edges $t=\pm 1$ are not points of $M$.