The riemannian volume form exists on every riemannian manifold

Write $T_m(t,s)=(t+m,(-1{)}^{m}s)$. Saturations of open sets are unions of their open translates, so $q$ is open. On any open rectangular box of $t$-width less than $1$, $q$ is injective and is a homeomorphism onto its open image. The transition maps on overlap components are restrictions of some $T_m$, hence are smooth with invertible diagonal derivative $(1,(-1{)}^m)$.

The quotient is Hausdorff: for distinct orbits choose representatives $z,w$. Only finitely many integers $m$ can give $\mathrm{\parallel }z-T_{m}w\mathrm{\parallel }\le 1$, by the first coordinate. None gives zero. Thus the distances from $z$ to the orbit of $w$ have a positive lower bound $\delta$ (take the minimum of $1$ and those finitely many positive distances). Since all $T_m$ are Euclidean isometries, the saturations of radius-$\delta \mathrm{/}3$ balls around $z,w$ are disjoint. Their quotient images separate the orbits. Images of rational boxes form a countable basis because $q$ is open. The charts in step 1.1 therefore make $M$ a smooth two-dimensional manifold.

Each $T_m$ preserves $d{t}^2+d{s}^2$. These coordinate metrics consequently agree on overlaps and define a smooth positive-definite metric on $M$. The positive density $\mathrm{\mid }dtds\mathrm{\mid }$ also agrees, since the absolute transition determinant is $1$.

If $\omega$ were a nowhere-vanishing ordinary two-form on $M$, write $q^{\ast }\omega =f(t,s)dt\wedge ds$. The local diffeomorphism property implies $f$ is smooth and never zero. Since $q{T}_1=q$, pullback invariance gives $f(t+1,-s)=-f(t,s)$. In particular the nonzero real numbers $f(0,0)$ and $f(1,0)$ have opposite signs. Continuity on the segment $\{(t,0):0\le t\le 1\}$ forces a zero by the intermediate value theorem, a contradiction. A Riemannian volume form would be such a nowhere-vanishing top form. Thus this metric has a global density but no ordinary volume form.