False: density measures require an orientation

Let $T(s,t)=(s+1,-t)$ act on $X=\mathbb{R}\times (-1,1)$ and let $q:X\to M=X\mathrm{/}⟨T⟩$ be the quotient. It is open since $q^{-1}q(V)={\bigcup }_{k\in \mathbb{Z}}{T}^{k}V$ is open for every open $V$. Rectangles of $s$-width less than one have disjoint translates, so $q$ restricts to a homeomorphism from each such rectangle onto an open chart. The quotient is Hausdorff: for two inequivalent points, first choose bounded neighborhoods; only finitely many integer translates can intersect them because their $s$-coordinates are bounded, and shrink the neighborhoods to exclude each of these finitely many intersections. Their quotient images then separate the two orbits. Images of rational rectangles form a countable base. Transition maps are restrictions of powers of $T$, hence smooth. If M had an orientation, its pullback to X would be a smooth sign $\sigma :X\to \{+1,-1\}$ relative to the coordinate frame. This sign is constant because X is connected (any two points are joined by a straight segment). But $DT=\mathrm{diag}(1,-1)$ implies $\sigma (Tp)=-\sigma (p)$, contradicting constancy. Thus M is a smooth nonorientable surface.

The explicit local coefficient one gives $\mathrm{\mid }dsdt\mathrm{\mid }$. Since $DT=\mathrm{diag}(1,-1)$ has absolute determinant one, all transition powers preserve this density. It descends to a positive smooth density on $M$, an explicit instance of the general existence theorem. Its coefficients are finite, so it defines a Radon measure. This nonorientable witness refutes the assertion.