Let $M$ be a complete n dimensional Riemannian manifold. $vol$ denotes the n dimensional Hausdorff measure. Let
$$
SM=\{(x,v)|x\in M, v\in T_xM, \|v\|=1\}
$$
be the unit tangent bundle of $M$. Then $SM$ will be equipped with the Liouville measure $\nu$. Given a subset $A=(U,A_x)\subset SM$, where $U\subset M$ is a subset of $M$, $A_x$ is a subset of the unit ball of the tangent space at $x\in U$, $\nu$ is defined by
$$
v(A)=\int_U \int_{A_x} dS^{n-1} dvol(x)
$$
where $dS^{n-1}$ is the usual Lebesgue measure on the unit sphere.


Then $\nu$ is invariant under the geodesic flow on $SM$. By the comments below, I know what it means: Let $y=(x,v)\in A$,
set $\gamma_y(s)=\exp_x(sv)$, then the geodesic flow is defined by
$$
\Phi_t(y)=(\gamma_y(t),\dot{\gamma}_y(t))
$$
And $\Phi_t(A)=\{\Phi_t(y)|y \in A \}$. We have $\nu(\phi_t(A))=\nu(A)$. 


Can you give a direct proof without introducing cotangent bundle, 1-form, 2-form?