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 sphere 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?