I recently read Khinchin's derivation of Liouville's theorem. I was able to follow the math for the most part, however I was hoping for an intuitive understanding about why the form of the measure on an energy surface in phase space is different than the form of the measure on the whole phase space
If we have an $N$-dimensional phase space then the measure on that phase space is simply $dV$. I was able to follow the proof about why that measure is conserved under the natural motion. However, if we restrict our analysis to an energy surface to that phase space then the measure on that surface becomes $\frac{d \Sigma}{| \nabla H |}$, and not just $d \Sigma,$ the area element on that energy surface. From what I understand, the stated reason for this is that $d \Sigma \cdot dn = \frac{d \Sigma}{| \nabla H |}$ (where $dn$ is the normal to the energy surface) is just a volume element in the phase space, and we can then appeal to the fact that we already know that a differential volume element in phase space is conserved. However, if we view the energy surface itself as an $N-1$ dimensional phase space then your old $d \Sigma$ essentially becomes a volume element in this new $N-1$ dimensional space. Why, then, does Liouville's theorem as derived for the higher dimensional phase space not also apply to the $N-1$ dimensional space, thus making $d \Sigma$ the correct measure?
Edit to clarify what my quantities mean: $dV$ is a differential volume element in the phase space. $d \Sigma$ is a differential area element on an energy surface in the phase space. $H$ is my Hamiltonian, and for my $N$ dimensional phase space
$$| \nabla H | = \sqrt{ \sum_{i=1}^{N/2} \left( \frac{\partial H}{\partial q_i} \right)^2 + \left( \frac{\partial H}{\partial p_i} \right)^2}.$$
$dn$ is a differential normal vector to the energy surface.