Theorem 1 (Liouville's theorem)   If $D\subseteq \mathbb{R}^n$ is a bounded measurable domain. Then $$\dot{V}(t)= \int_{\Phi_t(D)} \operatorname{div}\, f(x) dx$$

Proof. Let $V(t)$ be defined as above then \begin{eqnarray*} V(t_0+h) & = & \int_{\Phi_{t_0+h}(D)}dy\\ & = & \int_{\Phi_h(\Phi_{t_0}(D))}dy\\ & = & \int_{\Phi_{t_0}(D)} \operatorname{det}\left(\frac{\partial\Phi_h}{\partial x}(x)\right) dx. \end{eqnarray*} We claim that, for $x\in \Phi_{t_0}(D)$ , $$\frac{\partial\Phi_t}{\partial x}(x) = I + t\frac{\partial f}{\partial x}(x) + o(t)$$ as $t\to 0$ .

In fact, $$\Phi_t(x) = x + \int_{0}^t f(\Phi_s(x))ds,$$ and by the Leibniz integral rule $$\frac{\partial \Phi_t}{\partial x}(x) = I + \int_{0}^t \frac{\partial}{\partial x}f(\Phi_s(x)) ds,$$ so that $$\frac{\partial}{\partial t} \frac{\partial \Phi_t}{\partial x}(x) = \frac{\partial}{\partial x}f(\Phi_t(x))$$ and evaluating at $t=0$ we get $${\frac{\partial}{\partial t} \frac{\partial \Phi_t}{\partial x}(x)}\Big|_{t=0} = \frac{\partial}{\partial x}f(\Phi_0(x)) = \frac{\partial f}{\partial x} (x).$$ Our claim follows from this and from the definition of derivative.

Hence \begin{eqnarray*} \operatorname{det}\left(\frac{\partial\Phi_t}{\partial x}(x)\right) & = & \operatorname{det}\left(I + t\frac{\partial f}{\partial x}(x)\right) + o(t)\\ & = & \prod_{i=1}^n(1 + \frac{\partial f_i}{\partial x_i}(x)) + o(t)\\ & = & 1+t\sum_{i=1}^n\frac{\partial f_i}{\partial x_i}(x) +o(t)\\ & = & 1 + t\operatorname{div}\, f(x) + o(t) \end{eqnarray*}as $t\to0$ . It follows that $$V(t_0+h) = \int_{\Phi_{t_0}(D)} 1 + h\operatorname{div}\, f(x) + o(h) dx$$ and \begin{eqnarray*} \dot{V}(t_0) & = & \lim_{h\to 0}\frac{V(t_0+h)-V(t_0)}{h}\\ & = & \frac{\int_{\Phi_{t_0}(D)} 1 + h\operatorname{div}\, f(x) + o(h) dx -V(t_0)}{h}\\ & = & \frac{V(t_0) + h\int_{\Phi_{t_0}(D)}\operatorname{div}\, f(x)dx + o(h) -V(t_0)}{h}\\ & = & \int_{\Phi_{t_0}(D)}\operatorname{div}\, f(x)dx + \lim_{h\to 0}\frac{o(h)}{h}\\ & = & \int_{\Phi_{t_0}(D)}\operatorname{div}\, f(x)dx. \end{eqnarray*}

Corollary 1   The flow of an Hamiltonian system preserves volume.

Proof. It follows directly since the vector field of an Hamiltonian system has divergence equal to zero. Hence $\dot{V}=0$ implies that the volume is constant.