Time derivative of an invariant probability measure

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Consider a dynamical system defined through a vector field $F$ in $M \subset \mathbb{R^n}$ that generates a flow $\Phi^t$ of the form $$\bf{\Phi^t X_0 = X} \ , $$ being $X_0 \in \mathbb{R}^n$ the initial condition.

Let's say that an invariant probability measure $\mu$ is defined in $M$. So it satisfies: $$\mu(B) = \mu(\mathbf{\Phi^{-t}}(B)) ,$$ being $B$ any set in the $\sigma$-algebra of $M$. Considering that $\mu$ is a probability measure it should satisfy the principle of conservation of probability, which I am going to write as $$\mu(B) = \mu(\mathbf{\Phi^{-t}}(B)) = c, $$ with $0 < c \leq 1$. My question is, can we rewrite this last equation using a total derivative with respect to $t$? I am thinking in something as $$\frac{d}{dt}\mu = 0 \ ?$$ Is this correct? I am not an expert with measures, so any help would be useful.

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I think you are looking for Liouville equations. See here, for example http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28Hamiltonian%29

Basically if you want to understand what evolution $\phi(X,t)$ of a certain function $\phi_0 = \phi(X,0)$ under the action of the system you take its derivative with respect to the system i.e. $ d \phi(\Phi^t X_0,t) / dt $. If you know equations of the flow you can write it explicitly using chain rule. Then you can write an equation (you solve for $\phi$ such that $ d \phi(\Phi^t X_0,t) / dt = 0$) which corresponds to what you want if you then consider measure $\phi d Leb$, where $Leb$ is Lebesgue measure.

Of course it works well for measures that have densities wrt Lebesgue measure, otherwise you may have to use weak formulations.