Conservation of probability in phase space flow In J.Binney's notes on classical mechanics, under the section 'Liouville's theorem', he states that (paraphrasing): 

the conservation of probability requires that $\frac{df}{dt} = 0.$

where $f$ is the phase space density. 
I'm not sure why the probability is conserved. If we fix some initial phase space region $V_0$, which in time $t$ will become $V_t$, the probability of the system occupying that region is given by $$P_t = \int_{V_t} dxdp \ f(x,p;t).$$
But why is this constant in time? I know that the phase space volume is conserved, so at least the volume of the region won't change, but its shape will. What allows me to conclude that the function $f$ evolves in time in such a way that $P_t$ in fact remains constant?
 A: No, no, you miss something. Let me show you in a simple way. Since your density function $f$ is a function of $x, p$ and $t$, its total derivative by time looks as follows,
(1) $\frac {df}{dt} = \frac {∂f}{∂t} + \frac{∂f}{∂x} \frac {dx}{dt} + \frac{∂f}{∂p} \frac {dp}{dt}$.
Do you agree?
This is what says Liouville's theorem, i.e. that if the left-hand-side (LHS) is zero, then the right-hand-side (RHS) is zero. 
Now let me integrate (1) on both sides
(2) $\frac {d}{dt} \int_{V_t} f(x, p; t) dx dp = \frac {∂}{∂t} \int_{V_t} f(x, p; t) dx dp + \int_{V_t} \frac{∂f(x, p; t)}{∂x} \frac {dx}{dt} dx dp + \int_{V_t} \frac{∂f(x, p; t)}{∂p} \frac {d p}{dt} dx dp$
$ = \frac {∂}{∂t} \int_{V_t} f(x, p; t) dx dp + \int_{V_t} \nabla f(x, p, t) \  \vec v \ dx dp$
Here, you apply a variant of the Divergence theorem, in fact, the energy conservation: the change of the total probability in a volume, due to transport with velocity $\vec v$ is equal to the total flux through the surface of the volume
$\int_{V_t} (\nabla f) \ \vec v \ dpdx = \int_{S_t} f \vec v$ • $\vec n \ dS$,
In all, we have similarity with Gauss' theorem : nothing "disappears". If the amount of energy confined to some volume decreases, then there is some outwards flux. Alternatively, if it increases, then some inwards flux exists there.
(I notice that you speak of a volume $V_t$ and I went with that, but in 1 dimension you have a surface, and my $S_t$ is then a contour - i.e. the flux is through the contour of the surface.)
