Gravitational potential outside Lagrangian points or Lagrange points The diagram in Why are L4 and L5 lagrangian points stable? shows that the gravitational potential decreases outside the ring of Lagrange points — this image shows it even more clearly:

If I understand correctly, using the rubber-sheet model analogy, an object placed in the field moves as if a marble rolls on the sheet with downward gravity. That's fine for objects inside the Lagrange ring: They move towards either mass.
But outside it, it implies that they move away from both masses. Is that really what happens? If so, why? If not, why does the surface slope downwards?
 A: Yes the free body moves outward, but there are two critical things you have to know to interpret this statement correctly.
First, this is the effective potential, taking into account gravity and centrifugal force. It has this form because we went into the non-inertial frame co-rotating with the two masses. Mathematically, the potential is
$$ \Phi_\mathrm{eff}(\vec{r}) = -G \left(\underbrace{\frac{M_1}{\lvert \vec{r}-\vec{r}_1 \rvert}}_\text{potential from mass 1} + \underbrace{\frac{M_2}{\lvert \vec{r}-\vec{r}_2 \rvert}}_\text{potential from mass 2} + \underbrace{\frac{M_1+M_2}{2\lvert \vec{r}_1-\vec{r}_2 \rvert^3} \lvert \vec{r} \rvert^2}_\text{centrifugal component}\right), $$
and it only decreases far away because of that last term.
Physically, this is because placing an object "at rest" in this frame corresponds to having it move with the same angular frequency as $M_1$ and $M_2$ about the center of mass. If you initialize an object $5\ \mathrm{AU}$ on a tangential path having the same angular velocity as the Earth, it will be moving too fast for a circular orbit at that distance, and so it will move away from the Sun.
This does not mean the object will go away forever, and that brings us to the second point, explained in Chay's response: Not all effective forces have been accounted for; in particular, the Coriolis force does not arise from $\Phi_\mathrm{eff}$. The Coriolis force depends on velocity, so it has no scalar potential depending solely on position, and so it is not included in the analysis so far. Once your test object starts moving in your rotating frame, it will experience a perpendicular deflection that will eventually force it to turn around.
A: The stability of L4 and L5 isn't at all obvious from looking at the scalar gravitational potential, because the stability of orbits near those points is very much dynamical: in a rotating frame like the one in the image, you have an additional velocity-dependent Coriolis potential as well as the pictured position-dependent scalar gravitational and centripetal one. So there's a second potential here, a vector field that goes as $\Omega\hat{z}$ (up to a sign).
When an object is perturbed radially from L4, it accelerates outwards, but this isn't the same as moving outwards monotonically. Because we're in a rotating frame, moving radially results in a Coriolis acceleration acting normally to it's velocity, which eventually pulls it back around to where it started.
I don't know of any proof that this definitely stabilises the problem in the case of large test masses, but for small test masses it can be linearised and shown to be dynamically stable.
