Why is $d^4p$ "obviously" Lorentz invariant? Is there a mathematical and a physical reason? I believe it follows from the jacobian of the Lorentz transformation. But, if that is so, I don't see why it would be "obvious" unless you can do such calculations instantly in your head. What is the physical meaning of the invariance of Lorentz measure? Is it just conservation of 4-momentum?
 A: The defining property of the Lorentz transform is 
$$\Lambda^T\eta\Lambda = \eta$$
where $\eta_{\mu \nu}$ is the flat space-time metric $diag(-+++)$. Since the determinant is linear and the transformations are non-degenerate (as is also the metric itself), we have 
$$\mathrm{det}\Lambda^T \, \mathrm{det}\eta \;\mathrm{det}\Lambda=\mathrm{det}\eta\,,\implies \mathrm{det}\Lambda^T \mathrm{det}\Lambda= (\mathrm{det}\Lambda)^2=1$$
I.e., very straightforwardly from the defining properties of the Lorentz transform, the determinant of it can be only $\pm 1$.
One way to intuitively expect this is from the understanding that the Lorentz transform is "orthogonal" or a "space-time rotation" in a certain sense and thus it is natural that it's determinant is $\pm 1$ similarly to the spatial rotation. 
Another way to see this is the equivalence principle. By Lorentz-transforming into another frame, you should get a fully equal and emancipated set of coordinates. But $\mathrm{det} \Lambda < 1$ would mean that you are somehow "loosing information per coordinate unit". You could then repeat a series of $\mathrm{det} \Lambda < 1$ transformations to reach a set of degenerate coordinates, or conversely by $\mathrm{det} \Lambda > 1$, a blown-up set of coordinates. This is simply against the spirit of special relativity, the transformation should not prefer the "initial" or "latter" reference frame, they should all represent an equivalent description of the physical situation.
The transformation is linear, so the Jacobian is just $\Lambda$ itself, so the $\mathrm{d}^4 p$ (where we mean the momenta in Cartesian coordinates!) is trivially Lorentz-invariant. 
A: Under a Lorentz transformation $p \to p' = \Lambda p$, we have
$$
d^4 p' = d^4 p | \det \Lambda | = d^4 p 
$$
since $\det \Lambda = \pm 1$. 
