Lorentz invariance of the integration measure This is regards to the lorentz invariance of a classical scalar field theory. We assume that the action which is $S= \int d^4 x \mathcal{L}$, is invariant under a Lorentz transformation. How do you prove that the integration measure $d^4 x$ is Lorentz invariant. 
 A: $d^3 x$ is Lorentz contracted and $dt$ is Einstein dilated by the same factor, so these factors disappear in the new $d^4 x'$.
A: It's invariant because the Lorentz group is $SO(3,1)$ and the letter "S" stands for "special" which mathematically means the condition
$$\det M = +1.$$
But the determinant is exactly the coefficient by which the volume form gets multiplied when the coordinates are Lorentz-transformed:
$$ x \to M\cdot x\quad \Rightarrow \quad d^4 x \to \det M \cdot d^4 x $$
(this determinant-based transformation rule may also be derived if one views the volume form as an antisymmetric tensor with 4 indices) so if the determinant is equal to $+1$, the measure doesn't change. Well, $d^4 x$ is usually interpreted as $|d^4 x|$, so it's actually invariant under the whole $O(3,1)$, including the metrices with $\det M =-1$. And the condition $\det M=\pm 1$ (with "OR") follows from the orthogonality condition itself, so the adjective "special" is really unnecessary when we're already focusing on pseudoorthogonal matrices.
