Lorentz invariance of Action or Lagrangian? For what I know, the Action should be Lorentz-invariant in accordance to the axioms of SR. For example for a free particle the Action would be the integral of the proper time, which of course is invariant. Now my doubt is the following: does that mean that the Lagrangian should be Lorentz-invariant as well? In some texts I've read it is often stated that the Lagrangian should be a Lorentz scalar, but how does this assures the Action to be Lorentz-invariant too? For example, once one gets the Lorentz-invariant Lagrangian for the EM field $\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$, why should its action:$$\int -\frac{1}{4}F^{\mu\nu}F_{\mu\nu} d^4x$$ be Lorentz invariant too? Aren't we integrating a scalar over a specific set of  coordinates which are not Lorentz-invariant? 
 A: It's true that the specific set of coordinates that we choose is not Lorentz invariant, but when you integrate over all space, it doesn't matter. One however has to check that the measure is invariant.
It's a nice excercise to show that, under a Lorentz transformation $\Lambda$ the integral transforms are
$$
\int \mathrm{d}^4x\, f(x) \mapsto \int \mathrm{d}^4x' \,|\det\Lambda|\, f(\Lambda x')\,.
$$
Since, by definition, Lorentz transformations have determinant $1$ or $-1$, the integral is invariant.
A: Action is constructed as Lorentz invariant dencity integrated over all Minkowski space. $\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$ is Lorentz invariant dencity.
$d^4x$ is Lorentz invariant measure of integration in rectangular coordinate system.
If you wanna to choose some another coordinate system (curvilinear coordinate system), in general you need consider another measure of integration $d^4y \sqrt{-g}$, $g_{\mu\nu}$ is Minkowski metric in specific coordinates, that one can choose.
$$
g_{\mu\nu}(y) = \eta_{\rho\sigma}\frac{\partial x^{\rho}}{\partial y^{\mu}}\frac{\partial x^{\sigma}}{\partial y^{\nu}}
$$
