Is 4-volume element a scalar or a pseudoscalar in special relativity? In general relativity 4-volume element $\mathrm{d}^4 x = \mathrm{d} x^0\mathrm{d} x^1 \mathrm{d} x^2\mathrm{d} x^3$ is clearly a pseudoscalar (or scalar density) of weight 1 since it transforms as $\mathrm{d}^4 x \to \mathrm{d}^4 x' = J \,\mathrm{d}^4 x$, with $J$ Jacobian determinant of the coordinates transformation.
In special relativity we consider only Lorentz transformations, which have $J = \pm 1$, so shouldn't $\mathrm{d}^4 x$ be a pseudoscalar also in special relativity?  Landau states ("The Classical Theory of Fields", § 6) that «the element is a scalar: it is obvious that the volume of a portion of four-space is unchanged by a rotation of the coordinate system», but this doesn't proof that $\mathrm{d}^4 x$ is a scalar rather than a pseudoscalar since a pseudoscalar looks like a scalar under a proper rotation of the coordinate system ($J = 1$), differences arise only for an axis inversion ($J = -1$).
 A: I) Consider an arbitrary coordinate transformation
$$x^{\mu}\longrightarrow x^{\prime \nu}~=~f^{\nu}(x).$$
Let  
$$J ~:=~\det(\frac{\partial x^{\prime \nu}}{\partial x^{\mu}})$$ 
denote the corresponding Jacobian.
Traditionally in physics, 


*

*a scalar $\sigma$ transforms as
$$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~\sigma, $$    

*a pseudo-scalar $\sigma$ transforms as
$$ \sigma ~\longrightarrow~ \sigma^{\prime}~=~{\rm sgn}(J) \sigma,$$    

*a density $\rho$ transforms as
$$ \rho ~\longrightarrow~ \rho^{\prime}~=~\frac{\rho}{J}, $$     

*a pseudo-density $\psi$ transforms as
$$ \psi ~\longrightarrow~ \psi^{\prime}~=~{\rm sgn}(J)\frac{\psi}{J}, $$     

*a density $\rho$ of (integer) weight $w$ transforms as
$$ \rho ~\longrightarrow~ \rho^{\prime}~=~\frac{\rho}{J^w}, $$

*a pseudo-density $\psi$ of (integer) weight $w$ transforms as
$$ \psi ~\longrightarrow~ \psi^{\prime}~=~{\rm sgn}(J)\frac{\psi}{J^w}. $$ 
For tensor, pseudo-tensor, tensor-density, pseudo-tensor-density, etc, see the linked Wikipedia page.
Examples:


*

*On a Lorentzian manifold $(M,g)$ of signature $(-,+,\ldots,+)$, the square root $\sqrt{-\det(g_{\mu\nu})}$ is a density. 

*In General Relativity (GR), the four-form $\mathrm{d} x^0\wedge \mathrm{d} x^1\wedge \mathrm{d} x^2\wedge\mathrm{d} x^3$ transforms as an inverse density, i.e. a density of weight $w=-1$.
II) Within Special Relativity (SR), the Jacobian $J=\pm1 $ is plus/minus one, as OP correctly notes, so that $\mathrm{d} x^0\wedge \mathrm{d} x^1\wedge \mathrm{d} x^2\wedge\mathrm{d} x^3$ transforms as a pseudo-scalar.
III) Landau and Lifshitz (L&L), The Classical Theory of Fields, $\S 6$ p. 21 around eq. (6.13), indeed states that the element of integration
$$dx^0 dx^1 dx^2 dx^3 $$
is a scalar. Here are some suggestions:


*

*Perhaps L&L are only considering proper Lorentz transformations $\Lambda \in SO(3,1)$ where $J=1$ by definition? 

*Perhaps L&L are viewing $dx^0 dx^1 dx^2 dx^3$ not as a four-form but as a manifestly positive infinitesimal volume element, which by definition transforms with the absolute value $|J|$ of the Jacobian $J$?
However, neither of the two above interpretations (1 and 2) seem to fit particularly well with what is said in the rest of $\S 6$, in particular the footnote on p.21.
A: Regarding Landau's book, I am reading it in the Russian edition, and in that paragraph when he talks about elements of length/area/volume he neglects the pseudo- prefix. The explanation that the 4-volume element is a scalar he explicitly only considers transformations with unit determinant, and it can be easily deduced from what is written that in general, it is a pseudoscalar.
