Derivation of 4-current $j$ being a 4-vector in Landau-Lifschitz: Formulation with rigorous mathematical treatment? Here on Stack exchange, there appeared the question on how to derive the 4-current actually being a Lorentz-tensor. One of the answers (How do we prove that the 4-current $j^\mu$ transforms like $x^\mu$ under Lorentz transformation?) uses the sole assumption of Lorentz invariance of the charge. The answer contains a citation of "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz", §28).
The argument roughly is along the lines that the charge contained in an "infinitesimal Volume" dV, which is denoted as $\rho$, is supposed to be independent of the reference frame. The calculation then is
\begin{align}
dV \rho dx^{\mu} = dV \rho dt \frac{dx^{\mu}}{dt} 
\end{align}
And since $dV \rho$ is supposed to be "a Lorenz-scalar", and "dt dV" is as well, one concludes that $\rho \frac{dx^{\mu}}{dt}$ must be a 4-vector.
Initially, I'd only as for wether somebody has ever formulated this proof in the mathematical language of tensors on the spacetime. If somebody could present that to me, I'd be fine.
To clarify: There are several things about this proof that I don't understand:
Q1: What kind of objects are dV and dt? I'm used to dx to be a differential-form, which is an element from the cotangent space on the spacetime. Given the coordinate-base, $ {e_i}$, $dx^j$ is a dual vector so that $dx^j(e_i)= \delta^j_i$ holds). If I view it this way, I fail to make sense of the multiplication of those differential forms (I assume it's meant to be either as a tensor- or as a wedge-product, but I can't figure out what is meant exactly). Besides that, I don't understand that one is able to "divide differential forms" - what is really going on there? And in case the said quantities are just supposed to mean "small numbers", how could one write down the proof using differential forms and tangent vectors, that are elements of the space time?
Q2:What exactly is meant by "invariance" or "Lorentz-scalar" here?
I assume the author wants to make a statement on how the quantity changes when one performs a passive transformation in space time (that's what Lorentz transformations are: One looks at the same space-time point in a different basis, so the coordinates with respect to this base change).
In case the above mentioned quantities $dVdt$ are indeed differential forms, then they already live on the space time. They don't care about a change of the basis - Why would any of those change anyways?
In case the above mentioned quantities $dVdt$ are indeed just meant as "small numbers", and one talks about there transformation properties, what transformation is meant here?
If I imagine the "differential volume" dV to be spanned by 3 little "vectors", pointing in spatial directions $e_1$, $e_2$, $_3$ with respective length of $dx^1$, $dx^2$ and $dx^3$, then in a boosted frame of reference
those 3 vectors might have different time components. Of course one can still calculate their "new spatial lengths", but since in the new frame the volume will be moving, one would calculate a wrong quantity for the volume, it would appear larger by the factor $\gamma$.
In case one defines the volume not just spanned by 3 vectors, but instead by 3 world-lines (which would give Volume travelling in time), one could transform those, and choose 3 vectors in the new reference system that span the moving volume at equal times. Then one could look at their lengths $d\tilde{x}^1$, $d\tilde{x}^2$, $d\tilde{x}^3$, and find that their product in fact has diminished by the factor $\frac{1}{\gamma}$.  If I do this, then transformation that connects $dx^i$ and $d\tilde{x}^j$ is not a Lorenz transformation anymore, which leaves me puzzled, which of them to choose now: The first one is the transformation I would expect to be talking about when one talks about "transformation laws", while the second transformation yields the right results (the volume diminishing by the factor $\frac{1}{\gamma}$).
Q3: What is meant by $dx^{\mu}$ in the first place? Since it is divided by $dt$ to become a derivation, I assume that one previously assumes the volume talked about moves by a function $x(t)$
in space. Is that it? And in that case, is dx just something like the "infinitesimal displacement" of the position of the volume in space time? I'm asking, because in the proof, it is not mentioned at any
point before that the volume actually moves by such a function.
 A: First of all, $dx^\mu$ is a bunch of differentials as $cdt,dx,dy,dz$ that form a basis on the manifold. It is a linear map that swallows a vector in the space and spits the component on the direction of the coordinate $x^\mu$ (the $\mu$-th direction). So, you can write $\mu$-th components of every tensor as the projections on the $\mu$-th differential, $dx^\mu$.
The four-volume element, $dV dt$, is not a direct multiplication of four differential elements but is a four-form, sometimes called volume form, and they are the top degree forms in a spacetime manifold (4 in this case). Four-forms in 4D are not scalars but pseudo-scalars, that is, they have only one component, like scalars, but do not transform like a scalar.
Let $d \mathcal{V}$ be a volume form in an N dimensional space as follows:
\begin{align}
d \mathcal{V} &= {\rho} \, \epsilon_{\mu_1 \cdots \mu_N} \, dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_N} \\
&= \rho \, dx^0 dx^1 dx^2 dx^3 \\
& \equiv \rho \, dx^4
\end{align}
where $\rho$ is a scalar density, and $\epsilon$ is the Levi-Civita symbol (not the tensor). Most of the cases, the determinant of a 2-rank tensor will be the scalar density there. And mostly, the determinant of the metric tensor, $g_{\mu\nu}$, which is widely used in General Relativity.
Here, the scalar density is the component of the volume form. It is not a scalar, it transforms differently. The transformation rule for densities are $\rho \rightarrow \det|J| \, \rho$, where $J$ is the Jacobian matrix of the transformation in subject (e.g. Lorentz, translation, etc). 
On the other hand, the basis of the volume form transform exactly in the inverse way since the Levi-Civita symbol is just a bunch of numbers and $dx^\mu$ transform like a vector, which provides the inverse of the determinant of Jacobian and cancel each other, i.e., $dx^4 \rightarrow \frac{1}{\det|J|} \, dx^4 $
Let's make it simple and take the flat plane as an example. Let's think the volume element in Cartesian coordinates, i.e., 
$$d \mathcal{V} = \epsilon_{ij} dx^i \wedge dx^j = dx dy$$ 
If we switch to polar coordinates, $r$ and $\theta$, a coordinate basis could be chosen to be $dr$ and $d\theta$, however, their product would not be equal to the volume element, that is, $dx dy \neq dr d\theta$. According to the transformation from Cartesian to polar, one needs to insert the determinant of the Jacobian, which is $r $. Of course, you can define the basis vectors as $\hat{e}^r = dr$ and $\hat{e}^\theta = r d \theta$, then $dV = \epsilon_{ab} \, \hat{e}^a \wedge \hat{e}^b = \hat{e}^r \, \hat{e}^\theta$ would be equal to the Cartesian one. These kind of bases are called anholonomic or non-coordinate bases (since they are not closed forms). So, actually it is like swallowing the Jacobian.

As an example, in General Relativity (curved spacetime), the volume element becomes as follows:
$$
d\mathcal{V} = \sqrt{|\det g|} d^4 x
$$
where the metric $g_{\mu\nu}$ can become "flat" Minkowskian metric, $\eta_{\mu\nu}$, in a such a way that $\det \eta = 1$. 
In Affine Gravity, where there are no metric but only the curvature, the volume element is like the following:
$$
d\mathcal{V} = \sqrt{|\det R|} d^4 x
$$
where $R_{\mu\nu}$ is the Ricci curvature (see Eddington formalism).
