Special relativity: I arrive at a contradiction regarding the Lorentz invariance of certain quantity I want to show the Lorentz invariance of $d^3 p/E$ (Eq. 8.11 of Mandl-Shaw), where $E$ is the relativistic energy. Peskin-Schroeder gives sort-of, a proof in section 2.3 which I am convinced of. But my independent line of reasoning leads me to a contradiction.
$d^4$ p is Lorentz invariant $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$(of course)
$\implies d^3p~dp^0$ is Lorentz invariant
$\implies d^3p~E$ is Lorentz invariant because $E$ and $p^0$ are both time-components of 4-vectors $(E, p^1, p^2,p^3)$ and $(dp^0, dp^1, dp^2,dp^3)$ respectively.
So finally, I arrive at the conclusion that it's $d^3p~E$ which is Lorentz invariant, not $ d^3p/E$. What did I do wrong above? 
 A: I think that your assertion "...$d^3p~E$ is Lorentz invariant because $E$ and $p_0$ are both time-components of 4-vectors..." is wrong.
Replace the 4-vector $\mathbf p=\left(p^0,p^1,p^2,p^3\right)$ by the space-time position 4-vector $\mathbf x=\left(x^0\equiv t,x^1,x^2,x^3\right)$. The infinitesimal 4-volume $\mathrm dt\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3$ is Lorentz invariant
\begin{equation}
\mathrm dt'\,\mathrm d x'^1\,\mathrm d x'^2\,\mathrm d x'^3=\mathrm dt\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3
\tag{01}\label{01}    
\end{equation}
but I don't think you could prove that  $t\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3$ is Lorentz invariant
\begin{equation}
t'\,\mathrm d x'^1\,\mathrm d x'^2\,\mathrm d x'^3=t\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3
\tag{02}\label{02}    
\end{equation}
It's not permitted to mix the components of various 4-vectors in infinitesimal Lorentz invariant scalars like $\mathrm dx^0\,\mathrm d x^1\,\mathrm d x^2\,\mathrm d x^3$. The relation between the infinitesimal 4-volumes in Minkowski space is
\begin{equation}
 dx'^{1}dx'^{2}dx'^{3}dx'^{0} =\begin{vmatrix}
  \dfrac{\partial x'_{1}}{\partial x_{1}}& \dfrac{\partial x'_{1}}{\partial x_{2}}&\dfrac{\partial x'_{1}}{\partial x_{3}}&\dfrac{\partial x'_{1}}{\partial x_{0}}\\
 \dfrac{\partial x'_{2}}{\partial x_{1}}& \dfrac{\partial x'_{2}}{\partial x_{2}}&\dfrac{\partial x'_{2}}{\partial x_{3}}&\dfrac{\partial x'_{2}}{\partial x_{0}}\\
  \dfrac{\partial x'_{3}}{\partial x_{1}}& \dfrac{\partial x'_{3}}{\partial x_{2}}&\dfrac{\partial x'_{3}}{\partial x_{3}}&\dfrac{\partial x'_{3}}{\partial x_{0}}\\
\dfrac{\partial x'_{0}}{\partial x_{1}}& \dfrac{\partial x'_{0}}{\partial x_{2}}&\dfrac{\partial x'_{0}}{\partial x_{3}}&\dfrac{\partial x'_{0}}{\partial x_{0}} \end{vmatrix}
dx^{1}dx^{2}dx^{3}dx^{0}=\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{0}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{0}\right)}\right\vert dx^{1}dx^{2}dx^{3}dx^{0}
\tag{03}\label{03}
\end{equation}
where $\:\left\vert\partial\left(x'^{1},x'^{2},x'^{3},x'^{0}\right)/\partial\left(x^{1},x^{2},x^{3},x^{0}\right)\right\vert\:$ the Jacobian, that is determinant of the Jacobi matrix. But the Jacobi matrix is the Lorentz matrix $\:\Lambda\:$ with $\:\det(\Lambda)=+1$,
that is 
\begin{equation}
 \left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{0}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{0}\right)}\right\vert=\det(\Lambda)=+1 
\tag{04}\label{04}
\end{equation}
so
\begin{equation}
 dx'^{1}dx'^{2}dx'^{3}dx'^{0} =dx^{1}dx^{2}dx^{3}dx^{0}=\text{scalar invariant}
\tag{05}\label{05}
\end{equation} 
There is no similar way to prove that
\begin{equation}
x'^{0} dx'^{1}dx'^{2}dx'^{3} =x^{0}dx^{1}dx^{2}dx^{3}=\text{scalar invariant}
\tag{06}\label{06}
\end{equation}
because simply it's incorrect.
