Why does it follow that the Dirac delta function is a scalar "because determinant of the Lorentz transformation is 1" In Weinberg's Gravitation and Cosmology it's stated that

$\delta^4(x - x(\tau))$ is a scalar (because Det $\Lambda$ = 1)

(just after eqn 2.6.5, for reference.)
I cannot make sense of this statement. What is the relationship between the determinant of a Lorentz transformation and the invariance of the 4-delta function?
 A: So we want to prove that the expression $\,\delta^4\left[\boldsymbol{x-x}\left(\tau\right)\right]\,$  is a Lorentz invariant scalar, that is
\begin{align} 
& \delta^4\left[\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\right]  \boldsymbol{=}  \: \delta^4\left[\boldsymbol{x-x}_n\left(\tau\right)\right]
\tag{01a}\label{01a}\\
  & \boldsymbol{x'} \boldsymbol{=} \Lambda \boldsymbol{x}\,,\quad  \boldsymbol{x'}_n\left(\tau\right) \boldsymbol{=} \Lambda \boldsymbol{x}_n\left(\tau\right)
\tag{01b}\label{01b}\\
 & \Lambda \boldsymbol{=}  \texttt{ proper homogeneous Lorentz transformation } \left(\det \Lambda \boldsymbol{=+} 1\right) 
\tag{01c}\label{01c}
\end{align}
Note that in above equations
\begin{align} 
\boldsymbol{x} & \boldsymbol{=}\left(x_1,x_2,x_3,x_4\right)\boldsymbol{=}\left(x_1,x_2,x_3,c\,t\right)
\tag{02a}\label{02a}\\
 \boldsymbol{x'} & \boldsymbol{=}\left(x'_1,x'_2,x'_3,x'_4\right)\boldsymbol{=}\left(x'_1,x'_2,x'_3,c\,t'\right)
\tag{02b}\label{02b}
\end{align}
are the space-time position 4-vectors of a point while
\begin{align} 
\boldsymbol{x}_n\left(\tau\right) & \boldsymbol{=}\left[x_{n1}\left(\tau\right),x_{n2}\left(\tau\right),x_{n3}\left(\tau\right),x_{n4}\left(\tau\right)\right]\boldsymbol{=}\left[x_{n1}\left(\tau\right),x_{n2}\left(\tau\right),x_{n3}\left(\tau\right),c\tau\right]
\tag{03a}\label{03a}\\
\boldsymbol{x'}_n\left(\tau\right) & \boldsymbol{=}\left[x'_{n1}\left(\tau\right),x'_{n2}\left(\tau\right),x'_{n3}\left(\tau\right),x'_{n4}\left(\tau\right)\right]\boldsymbol{=}\left[x'_{n1}\left(\tau\right),x'_{n2}\left(\tau\right),x'_{n3}\left(\tau\right),c\tau\right]
\tag{03b}\label{03b}
\end{align}
is the parametric representation of the 4-dimensional path of the $n-$particle with parameter the proper time $\tau\,$ of this same particle.
As a general remark, equality between Dirac $\delta-$functions like \eqref{01a}  is not literally  valid as it is since here we have improper functions. For example the well-known equation
\begin{equation}
\delta\left(\alpha x\boldsymbol{+}\beta\right) \boldsymbol{\doteq}  \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\,\qquad x,\alpha,\beta \in \mathbb{R}
\tag{04}\label{04}
\end{equation}
is valid in the following  sense
\begin{equation}
\int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(\alpha x\boldsymbol{+}\beta\right)\mathrm{h}\left(x\right)\mathrm dx\boldsymbol{=}\dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\mathrm{h}\left(x\right)\mathrm dx
\tag{05}\label{05}
\end{equation}
where $\,\mathrm{h}\left(x\right)\,$ any proper real scalar function.
So, we must view equation \eqref{01a} in the following sense
\begin{equation}
\!\!\!\!\!\!\!\iiiint\limits_{\texttt{space-time}}\!\!\!\!\delta^4\left[\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x} \boldsymbol{=}\!\!\!\!\iiiint\limits_{\texttt{space-time}}\!\!\! \delta^4\left[\boldsymbol{x}\boldsymbol{-}\boldsymbol{x}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x}\boldsymbol{=}\mathrm{M}\left[\boldsymbol{x}_n\left(\tau\right)\right]
\tag{06}\label{06}
\end{equation}
where $\,\mathrm{M}\left(\boldsymbol{x}\right)\,$ any proper real scalar function of the 4-vector $\,\boldsymbol{x}$.
In the first integral in the lhs of above equation we apply the following 4-dimensional variable change from $\,\boldsymbol{x}\,$ to $\,\boldsymbol{u}\,$(1)
\begin{equation}
\boldsymbol{u} \boldsymbol{=}\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\boldsymbol{=}\Lambda \boldsymbol{x}\boldsymbol{-}\Lambda \boldsymbol{x}_n\left(\tau\right)
\tag{07}\label{07}
\end{equation}
Note that this variable change is invertible since
\begin{equation}
\boldsymbol{x} \boldsymbol{=}\Lambda^{\boldsymbol{-}1}\boldsymbol{u}\boldsymbol{+} \boldsymbol{x}_n\left(\tau\right)
\tag{08}\label{08}
\end{equation}
Now, the relation between the infinitesimal 4-volumes is(1)
\begin{equation}
\!\!\!\!\!\!\!\mathrm d^4 \boldsymbol{x}\!\boldsymbol{=}\!\mathrm d x_1\mathrm d x_2\mathrm d x_3\mathrm d x_4\!\boldsymbol{=}\!\left\vert\dfrac{\partial\left(u_1,u_2,u_3,u_4\right)}{\partial\left(x_1,x_2,x_3,x_4\right)}\right\vert^{\boldsymbol{-}1}\!\!\!\mathrm d u_1\mathrm d u_2\mathrm d u_3\mathrm d u_4\boldsymbol{=}\left(\det\Lambda\right)^{\boldsymbol{-}1}\mathrm d^4\boldsymbol{u}\boldsymbol{=}\mathrm d^4\boldsymbol{u}
\tag{09}\label{09}
\end{equation}
Inserting the expressions \eqref{07} and \eqref{09} in the lhs of \eqref{06} we have
\begin{align}
\iiiint\limits_{\texttt{space-time}}\delta^4\left[\boldsymbol{x'}\boldsymbol{-}\boldsymbol{x'}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x} & \boldsymbol{=}\iiiint\limits_{\texttt{space-time}} \delta^4\left(\boldsymbol{u}\right)\mathrm{M}\left[\Lambda^{\boldsymbol{-}1}\boldsymbol{u}\boldsymbol{+} \boldsymbol{x}_n\left(\tau\right)\right] \mathrm d^4 \boldsymbol{u}
\nonumber\\
 & \boldsymbol{=}\mathrm{M}\left[\Lambda^{\boldsymbol{-}1}\boldsymbol{0}\boldsymbol{+} \boldsymbol{x}_n\left(\tau\right)\right]\boldsymbol{=}\mathrm{M}\left[ \boldsymbol{x}_n\left(\tau\right)\right]
\nonumber\\
& \boldsymbol{=} \iiiint\limits_{\texttt{space-time}} \delta^4\left[\boldsymbol{x}\boldsymbol{-}\boldsymbol{x}_n\left(\tau\right)\right]\mathrm{M}\left(\boldsymbol{x}\right) \mathrm d^4 \boldsymbol{x}
\tag{10}\label{10}
\end{align}
In summary : the proof is based on the invariant 4-dimensional infinitesimal volume under a proper homogeneous Lorentz transformation, equation \eqref{09}.
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
NOTE :
By equations (2.6.1)-(2.6.5) in Weinberg's $\boldsymbol{\S\:2.6 }\textbf{  Currents and Densities}$ it is proved that in the discontinuous case of finite charged moving particles  the 4-dimensional current charge density is a Lorentz 4-vector.
For the continuous case we refer to "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition (see my $\color{blue}{\textbf{ANSWER A}}\:$ here How do we prove that the 4-current  jμ  transforms like  xμ  under Lorentz transformation?).
$\boldsymbol{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$
(1)
REFERENCE : Physical meaning of the Jacobian in relation to Dirac delta function.

