$\varphi^*(dx)$ of $dx$ makes sense. Suppose also that there is a $G$-invariant smooth measure $d\mu$ on $G/K$, then the product measure $d\mu dy$ on $G/K\times Y$ is smooth, so $$\label{F:J(g,y)}
\varphi^*(dx)=J([g],y)d\mu dy$$ for some $J\in C^\infty(G/K\times Y)$ with $J\geq 0$.

The $G$-invariant smooth measure $d\mu$ on $G/K$ exists if and only if $\Delta_G|_K=\Delta_K$, where $\Delta_G$ and $\Delta_K$ are the modular functions on $G$ and $K$, respectively, see, e.g., Knapp [@Kn], Section 8.3. For concrete examples in the following sections, this condition always hold.

\[P:J\] The smooth function $J\in C^\infty(G/K\times Y)$ is independent of the first variable $[g]\in G/K$. So we can rewrite formula as $$\label{F:J}
\varphi^*(dx)=J(y)d\mu dy$$ where $J\in C^\infty(Y)$ with $J\geq 0$.

We denote the natural action of $h\in G$ on $G/K$ also by $l_h$, then one can easily verify that $\sigma_h\circ\varphi=\varphi\circ(l_h\times id)$. By the $G$-invariance of $dx$ and $d\mu$, we have $$\begin{aligned}
&J([g],y)d\mu dy\\
=&\varphi^*(dx)\\
=&\varphi^*\circ\sigma_h^*(dx)\\
=&(l_h\times id)^*\circ\varphi^*(dx)\\
=&(l_h\times id)^*(J([g],y)d\mu dy)\\
=&J(h[g],y)(l_h^*(d\mu)\times id^*(dy))\\
=&J([hg],y)d\mu dy.\end{aligned}$$ So $J([g],y)=J([hg],y)$ for all $g,h\in G$, which means $J$ is independent of the first variable.

\[C:J\] There exists a quasi-smooth measure $d\nu$ on $Y$ such that $$\label{F:dmudnu}
\varphi^*(dx)=d\mu d\nu.$$ The measure $d\nu$ is given by $$\label{F:dnu}
d\nu(y)=J(y)dy.$$

The factor $J(y)$ can also be 