For the stereographic projection of the plane on the $2-$sphere $S^2$, we have the following formula between the meassures: \begin{equation} dA=\frac{4}{(1+X^2+Y^2)^2}dXdY \end{equation}
where $dA$ is the usual surface Lebesgue measure on $S^2$ and $X,Y$ are the Cartesian coordinates of $\mathbb{R}^2$.
I wonder how this formula generalizes to arbitrary $\mathbb{R}^n$ and $S^n$. Especially, in the case of $n=4$ could anyone please provide the explicit formula?
Let $(x^1,\dots, x^n,x^{n+1})$ denote the standard cartesian coordinates in $\Bbb{R}^{n+1}$, restricted to $S^n$, and let $(\xi^1,\dots, \xi^n)$ denote the coordinates for stereographic projection from the north pole. These are related as \begin{align} (x^1,\dots, x^n,x^{n+1})&= \left(\frac{2\xi^1}{1+\|\xi\|^2},\dots, \frac{2\xi^n}{1+\|\xi\|^2},1-\frac{2}{1+\|\xi\|^2}\right). \end{align} We thus have \begin{align} \begin{cases} dx^i&=\frac{2d\xi^i}{1+\|\xi\|^2}-\frac{4\xi^i\langle \xi,d\xi\rangle}{(1+\|\xi\|^2)^2}&\text{for $i\in\{1,\dots, n\}$}\\\\ dx^{n+1}&=\frac{4\langle \xi,d\xi\rangle}{(1+\|\xi\|^2)^2}, \end{cases} \end{align} where $\langle \xi,d\xi\rangle:=\sum_{j=1}^n\xi^j\,d\xi^j$. So, plugging this into the Riemannian metric on the sphere, we get \begin{align} g_{S^n}&:=\sum_{i=1}^{n}(dx^i)^2+ (dx^{n+1})^2\\ &=\frac{4}{(1+\|\xi\|^2)^2}\sum_{i=1}^n(d\xi^i)^2\\ &-\frac{16}{(1+\|\xi\|^2)^3}\langle \xi,d\xi\rangle^2+ \frac{16\|\xi\|^2}{(1+\|\xi\|^2)^4}\langle \xi,d\xi\rangle^2+ \frac{16}{(1+\|\xi\|^2)^4}\langle \xi,d\xi\rangle^2\\\\ &=\frac{4}{(1+\|\xi\|^2)^2}\sum_{i=1}^n(d\xi^i)^2, \end{align} where we note that the last three terms in the second equal sign cancel out. Also, if $\omega,\eta$ are $1$-forms, then $\omega\eta$ denotes the symmetrized tensor product $\frac{\omega\otimes \eta+\eta\otimes\omega}{2}$, so all the algebra done above makes perfect sense. So, we have that the Riemannian metric of the sphere, in sterographic coordinates is conformal to the Riemannian metric on the plane $\Bbb{R}^n$. From here, the volume measure is easily computed via the square root of the determinant: \begin{align} dV_{S^n}&=\sqrt{\left|\det g_{ab}(\xi)\right|}\,d\mu_n=\sqrt{\left(\frac{4}{(1+\|\xi\|^2)^2}\right)^n}\,d\mu_n= \left(\frac{2}{1+\|\xi\|^2}\right)^n\,d\mu_n, \end{align} where $d\mu_n$ denotes the pullback of the Lebesgue measure on $\Bbb{R}^n$ to $S^n\setminus\{\text{north pole}\}$, via the stereographic projection map. In the case $n=2$, this coincides with what you wrote.
