Define \begin{equation*} I(e)=\int_{\mathbb{S}^{n-1}}|e\cdot \omega|\,\mathrm{d}S(\omega),\quad e\in \mathbb{S}^{n-1}, \end{equation*} where $\mathbb{S}^{n-1}\subset \mathbb{R}^n$ with $n\geqslant 2$ is the sphere of the unit ball and $e\cdot \omega$ stands the inner product of $e$ and $\omega$. Then $I(e)$ is orthogonal invariant. Indeed, Let $T$ be an arbitrary orthogonal matrix. Then \begin{align*} I(Te) &=\int_{\mathbb{S}^{n-1}} |Te\cdot \omega|\,\mathrm{d}S(\omega)\\ &=\int_{\mathbb{S}^{n-1}} |e\cdot T^\mathrm{T}\omega|\,\mathrm{d}S(\omega). \end{align*} Let $\eta=T^\mathrm{T}\omega$, then $\omega=T\eta$. Hence \begin{align*} I(Te) &=\int_{\mathbb{S}^{n-1}} |e\cdot \eta| \,\mathrm{d}S(T\eta) =\int_{\mathbb{S}^{n-1}} |e\cdot \eta| \,\mathrm{d}S(\eta) =I(e). \end{align*}
My question is, I do not know exactly why $\mathrm{d}S(T\eta)=\mathrm{d}S(\eta)$. But if the sphere integral $\mathrm{d}S(T\eta)$ is replaced by the usual integral $\mathrm{d}(T\eta)$, I am sure that \begin{equation*} \mathrm{d}(T\eta)=|T|\mathrm{d}\eta=\mathrm{d}\eta. \end{equation*} So could you provide me with more details? Thank you!
