Integration with spherically symmetric measure in $\mathbb R^d$

Question

Let $\mu$ be a finite spherically symmetric measure over $\mathbb R^d$, so that $\mu(TB) = \mu(B)$ for all orthogonal transformations $T: \mathbb R^d \rightarrow \mathbb R^d$ and Borel set $B$. Let $g: \mathbb R \rightarrow \mathbb R $ be Borel-measurable function such that $\int| g(z \cdot x) |\mu(dz) < \infty$ for all $x \in \mathbb R^d$. Let $e_1 = (1,0,...,0) \in \mathbb R^d$. (The dot $\cdot$ denotes the usual inner product in $\mathbb R^d$ and $\| \cdot \|$ denotes the usual Euclidan norm.)

Prove that for any $x \in \mathbb R^d$ we have

$$ \int g(z \cdot x) \mu(dz) = \int g( \| x \|z \cdot e_1 ) \mu(dz). $$

This is a transformation that I got from a paper by Iosif Pinelis, but I don't know how it's done. It seems trivial but I couldn't see why (though it is intuitively clear).

Answer 1

You don’t actually need $\mu$ to be a finite measure; it can be any positive measure.

If $x=0$, then both sides are equal to $g(0)\cdot \mu(\Bbb{R}^d)$. Otherwise, fix an orthogonal map $T:\Bbb{R}^d\to\Bbb{R}^d$ such that $T(e_1)=\frac{x}{\|x\|}$. Then, \begin{align} g(\langle\cdot,x\rangle)&=g(\langle\cdot,T(\|x\| e_1)\rangle)=g(\langle T(\cdot),\|x\|e_1\rangle)=\underbrace{g(\langle\cdot,\|x\|e_1\rangle)}_{:=\gamma}\circ T. \end{align} The first equal sign is by definition of $T$, the second is because $T$ is orthogonal and the last is just notation. So, we have \begin{align} \int_{\Bbb{R}^d}g(\langle z,x\rangle)\,d\mu(z)&=\int_{\Bbb{R}^d}(\gamma\circ T)\,d\mu\\ &=\int_{\Bbb{R}^d}(\gamma\circ T)\,d(T^*\mu)&\tag{$\mu$ spherically-symmetric}\\ &=\int_{T(\Bbb{R}^d)}\gamma\,d\mu\tag{CoV}\\ &=\int_{\Bbb{R}^d}\gamma\,d\mu\\ &\equiv\int_{\Bbb{R}^d}g(\|x\|\langle z,e_1\rangle)\,d\mu(z), \end{align} where I have used the general change of variables theorem, e.g as mentioned in the beginning of this answer.