Not sure if I can ask such fundamental problem here.

Let

- $G$ be a finite group, $\sigma \in G$. Consider linear group actions fo $G$ on $\mathbb{R}^n$.
- $\sigma(K)=\{x\in \mathbb{R}^n: \sigma^{-1}(x) \in K\}$.
- $K\subseteq \mathbb{R}^n$ be $G$-invariant, i.e., $\bigcap_{\sigma\in G} \sigma(K) = K$.

Suppose

- $K\subseteq \mathbb{R}^n$ is $G$-invariant.
- polynomial $h(x)\in \mathbb{R}[X]$ is $G$-invariant, i.e., $h(\sigma^{-1}(x))=h(x), \, \forall \sigma \in G$.

Note that

- Let $\mathcal{P}(K)$ be the set of probability measures supported on $K$.
- For $\mu\in \mathcal{P}(K)$, $\sigma^{-1}(K)\subseteq K, \, \forall\sigma$. $\mu$ is $G$-invariant if $\mu (B) = \mu(\sigma^{-1}(B))$ for any Borel set $B\subseteq K$.

Consider the following

\begin{equation} \begin{aligned} \int_K h(x) \mu(dx) &=_1 \int_K h(\sigma(x)) \mu(dx) \\ &=_2 \int_K h(x) \mu(\sigma^{-1}(dx)) \end{aligned} \end{equation}

- comes from the fact that $h(x)$ is $G$-invariant and $G$ is a group.

My questions are

- Is the changing variable of step 2. correct? (Note that $\sigma(K)=K,\, \forall \sigma \in G$ from assumption.)
- Can I say that since the $2$nd equality holds for all $dx$, so $$\mu(dx) = \mu(\sigma^{-1}(dx)),$$ i.e., $\mu$ is $G$-invariant.

I appreciate any help or comment!