Just ask a very fundamental problem of changing variable when doing integration. I am a bit confused about the following:
Suppose I want to do the integral $$\int_X v(x) d\mu(\gamma^{-1}x)$$
where $\gamma: X\mapsto X$ and $\gamma$ is invertible, i.e., $\gamma^{-1}$ exists. You can think $\gamma \in O(n)$, the orthogonal gruop. $\mu$ is a nonnegative measure. $\,v:X\mapsto \mathbb{R}$ is continuous.
- Let $y=\gamma^{-1}x$. So $x=\gamma y$.
- Integration over $X$ becomes over $\gamma^{-1}X$.
So we obtain $$\int_{\gamma^{-1}X} v(\gamma y) d\mu(y)$$
Now, change $y$ back to $x$, $$\int_{\gamma^{-1}X} v(\gamma x) d\mu(x)$$
I am not sure if my derivation makes any sense. In particular, the part of $\gamma^{-1}X$.
Thanks!
If $(X,\mu, \Sigma)$ is a measurable space, and $\gamma:X\to X$ is measurable, then $\gamma_*(\mu)=\mu \gamma^{-1}$ is the pushforward measure defined in the obvious way by by $\gamma_*(\mu)(A)=\mu (\gamma^{-1}(A)).$ If $v:X\to \mathbb R$ is measurable, then
$$\int_X vd\mu (\gamma^{-1})=\int_X v\circ \gamma d\mu.$$
You can check this formula by first showing that $v\circ \gamma$ is measurable, then assuming that $v$ is the characteristic function of a measurable set, which implies that it is true for non-negative simple functions, and then the general result follows first by the monotone convergence theorem, and finally by considering positive and negative parts.