The expectation of a random variable $X$ is defined by
$$\operatorname{E}[X] = \int_\Omega X \, dP = \int_\Omega X(\omega)\,dP(\omega)$$
$X$ defines a probability measure $\mu_X(B) = P(X^{-1}(B))$ and that implies
$$\operatorname{E}[X] = \int_{\mathbb{R}^n} x \, d\mu_X(x)$$
Why does $\int_\Omega X(\omega)\,dP(\omega) = \int_{\mathbb{R}^n} x \, d\mu_X(x)$ hold? What kind of mechanism transforms the two?
Thank you very much!
edit: thank you for the comment! Wikipedia told me the following:
$X: \Omega \rightarrow \mathbb{R}^n$, so $(X_*(P))(B) :=_{} P(X^{-1}(B)) =: \mu_X(B)$ is the pushforward of $P$ w.r.t. $X$.
How do I apply the Change of variables formula?
Thank you, Did, for showing me the right wikipedia-page!
I reckon it is just:
$\int_{\mathbb{R}^n} x \, d\mu_X(x) = \int_{\mathbb{R}^n} x \, (dX_*(P)) = \int_\Omega X(\omega)\,dP$
By definition and the change of variables theorem that is cited from the book from Bogachev.