Why is are the definitions of Expectation w.r.t. Probability $P$ and density $\mu$ equivalent?

Question

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. $X$.

https://en.wikipedia.org/wiki/Pushforward_measure#Main_property:_Change-of-variables_formula — Did, Jun 05 '17 at 17:05

score 0 · Answer 1 · answered Jun 05 '17 at 17:37

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.

1 Answers1