Two Definitions for $E(X)$

Question

Given that $\int_{\Omega}|X(\omega)|dP(\omega)<\infty$ and $\mu_X$ to be the induced probability measure of $X$ on $\mathbb{R}^n $. Why is it that

\begin{equation*} \int_\Omega X(\omega)dP(\omega) \end{equation*}

and

\begin{equation*} \int_{\mathbb{R}^n}xd \mu_X(x) \end{equation*}

are equivalent definition for the expected value of $X$, $E(X)$?

This might seem trivial to some. But none is trivial to a beginner. Can someone elaborate please. Thank you in advance.

Answer 1

$\mu_X$ is defined by $\mu_X(A)=P(X^{-1}(A))$. This can be written as $\int I_A d\mu_X=\int I_A(X) dP$. [ Because $I_A(X(\omega)) =I_{X^{-1}(A)}(\omega)$]. For any simple function $f$ we get $\int fd\mu_X=\int f(X) dP$. From here you can use a standard measure theoretic argument to say that the equation hods for any non-negative measurable function $f$ as also for any $f$ which is integrable w.r.t. $\mu_X$. Taking $f(x)=x$ you get the equation you want provided the integral exists (which is true iff $X$ has finite mean).

Answer 2

I would think of this in the binary case. Say we have only two sets, $\omega_0$ and $\omega_1$. Say $X(\omega_0) = 10$ and $X(\omega_1)=20$. Let $P(\omega_0)=.2$ and $P(\omega_1)=.8$. Now compute the expectation of $X(\omega)$ where we consider $\omega$ to be a random event, i.e. $\int X(\omega)dP(\omega)$. The induced measure on the set 10, 20, puts the identical weights of .2 and .8 on these values and therefore the expectation, $\int x d\mu(x)$ is the same.