I've seen some post addressing this but still I didn't find the answer I looking for, if there's some already existing question where this is discussed in detail please let me know.
Let $(\Omega,\mathscr F,P)$ be a probability space and let $(\mathbb R,\mathscr B,\lambda)$ the a measure space where the $\sigma$-algebra is the Borel $\sigma$-algebra, and $\lambda$ stands for the Lebesgue measure.
Let be $X:\Omega\rightarrow \mathbb R$ be a $\mathscr F/\mathscr B$-measurable function, aka. a random variable.
My aim is to prove rigorously that
$$\int_{\Omega}XdP=\int_{\mathbb R}Xfd\lambda$$
where $f$ is the density function (assume for now that it exists).
I have that the distribution of my random variable $X$ can be written as the image measure of $P$ under $X$.
$$P(X\in B)=P(X^{-1}(B)), \forall B\in\mathscr B$$
Assume furthermore that
$P\circ X^{-1}<<\lambda$ and hence we can find the Radon Nikodym derivative and call it the density function $f$ of $X$.
Now in this set of slides it's stated that (I am changing the notation to make it compatible with mine):
Let be $P\circ X^{-1}:=\nu=f\cdot\lambda $ be a measure on $(\mathbb R,\mathscr B)$, then forall $g\in\mathcal M(\mathscr B)$ it holds that if
$$\int |gf|d\lambda<\infty$$
then $$\int gd\nu=\int g P\circ X^{-1} (dx)=\int gf\lambda(dx)$$
(I assume from now on that the condition holds)
I can take $g(x)=1_Bx$,$B\in\mathscr B$ which is a Borel measurable function
$$\int_B x P\circ X^{-1} (dx)=\int_B xf\lambda(dx)$$
But now I can see $x$ as a particular realization of $X$, i.e. $x=X(\omega)$
$$\int_{X^{-1}(B)} X(\omega) P\circ X^{-1} (dX(\omega))=\int_B xf\lambda(dx)$$
Now if I let $B=\mathbb R$
$$\int_{\Omega} X(\omega) P\circ X^{-1} (dX(\omega))=\int_{\mathbb R} xf\lambda(dx)$$
And there is where I get stucked, I don't really know how to go from the integral w.r.t. the image measure to the integral w.r.t the probability measure.
I feel that I might be close to the solution (or completely off).
Any hint or suggestion will be greatly appreciated.
Best Answer
By the definition of the density $f$ we have $P(X^{-1}(A))=\int_A f(x)d\lambda(x)$ for every Borel set $A$ in $\mathbb R$. This is where the connection between integrals w.r.t. $P$ and $\lambda$ comes in.
This can be written as $\int I_A (X) dP=\int_A f(x)d\lambda(x)$. This gives $\int h (X) dP=\int h(x) f(x)d\lambda(x)$ for any simple function $h$. The identity function $h(x)=x$ on $\mathbb R$ is a Borel measurable function. Hence there exists simple functions $h_n$ converging to $h$. With suitable assumptions on the existence of the integrals we can pass to the limit and get $\int XdP=\int xf(x)dx$. [Finiteness of $E|X|$ is sufficient for this].