Equality between two integrals

Question

Let $(\Omega , \mathcal{A} , P)$ be a measure space of probability and let $X : (\Omega , \mathcal{A}) \to (\mathbb{R} , {\mathcal{B}}_{\mathbb{R}})$ be a random variable (${\mathcal{B}}_{\mathbb{R}}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$). By definition, \begin{equation} E(g(X)) = \int_{\Omega} (g \circ X) \, d P \end{equation} if $g : \mathbb{R} \to \mathbb{R}$ is a Borel-measurable function, and I also have been that \begin{equation} E(g(X)) = \int_{\mathbb{R}} g \, d P_X\mbox{,} \end{equation} where $P_X = P \circ X^{- 1} : {\mathcal{B}}_{\mathbb{R}} \to [0 , 1]$ is the probability induced by $X$, so I we can obtain that \begin{equation} \int_{\Omega} (g \circ X) \, d P = \int_{\mathbb{R}} g \, d P_X\mbox{.} \tag{1} \end{equation} How can I show that? On the other hand, I have studied measure theory and I know the next statement: if $f , g : \Omega \to [0 , \infty)$ are two measurable functions and we consider ${\nu}_f : \mathcal{A} \to [0 , \infty)$, given by \begin{equation} {\nu}_f(A) = \int_A f \, d P \end{equation} for all $A \in \mathcal{A}$, then ${\nu}_f$ is a measure on $(\Omega , \mathcal{A})$ and \begin{equation} \int_{\Omega} g \, d {\nu}_f = \int_{\Omega} (f g) \, d P\mbox{,} \end{equation} but I think that I can't use this to prove $(1)$. Can you help me please? Thank you very much in advance.

Answer 1

As often fruitful in integration theories, start with indicator/step functions, then try to pass to the limit. Your construction of the probability measure $P_X$ on $(\mathbb R,\mathcal B_{\mathbb R})$ is $$ A\mapsto P_X(A):= A\mapsto \int_\Omega1_A(X(\omega))P(d\omega),\quad \text{where }A\in\mathcal B_{\mathbb R}.$$ Now take the indicator function $g=1_A$. Then $$ \int_{\mathbb R}g(y)d P_X(dy)=P_X(A)=\int_\Omega1_A(X(\omega))P(d\omega)=\int_\Omega g\circ X (\omega)P(d\omega).$$ Show the identity for linear combination of indicator functions, then show the result taking limits for an appropriate class of Borel-measurable functions $g$ (do you need some other condition to pass to the limit?).