Let $X$ be a random variable with values in $\mathbb{R}$ and let $g: \mathbb{R} \to \mathbb{R}$ a measurable function such that $ E g(X) $ exist.
If we define the expected value as $E(X)= \int_{x \in \mathbb{R}}{x \cdot f_X(x) dx}$, how can I deduce that $E(g(X))= \int_{x \in \mathbb{R}}{g(x) \cdot f_X(x) dx}$?
I know that this was asked before ( discrete case). Here Discrete case
My problem is that I cannot use the same technique here.
A fundamental theorem in measure theory is the key. Let $(\Omega, F, P)$ be a probability space. By definition we have $\mathbb{E} g(X) = \int_{\Omega} g(X) dP$. We are aware that $g(X) = g \circ X$. If $X: \Omega \to \mathbb{R}$ is a random variable measurable-$F$ and if $p_{X}$ is the probability measure induced by $X$, then a fundamental theorem in measure theory (see, say the chapter on probability in Folland's real analysis) shows that $$ \int_{\Omega} g\circ X dP = \int_{\mathbb{R}} g(x) dp_{X}(x). $$ If $p_{X}$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$, then by the Radon-Nykodym theorem the probability density $f_{X}$ of $X$ exists, and a basic property of Lebesgue-Stieltjes integration asserts that $$ \int_{\mathbb{R}} g(x)dp_{X}(x) = \int_{\mathbb{R}} g(x)f(x) dx. $$