Let $(\Omega,\mathcal B,P)$ be a probability space. I have two (related) questions. Assuming that $g:\mathbb{R}\to\mathbb{R}$ is Borel measurable, and understanding that
$$E(g(X)) = \int_{\Omega}g(X(\omega))dP(\omega),$$
how do I prove that these equalities hold for the two following circumstances?
First, $X$ is discrete, with range $\{x_{i}:i\in \mathbb N\}$. Then
$$E(g(X))=\sum_{i=0}^{\infty}g(x_{i})P(X=x_{i}),\ \ \text{ provided } \sum_{i=0}^{\infty}|g(x_{i})|P(X=x_{i})<\infty.$$
Second, $X $ is absolutely continuous with density $f.$ Then
$$E(g(X))=\int g(x)f(x)dx\ \ \text{ provided }\int |g(x)|f(x)dx<\infty.$$
and my try:I think about number one I can write $$E(g(X))=\int_{\mathbb R}g(X)dP$$ As regards $X$ is discrete I think $$E(g(X))=\sum_{i=1}^{\infty}g(x_{i})\int dP=\sum_{i=1}^{\infty}g(x_{i})P(X=x_{i}) $$ about number two i just now $X$ is absolutely continuous hence$$P_{X}(E)=\int_{E}f(x)dx \qquad \forall E \in \mathcal B_{\mathbb R}$$ thanks for any help.
In the discrete case $g(X)$ is a discrete RV and you can write $g(X)$ as the nearly-simple function $$g(X) = \sum_i g(x_i)1_{\{X =x_i\}}$$ If $\sum_i |g(x_i)|< \infty$ approximate $g(X)$ with $Y_n := \sum_{i=1}^n g(x_i)1_{\{X=x_i\}}$. Notice that for every $n$, $|Y_n| \leq \sum_{i}|g(x_i)|$ and $Y_n \to g(X)$. This will allow you to apply DCT to $$E[Y_n] = \sum_{i=1}^n g(x_i)\int_{\Omega} 1_{\{X=x_i\}} dP = \sum_i^n g(x_i)P(X=x_i).$$
A similar argument will work in the non-discrete case. Approximate $g$ by simple functions $\varphi_n = \sum_{i=1}^{N_n} \lambda^n_i 1_{A_i}$ (here $A_i$ are measurable sets). Then $\varphi_n(X) \to g(X)$ and $$ E[\varphi_n(X)] = \sum_{i=1}^{N_n} \lambda^n_i \int_{\Omega} 1_{A_i}(X) dP = \sum_{i=1}^{N_n} \lambda^n_i \int_{X \in A_i} dP = \sum_{i=1}^{N_n}\lambda_i^n P(X \in A_i) \\= \sum_{i=1}^{N_n} \lambda_i^n \int_{A_i}f(x)dx = \int \varphi_n(x) f(x)dx $$ Once again use the bound on $\int|g(X)|$ to use the DCT and finish off the proof!