I was staring at one of my questions at SE and realized that I do not really understand what I mean by $dP(\omega)$ when I write:
$$EX = \int_{\Omega} X(\omega) \, dP (\omega)$$
where $X: \Omega \to \mathbb{R}$ is a real-valued random variable defined on a probability space $(\Omega, \mathcal{F}, P)$. It seems $P(\omega)$ does not make much sense since $P: \mathcal{F} \to [0, 1]$. I started looking at what Durrett writes in his book, and, when it comes to integrals with respect to measures, there is no any argument next to $P$:
$$EX = \int X \, dP = \int_\Omega X(\omega) \, dP.$$
Question: What does one actually mean by $dP$ in this context?
If it is $P(\{ \omega \})$ then again I cannot make heads or tails of it since $P(\{ \omega \})$ is zero assuming that there are no atoms, and $\{ \omega \}$, $\forall \omega \in \Omega$, should be in $\mathcal{F}$.
Thank you!
Regards, Ivan