Let $F(x)=\mathbb P\{X\leq x\}$ and $X:\Omega \longrightarrow \mathbb R$ a random variable.
Q1) How can I prove that $$\int_\Omega X(\omega)\,d\mathbb P(\omega ) = \int_{\mathbb R}x \, dF(x) \text{ ?}$$ I tried to make the substitution $x=X(\omega )$ but I can't compute $d\mathbb P$.
Q2) Moreover, I don't really understand the notation $\mathbb P(\omega )$ in $d\mathbb P(\omega )$. Indeed, $\mathbb P$ is a function that take element of $\mathcal P(\Omega )$ (i.e. measurable subset of $\Omega $), and in particular $\omega \in \Omega $, i.e. $\omega \notin \mathcal P(\Omega )$. So what mean $\mathbb P(\omega )$ ? Should it be $\mathbb P\{\omega \}$ ? But if yes, this is $0$ when $X$ it's continuous.
The result itself is really just simple function approximation. No change of variable is required. The left side is a limit of simple function approximations. Each of those approximations corresponds exactly to a Riemann-Stieltjes approximation to the integral on the right side.
$d\mathbb{P}(\omega)$ is just notation: it tells you that the measure you are integrating with respect to is $\mathbb{P}$ and the variable of integration in the $\mathbb{P}$ integration is $\omega$. Sometimes people write the even more confusing $\mathbb{P}(d\omega)$ to mean the same thing.