We know that if $P$ is a probability in the measurable space $(\mathbb{R}, \mathcal{B})$, then we can define a distribuition function $F$ as following: $$F(x):= P((-\infty,x])), \quad \forall x \in \mathbb{R}$$ By the Carathéodory's extension theorem we can prove that the converse is also valid, that is: if $F$ is a distibuition function, then there is a unique $P_{\small{F}}$ in $(\mathbb{R}, \mathcal{B})$ that $$F(x)= P_{\small{F}}((-\infty,x])),\quad \forall x \in \mathbb{R}$$ In other words, there is a one-to-one correspondence between the set of probabilities in $(\mathbb{R}, \mathcal{B})$ and the set of all distribuition functions.
$Remake:$:: If $X:(\Omega,\mathcal{F}, \mu) \to \mathbb{R}$ is a random variable, we can define a probability in $(\mathbb{R}, \mathcal{B})$ and a distribuition function, respectively, as following: $$P_{\small{X}}(B) := \mu(X \in B),\quad B \in \mathcal{B}$$ $$F_{\small{X}}(x) := P_{\small{X}}((-\infty, x]))= \mu(X \leq x), \quad x \in \mathbb{R}$$
A random variable is continue (absolutly?) if there is a density (non negative) function $f$ such that
$$F_{\small{X}}(x) = \int_{-\infty}^{x}f(y)dy$$
I dont will write the Lebesgue integral definition, but it is supossed that the classical definition is well known. I would like to remember its notations
$$E(X) = \int_{\Omega}Xd\mu = \int_{\Omega}X^{+}d\mu - \int_{\Omega}X^{-}d\mu$$
Although I consider the one-to-one correspondence between $F_{X}$ and $P_X$, I can not find, besides the notations justificatives, good arguments for the following equivalences (for each iquality)
$$\int_{\Omega}Xd\mu = \int_{\mathbb{R}} xdF_{\small{X}}(x) = \int_{\mathbb{R}} xdP_{\small{X}}(x) = \int_{\mathbb{R}} x f(x)dx$$ Some justificatives?
Consider measurable spaces $(\Omega_1,\mathcal{A}_1)$ and $(\Omega_2,\mathcal{A}_2)$ and a measure $\mu$ on $(\Omega_1, \mathcal{A_1})$. Any measurable function $X:\Omega_1\to \Omega_2$ induces a measure $\mu_X$ on $\Omega_2$ through $\mu_X(A_2)=\mu( X^{-1}(A_2))$ for $A_2 \in \mathcal{A}_2$, the pushforward measure. If $g:\Omega_2 \to \mathbb{R}$ is an integrable function, then $\int_{\Omega_2} g ~\mathrm{d}(\mu_X) = \int_{\Omega_1} g \circ X ~\mathrm{d}\mu$. Take for $(\Omega_2,\mathcal{A}_2)$ the real numbers with the borel sigma algebra and for $g$ the identity. Then you have
$$\int_{\Omega_1}X~\mathrm{d}\mu = \int_ {\Omega_1} id(X) \mathrm{d}\mu = \int_{\mathbb{R}} id ~\mathrm{d}\mu_X.$$
This gives you the equivalence between the first and the third expression in your last serie of equations: you can calculate the expectation through integration with respect to the pushforward measure. The second and third expression just differ by notation: $dF_X$, stresses that this is the Lebesgue-Stieltjes integral. The last expression is a property of absolutely continuous measures: if $\mu=f dx$ is a absolutely continuous measure with density $f$, then $\int_{\mathbb{R}} g d\mu = \int_{\mathbb{R}} g \cdot f d\lambda$, for all integrable $g$, where $\lambda$ is the onedimensional Lebesgue measure.