Intuition/proof that $E(X)= \int X(w) dP = \int x d\alpha$, where $\alpha$ is the cumulative distribution function of X Looking for more intuition/help explaining the equivalence of the following two integrals. 
I know that the push-forward measure, or the CDF, of a random variable $X$ on a prob. space $(\Omega, \cal B, P)$ makes perfect sense as a probability measure, I'm just looking for more insight and an intuitive proof. 
Prove that 
\begin{equation*}
E(X)= \int X(w) dP = \int x d\alpha, 
\end{equation*}
where $\alpha$ is the cumulative distribution function of X
 A: Let $(\Omega,\mathcal A,P)$ be a probability space and let $\mathcal B$ denote the Borel $\sigma$-algebra on $\mathbb R$.
Any random variable $X:\Omega\rightarrow\mathbb R$ induces a probability $P_X$ on measurable space $(\mathbb R,\mathcal B)$. 
This by: $$P_X(B)=P(\{X\in B\})$$ for $B\in\mathcal B$. This $P_X$ is the distribution of $X$ and is characterized by a CDF $F_X$ prescribed by $x\mapsto P_X(\langle-\infty,x])$.
This equality can also be written as: $$\int 1_B(x)dP_X(x)=\int 1_B\circ X(\omega)dP(\omega)$$
Using the linearity of integrals and monotone convergence theorem it can be shown that this equality still holds when characteristic function $1_B:\mathbb R\rightarrow\mathbb R$ is replaced by a $P_X$-integrable function $f:\mathbb R\rightarrow\mathbb R$:$$\int f(x)dP_X(x)=\int f\circ X(\omega)dP(\omega)$$
If $\text{id}_\mathbb R$ is $P_X$-integrable, i.e. $\int|x|dP_X(x)<\infty$, then we can take $f=\text{id}_\mathbb R$ to come to:$$\int xdP_X(x)=\int X(\omega)dP(\omega)$$
Another notation for the LHS of this equation is $\int xdF_{X}\left(x\right)$.
