In measure theoretic probability, why is $P_X = P \circ X^{-1}$ the same as the CDF $F_X$ Suppose that we are given a random variable $X: \Omega \to \mathbb{R}$ in a probability space $(\Omega, \mathcal{A}, P)$. 

Then the expectation is given by,

$$
E(X) = \int X(w) dP(w)
$$

I am trying to make sense of the "change of variable" formula,

$$
E(X) = \int_\Omega X(w) dP(w) = \int_\mathbb{R} x dP_X(x) = \int_\mathbb{R}x dF_X(x)
$$

where $P_X$ is the distribution of $X$ and $F_X(x)$ is the CDF of $X$.

I am stuck at various points:



To start, let $x = X(w) \Leftrightarrow X^{-1}(x) = w$, 

Then $P(w) = P(X^{-1}(x)) = (P \circ X^{-1})(x)$ 

Then define $P_X := P \circ X^{-1}$. 

But then we claim that $P_X = F_X$. 

Recall that $F_X(x) = P(X \leq x) = P(\{w \in \Omega | X(w) \leq x)$. 

But $P \circ X^{-1}(x) =  P(w) \neq F_X(x)$          $\quad \quad \quad \quad \quad $ (?)

How do I proceed from here?
Start by assuming $X$ is a simple function and show the equalities in this case.  This case just amounts to $P_X$ having only point-masses and $F_X$ having finitely many jumps (and otherwise constant), so you can evaluate all the integrals directly.

Then assume $X \geq 0$ and approximate $X$ from below by simple functions (use the monotone convergence theorem).  

Finally write $X$ as the difference of two non-negative random variables and, assuming at least one is integrable, conclude the equations hold whenever they make sense.  (The only case not dealt with by this treatment is one in which the integrals don't exist)
Suppose that we are given a random variable $X: \Omega \to \mathbb{R}$ in a probability space $(\Omega, \mathcal{A}, P)$.
Then the expectation is given by,
$$ E(X) = \int X(\omega) dP(\omega)$$

No, it is:
$$\mathsf E(X)=\int_\Omega X(\omega)~\mathsf P(\mathrm d\omega)$$

I am trying to make sense of the &quot;change of variable&quot; formula,

$$
\mathsf E(X) = \int_\Omega X(\omega)~\mathsf P(\mathrm d\omega) = \int_\mathbb{R} x ~\mathsf P_X(\mathrm dx) = \int_{\mathbb{R}}x~\dfrac{\mathrm dF_X(x)}{\mathrm d x}\mathrm d x
$$