$F: \mathbb{R} \rightarrow [0,1]$ is a cumulative distribution function. After a theorem it defines an unique probability measure P with $F(x)=P((-\infty,x])$. $\lambda$ is the lebesgue measure of $ [0,1]$. Why the Pushforward measure of $ \lambda$ by $F^{-1}$ is P: $((F^{-1})_{*} \lambda)=P$?
We need this to write:
$$ \int_{\mathbb{R}} f dP=\int_0^1 (f \circ F^{-1}) (x) d \lambda (x)$$
Hear i use the change of variables theorem:
$f : \Omega \rightarrow \tilde{\Omega}, g: \tilde{\Omega} \rightarrow \mathbb{R} $ are measureable and $\mu$ a measure on $\Omega$. Define the pushforward $f_*\mu$ of $\mu$ by $f$ by the formula $f_*\mu(E):=\mu(f^{-1}(E))$. $g \circ f$ is integrable on P iff(and only iff) $g$ is integrable on $(f_{*} P)$ and $$\int_\Omega g\ d (f_* \mu)=\int_\Omega(g\circ f)\ d\mu. $$
Our definition: F is a cumulative distribution function of a random variable X. Then $F^{-1}:= \inf \{x | F(x) \ge t\}=\inf F^{-1} ([t, \infty))$ for $0<t<1$
We already discuss that $F^{-1} (F(x)) \le x, t \le F(F^{-1} (t)), \{x| F(x) \ge t\}= [F^{-1} (t), \infty) $
Our definition of the pushforwand measure: Is $f: \Omega \rightarrow \tilde{\Omega}$ measureable and $\mu$ a measure of $\Omega$, than $(f_{*} \mu)(A):=\mu(f^{-1} (A))$ is a measure on $ \tilde{\Omega}$.
My try:
$ P(X \le b)=F(b)= \lambda (\{t| t \le F(b)\})$
$ ((F^{-1})_{*} \lambda ) (A)= \lambda ((F^{-1})^{-1} (A))= \lambda (\{0 \le t \le 1| F^{-1}(t) \in A\})$