Exercise: We have $\lambda$ as Lebesgue measure on $\mathbb{R}$ and $F$ a strict monotone increasing, continuous function with limit $\lim_{x\rightarrow -\infty} F(x)=a$ and $\lim_{x\rightarrow +\infty} F(x)=b$
Show: $$ d(F_{*} \lambda)=d(F^{-1})$$- with $F^{-1}$ the inverse of $F$ and $d(F^{-1})$ a measure of $(a,b)$ to the cumulative distribution function $F^{-1}$.
If $F$ has a density $f>0$, then $$d(F_{*} \lambda)= \frac{1}{f(F^{-1})} 1_{(a,b)} \,d\lambda$$-
We already have the change of variables theorem.
$f : \Omega \rightarrow \tilde{\Omega},\ g: \tilde{\Omega} \rightarrow \mathbb{R} $ are measurable 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 $P$ if and only if $g$ is integrable on $(f_{*} P)$ and $$\int_\Omega g\ d (f_* \mu)=\int_\Omega(g\circ f)\ d\mu. $$
1) Why is there a $d$ in front of the pushforward measure?
$G:=\{(a,y] \subseteq [a,b]| y \in \mathbb{R}\}$ , $A\in G \Rightarrow A=(a,y]$
$(F_{*} \lambda) (A):= \lambda (F^{-1} (A))= \lambda(t \in \mathbb{R}| F(t) \in A)= \lambda (t \in \mathbb{R}| F(t) \le y)= \lambda ((- \infty, F^{-1}(y)])= F^{-1}(y)-F^{-1}(a)=dF^{-1}((a,y])-dF^{-1} ((a,a])= dF^{-1} ((a,y])=dF^{-1} (A)$
It follows that the two measures are the same because they're the same on $G$.
Is this right? What means $\frac{1}{f(F^{-1})} 1_{(a,b)} \,d\lambda$?