Let $G:[a,b]\to\mathbb{R}$ be continuous and be strictly increasing, with $G(a)=c$ and $G(b)=d$. Prove that:
(a) If $E \subset [c,d]$ is a Borel set, then $m(E)=\mu_G(G^{-1}(E)).$
(b) If $f\in L^1([c,d],m),$ then $$\int_c^d f(y)dy = \int_a^b f(G(x))dG(x).$$ In particular, if $G$ is absolutely continuous, then $$\int_c^d f(y)dy = \int_a^b f(G(x))G'(x)dx.$$
This is Problem 36 in chapter 3 of Folland's "Real Analysis."
I am stuck on part (a) and the second half of part (b). Here is what I have:
From part (a) of that exercise, we have $m(E) = \mu_G(G^{-1}(E))$ for any measurable $E \subset [c,d]$. Let $\chi_E$ be the indicator function $E$. Then $\int_{[c,d]} \chi_E \; dy = m(E) = \mu_G(G^{-1}(E))$. Also, observe that $\chi_E(G(y)) = \chi_{G^{-1}(E)}(y)$. Finally, note that $G^{-1}(E)$ lies in $[a,b]$ by continuity of $G$. $$\mu_G(G^{-1}(E)) = \int_{[a,b]} \chi_{G^{-1}(E)}(y) \; dG = \int_{[a,b]} \chi_{E}(G(y)) \;dG $$ and $$\int_{[a,b]} \chi_E(G(y)) \; dG = \int_{[c,d]}\chi_E\;dy$$
Everything we have done is linear so it applies just as well to simple functions. Now we may extend it to more general functions.
Now for the second part, we know that $G$ is differentiable a.e. Theorem 3.35 in Folland tells us that $G(t) - G(x) = \int_x^t G'(y)dy$ for any $(t, x) \subset [a,b]$. But $G(t) - G(x) = \mu_G((t,x))$.
$a).\ $For $c\le \alpha\le \beta\le d$ there are $a\le x\le y\le b$ such that $G^{-1}([\alpha,\beta])=[x,y]$. Therefore, $\mu_G(G^{-1}([\alpha,\beta]))=\mu_G ([x,y])=G(y)-G(x)=\beta-\alpha=\lambda ([\beta-\alpha]).$ This implies that $\mu_GG^{-1}$ agrees with $\lambda$ on $\mathscr B([c,d]).$
$b).\ $ Note that $\mu_GG^{-1}$ is a push forward measure, and so satisfies $\int fd(\mu_GG^{-1})=\int (f\circ G) d\mu_G.$ But the left hand side of this is just $\int fd\lambda$ and the right hand side is by definition $\int f(G(x))dG(x).$
You already have done the last part.