TITLE: What does it mean $\int_a^b f(G(x)) dG(x)$? - An exercise question on measure theory
QUESTION [5 upvotes]: I am reading Folland's book and definitions are as follows (p. 108).

Let $G$ be a continuous increasing function on $[a,b]$ and let $G(a) = c, G(b) = d$.

What is asked in the question is:

If $f$ is a Borel measurable and integrable function on $[c,d]$, then $\int_c^d f(y)dy = \int_a^b f(G(x))dG(x)$. In particular, $\int_c^d f(y) dy = \int_a^b f(G(x))G'(x)dx$ if $G$ is absolutely continuous.

As you can see from the title, I did not understand what does it mean $\int_a^b f(G(x))dG(x)$. Also, I am stuck on the whole exercise. If one can help, I will be very happy! Thanks.

REPLY [2 votes]: For these sorts of problems in integration, it's often helpful to start with indicator functions.
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 extend it to more general functions by studying the definition of the Lebesgue integral.
For the second part, we know that $G$ is differentiable a.e.  Theorem 3.35 tells us that $G(t) - G(x) = \int_x^t G'(y)dy$ for any $(t, x) \subset [a,b]$.  But $G(t) - G(x) = \mu_G((t,x))$.