# What does it mean $\int_a^b f(G(x)) dG(x)$? - An exercise question on measure theory

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.

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$G(x)$ defines the measure that you're taking. For example, the normal lebesgue measure can be interpreted with $G(x) = x$. Increasing means that the measure is positive (non-negative), and continuous means that your measure has no pure point / discrete measure. – Calvin Lin Jan 1 '13 at 21:50
OK I know the associated functions with measures. Then, say the measure associated with function $G$ is $\mu_G$. How can I associate this integral with the measure $\mu_G$? – oeda Jan 1 '13 at 21:55
$\mathrm dG(x)=G'(x)\mathrm d x$? – leo Jan 1 '13 at 22:04
leo, since author uses the phrase "in particular", your answer can be valid only for these conditions. Is there a more general definition? – oeda Jan 1 '13 at 22:10

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 $G(t) - G(x) = \mu_G((t,x))$.

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thank you very much! I suppose you denote $G(y)$ with $Gy$ and $G^{-1}(E)$ with $G^{-1} E$? – oeda Jan 2 '13 at 0:19
That's a very polite way of pointing out my sloppiness. I fixed it up some. – Adam Saltz Jan 2 '13 at 0:35
I think about a solution and hereby I want to ask whether it is true. Let $F$ be antiderivative of $f$. Then $\int_a^b f(y) dy = F(b) - F(a)$ naturally. Also, $$\int_c^d f(G(x)) G'(x) dx = \int_c^d \frac{dF(G(x))}{dx}dx = \int_c^d dF(G(x)) = F(G(d)) - F(G(c)) = F(b) - F(a)$$ Then, equality is satisfied. Is this reasoning valid? – oeda Jan 2 '13 at 0:38
One thing I find trick in analysis is keeping track of the facts I know from calculus and the facts I know from analysis. Here you are using the chain rule and the fundamental theorem of calculus. Both of these theorems can be found in Folland (the chain rule is "change of variables") but make sure that all their hypotheses are satisfied. – Adam Saltz Jan 2 '13 at 1:07
Moreover, how do you know that $\int_c^d \frac{dF(G(x))}{dx}\;dx = \int_c^d dF(G(x))$? I think you are begging the question. – Adam Saltz Jan 2 '13 at 1:07

This is likely either Riemann-Stieltjes or Lebesgue-Stieltjes integration (most likely the latter, given the context).

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OK, thank you very much! Do you have an idea about the whole exercise? :) – oeda Jan 1 '13 at 22:24
@John: I don't have a good intuitive grasp of the integration right now, but if I recall correctly, this should follow naturally from the definition. – tomasz Jan 1 '13 at 22:28