Justifying the change of variables formula $\int_{g(a)}^{g(b)} f(y)dy = \int_a^b f(g(x))g'(x)dx$ for Lebesgue Integration

Question

This is a problem from Royden & Fitzpatrick 4th ed, page 129 problem 59. I am struggling proving it and was wondering if someone can help prove it please? Thank you

For a nonnegative integrable function $f$ over $[c,d],$ and a strictly increasing absolutely continuous function $g$ on $[a,b]$ such that $g([a,b]) \subseteq [c,d],$ is it possible to justify the change of variables formula $$\int_{g(a)}^{g(b)} f(y)dy = \int_a^b f(g(x))g'(x)dx,$$ by showing that $$\frac{d}{dx} \left[\int_{g(a)}^{g(x)} f(s)ds - \int_a^x f(g(t))g'(t)dt \right] = 0 \text{ for almost all } x\in (a,b)?$$

I suspect you need $g(x)$ rather than $g(b)$ in the last integral. — copper.hat, Oct 20 '20 at 19:37

score 2 · Answer 1 · answered Oct 20 '20 at 20:23

2

Assuming you can use the Fundamental Theorem of Calculus(FTC) you can prove just as you said.

Define the functions $F(x) = \int_{g(a)}^x f(t)dt$ and $G(x) = \int_a^x f(g(t))g'(t) dt$, by the FTC $F$ and $G$ are differentiable almost everywhere and $F'(x) = f(x)$, $G'(x) = f(g(x))g'(x)$ a.e.

If $g$ is absolutely continuous it follows that $g$, and therefore $F\circ g$, are also differentiable a.e.

By the chain rule we have that $(F\circ g)'(x) = F'(g(x))g'(x)= f(g(x))g'(x)= G'(x)$ a.e.

Since $F\circ g - G$ is a absolutely continuous function and its derivative is $=0$ a.e., we conclude that $F\circ g - G$ is constant and therefore $\int_{g(a)}^{g(b)} f(t)dt- \int_a^b f(g(t))g'(t) dt=(F\circ g)(b) - G(b) = (F\circ g)(a) - G(a) = 0$

answered Oct 20 '20 at 20:23

Pierto

458

Obviously I mean the measure-theoretic version of the FTC. – Pierto Oct 20 '20 at 20:26
Where did you use that f is nonnegative? Or is the claim true without that requirement? – user0 Sep 19 '21 at 18:55
Where did you use that $g$ is increasing? is strictly increasing? Or is the claim true without one of those requirements? – user0 May 15 '22 at 18:10

score 0 · Answer 2 · answered May 10 '22 at 01:18

0

Actually, we cannot. The problem is that we need to first prove that $f(g(t))g'(t)$ is integrable.

The proof that this function is measurable is itself non-trivial. See here.

answered May 10 '22 at 01:18

Petra Axolotl

1,106

Since $f\ge 0$ and $g'>0$ we don't have to worry about integrability. – Kurt G. May 10 '22 at 06:00
@KurtG. We do, convergence is not a priori guaranteed. (By this I mean it may be infinity.) In fact, $f(g(t))$ is not even necessarily measurable, although we can prove $f(g(t)) g'(t)$ is. For a concrete example, see Exercise 20 in Chapter 3, Stein and Shakarchi. – Petra Axolotl May 10 '22 at 19:55
1

@user0 Of course we can prove $f(g(t)) g'(t)$ is integrable. But the proof cannot be based on the assumption that $f(g(t))$ is integrable. In fact, $f(g(t))$ does not even have to measurable. An example can be found in Stein and Shakarchi, Real Analysis, Chapter 5, Exercise 20(b) on Page 149. – Petra Axolotl May 14 '22 at 15:15

Justifying the change of variables formula $\int_{g(a)}^{g(b)} f(y)dy = \int_a^b f(g(x))g'(x)dx$ for Lebesgue Integration

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