# I Substitution in a Lebesgue integral

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1. Jul 15, 2016

### DavideGenoa

Hi, friends! I read that, if $f\in L^1[c,d]$ is a Lebesgue summable function on $[a,b]$ and $g:[a,b]\to[c,d]$ is a differomorphism (would it be enough for $g$ to be invertible and such that $g\in C^1[a,b]$ and $g^{-1}\in C^1[a,b]$, then $$\int_\limits{g([a,b])}f(x)\,d\mu_x=\int_\limits{[a,b]}f(g(t))|g'(t)|\,d\mu_t$$where $\mu$ is the linear Lebesgue measure.

I know that the function $F$ defined by $$F(x):=\int_\limits{[c,x]}f(\xi)\,d\mu_{\xi}$$is absolutely continuous, and that the derivative $\varphi$ of an absolutely continuous function $\Phi:[c,d]\to\mathbb{R}$, which exists almost everywhere on $[c,d]$, is such that $$\int_\limits{[c,d]}\varphi(\xi) \,d\mu_{\xi}=\Phi(d)-\Phi(c)$$but I cannot use these two facts alone to prove the desired result.
I do see, for ex. for a non-decreasing $g$, that $\frac{d}{dt}\int_\limits{[g(a),g(t)]}f(x)\,d\mu_x=F'(g(t))g'(t)$ exists and is equal to $f(g(t))g'(t)$ for almost every $g(t)$ (and therefore for almost every $t$, since I think that this implies that a homeomorphism like $g$ maps null measure sets to null measure sets), but I am not able to derived the desired identity from this.

How can it be proved? I thank you any answerer very much!

2. Jul 15, 2016

### mathman

Your question is a little unclear. Are you trying to prove g maps zero measure sets to zero measure sets or something else? The g property is trivially a result of g being C1.

3. Jul 16, 2016

### DavideGenoa

I am trying to prove that $\int_\limits{g([a,b])}f(x)\,d\mu_x=\int_\limits{[a,b]}f(g(t))|g'(t)|\,d\mu_t$.
What follows in my previous posts is just an exposition of what I tried, of my background knowledge. I wrote that in order for potential answerers to know my level: I have studied only Kolmogorov-Fomin's Элементы теории функций и функционального анализа ($\approx$ Introductory Real Analysys) and the absolute continuity of what I called $F$, together with the equality $\int_{[c,d]}\Phi'(\xi)\,d\mu_{\xi}=\Phi(d)-\Phi(c)$, are two results, which I know from that book, which I suppose to be related to the proof of what I am trying to prove: $\int_\limits{g([a,b])}f(x)\,d\mu_x=\int_\limits{[a,b]}f(g(t))|g'(t)|\,d\mu_t$.
Thank you for your comment, mathman!

4. Jul 16, 2016

### micromass

You can do this with Lebesgue theory, but I find the Henstock integral to give way neater and more general results of these things. Note that every Lebesgue integral is a special case of the Henstock integral.

Theorem: Let $f:[c,d]\rightarrow \mathbb{R}$ and let $\Phi:[a,b]\rightarrow [c,d]$ be continuous and strictly monotone and suppose that $\Phi'(x)$ exists for all points in $[a,b]$ except possibly countably many. Define $\varphi(x) = \Phi'(x)$ wherever defined and $\varphi(x) = 0$ on the countable set where it is not defined. Then
(a) $f$ is Henstock integrable on $\Phi([a,b])$ iff $(f\circ \Phi)\cdot \varphi$ is Henstock integrable on $[a,b]$.
(b) $f$ is Lebesgue integrable on $\Phi([a,b])$ iff $(f\circ \Phi)\cdot \varphi$ is Lebesgue integrable on $[a,b]$
(c) In both cases we have $\int_{\Phi(a)}^{\Phi(b)} f = \int_a^b (f\circ \Phi)\cdot \varphi$.

Proof: Bartle, a modern theory of integration, Theorem 13.5

If you want to stay inside Lebesgue theory and choose not to use the superiority of the Henstock integral, then check out Jones "Lebesgue integration on Euclidean space" Section 16.4

5. Jul 16, 2016

### DavideGenoa

Mmh... I have never studied the theory of the Henstock integral: just Kolmogorov-Fomin's as I said, so I think the proof you use is above my level...
As to Jones's Lebesgue integration on Euclidean space, I cannot find the exact part where it proves the desired result: what page(s)? I cannot find the 16.4 section... Thank you so much again!

6. Jul 16, 2016

### micromass

Henstock integration is actually surprisingly simple to define. It's a lot like Riemann integration, just more general.

As for the Jones book, it is in the chapter "Differentation for functions on $\mathbb{R}$", section "change of variables". I am using the revised edition though, maybe it's not in the original one.

7. Jul 16, 2016

### wrobel

if we know this theorem for smooth $f$ then we know it for $f\in L^1$ since the space of smooth functions is dense in $L^1[a,b]$

8. Jul 18, 2016

### DavideGenoa

Found. Section F of chapter 16. Thank you so much!
I follow the proof until it says that, since $\phi_1\le\phi_2\le\ldots\le g\le\ldots\le\psi_2\le\psi_1$, the function $g$ is measurable. Why?