# Lebesgue integration by substitution

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 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!

The integrand $f$ can (and should) essentially be ignored here. It's probably clearer to prove a more general substitution rule. If $g: X\to Y$ is a map between measure spaces and $\mu$ is a measure on $X$, then its image measure $g\mu$ is defined by $(g\mu)(B)=\mu(g^{-1}(B))$ (on a suitable $\sigma$-algebra on $Y$). This comes with a substitution rule $$\int_Y f(y)\, d(g\mu)(y) = \int_X f(g(x))\, d\mu(x) ;$$ to prove this, start out by checking it for $f=\chi_B$ (this holds by the definition of $g\mu$) and then for general measurable $f\ge 0$ by approximating by simple functions, and then we have it for general $f\in L^1$.
Now back to your case: it suffices to show that Lebesgue measure on $Y=[c,d]$ is the image measure of $d\mu=|g'|\, dt$ on $X=[a,b]$, under the map $g$ (let's call this measure $\nu$). This is obvious because it holds for intervals: $$\nu(p,q) = \mu(g^{-1}(p,q)) = \int_{g^{-1}(p)}^{g^{-1}(q)} g'(t)\, dt = g(g^{-1}(p)) - g(g^{-1}(q)) = p-q,$$ as desired (here I assumed $g'>0$, but of course the calculation is analogous in the other case).
• Thank you very much! In order to try to understand how to derive the general case from the simple functions case (a derivation that isn't trivial to my eyes), I'd like to be sure what conditions are needed for $g$ and $f$: $g\in L^1(X)$ with respect to the measure $f\mu$? Is $f$ subject to any measurability condition (under any definition of measurability)? Thank you a lot again! – Self-teaching worker Jul 15 '16 at 15:54
• @Self-teachingworker: Yes, that's right, the substitution rule holds for $g\in L^1(Y, f\mu)$ (that $g\circ f\in L^1(X,\mu)$ then is part of the statement). Also, it holds for all measurable (wrt to the $\sigma$-algebra $\{ B\subseteq Y: f^{-1} \in \mathcal M_X\}$) functions $g\ge 0$. – user138530 Jul 16 '16 at 0:02
• I think I've been able to prove that $\int_Y \gamma(y) d(\varphi\mu)(y)=\int_X \gamma(\varphi(x)) d\mu(x)$ for any $\gamma(y)\in L^1(Y,\varphi\mu)$ and any $(\mathfrak{S}_X,\mathfrak{S}_Y)$-measurable function $\varphi$. Nevertheless, although I see that $\int_{g^{-1}((p,q))}|g'(t)|dt$ $=q-p$ $=\mu((p,q))$, I am not sure which is which, among the $f$ and $g$ of my OP, in $\gamma$ and $\varphi$: is the diffeomorphism $g$ of the OP $\varphi:X\to Y$? Moreover I see what the image measure is, but I don't understand what "the image of" a notation like $d\mu=|g′|dt$ means... Thank you again! – Self-teaching worker Jul 16 '16 at 15:02
• @Self-teachingworker: There was indeed some inconsistency in my notation, between the two paragraphs. I've changed that now and also brought the notation in line with yours. The notation $|g'|dt$ just refers to the measure $B\mapsto \int_B |g'|\, dt$. – user138530 Jul 16 '16 at 15:20
• @Self-teachingworker: I'm not sure I completely understand your last comment, but I believe what is confusing you is the fact that I use the notation $h\rho$, $h$ a function, $\rho$ a measure, for two totally different things: (1) the image measure of $\rho$ under $h$; (2) the measure $B\mapsto \int_B h\, d\rho$, where now $h$ serves as a density. – user138530 Jul 16 '16 at 17:07