Integration by substitution explained from measure theory?

From Wikipedia:

Let $U$ be a measurable subset of $\mathbb{R}^n$ and $\varphi : U \to \mathbb{R}^n$ an injective function, and suppose for every $x$ in $U$ there exists $\varphi'(x)$ in $\mathbb{R}^{n,n}$ such that $\varphi(y) = \varphi(x) + \varphi'(x) (y − x) + o(||y − x||)$ as $y \to x$. Then $\varphi(U)$ is measurable, and for any real-valued function $f$ defined on $\varphi(U)$, $$\int_{\varphi(U)} f(v)\, dv \;=\; \int_U f(\varphi(u)) \; \left|\det \varphi'(u)\right| \,du$$ in the sense that if either integral exists (or is properly infinite), then so does the other one, and they have the same value.

Since I have learned the definition of Lebesgue integral, I have been trying to understand why integration by substitution works from the view of measure theory. I admit that I still don't quite understand Jonas Meyer's excellent answer.

But here is my current understanding (not sure if it is related to Jonas' answer), and hope it can be either finished in the same direction, or shown to be wrong:

1. The integrals on both sides of the quoted equation are integrals wrt the Lebesgue measure $m$ on $\mathbb{R}^n$.
2. Some extension $\varphi_e: \mathbb{R}^n \to \mathbb{R}^n$ extends $\varphi$ in such a way that it induces the Lebesgue measure $m$ on $\mathbb{R}^n$ from some measure $\mu$ on $\mathbb{R}^n$, i.e. $$m(A)=\mu(\varphi_e^{-1}(A)), \quad \forall A \in \mathcal{B}(\mathbb{R}^n).$$ So $$\int_{\varphi(U)} f(v)\, dv \;=\; \int_U f(\varphi) \, d\mu$$
3. Then by some unknown proof, $\mu$ can be shown to be absolutely continuous wrt the Legesgue measure and its Radon-Nikodym derivative is $\left|\det \varphi'(u)\right|$, i.e. $$\mu(A)=\int_A \left|\det \varphi'(u)\right| du, \quad \forall A \in \mathcal{B}(\mathbb{R}^n).$$ So $$\int_{\varphi(U)} f(v)\, dv \;=\; \int_U f(\varphi) \, d\mu \;=\; \int_U f(\varphi(u)) \; \left|\det \varphi'(u)\right| \,du$$

I especially wonder if "$\mu$ can be shown to be absolutely continuous wrt the Legesgue measure and its Radon-Nikodym derivative is $\left|\det \varphi'(u)\right|$" can be justified not necessarily rigorously in more details? Or better, are there some texts that can help?

Thanks and regards!

• Your sentence "$\mu$ is absolutely continuous, and its derivative is the Jacobian" is just a restatement of the transformation formula. So you have to look up the reference quoted in Wikipedia. – Florian Nov 10 '11 at 18:47
• Not necessarily rigorously? If you just want intuition, it's the same than the one from Riemann integration... What do you mean by that? – Patrick Da Silva Nov 10 '11 at 18:49
• @PatrickDaSilva: I mean if there is rigorous explanation, that will be great, and if there is a non-rigorous one, that is fine too. What is "the one from Riemann integration"? – Tim Nov 10 '11 at 19:02
• Sorry, made it into an answer and forgot to delete this comment. – Patrick Da Silva Nov 10 '11 at 19:22
• It's true, and the reason is that the transformation formula holds. And Wikipedia lists references where you can find a proof of the latter. – Florian Nov 10 '11 at 19:26

Well, I'll go for the intuition. For instance when you're in one variable, you have the usual $$du = f'(x) dx$$ where $f'$ is the linear transform that takes $dx$ to $du$. Since we are speaking of measure here, the measure induced by $dx$ and $f'$ clearly gives you $|f'| dx$ as a measure, $dx$ being the Lebesgue measure you started with and $du$ being the new one. The ideas carry over when $f'$ is a linear transform, so that you can write something like $dU = f' dX$, and when you're looking at the measure on both sides, the right notion of seeing "how the measure varies" is obviously the determinant in absolute value. How's that?
Take a small rectangle $U$, say in the plane, and let $\varphi : U \to \mathbb R^2$. The determinant of a matrix that maps this rectangle to another is precisely the ratio of the areas of the image rectangle over the domain rectangle. Taking areas and the determinant in absolute value, the measure of the new rectangle is the measure of the old one times the absolute value of the initial rectangle's area, i.e. if $\mu$ stands for Lebesgue measure, then $\mu(\varphi(U)) = \mu(U) \cdot |\det \varphi'|$.
The same ideas carry over when $f : \mathbb R^n \to \mathbb R^n$, but you need to work things out and I wasn't in the mood to check it if this would satisfy you enough...
• Thanks! +1 for trying to help. (1) "if $\mu$ stands for Lebesgue measure", in my question, $\mu$ does not mean Lebesgue measure, but the measure that induces it via $\varphi$. (2) Also does $\mu$ have a Radon-Nikodym derivative wrt Lebesgue measure? If yes, is $\left|\det \varphi'(u)\right|$ equal to its R-N derivative a.e.? (3) I don't think the idea can be carried over for $\varphi:\mathbb{R}^n \to \mathbb{R}^m$ when $m\neq n$. Integration by substitution can apply only when $m \equiv n$. Note: you use notation $f$ instead of $\varphi$ in my question. – Tim Nov 10 '11 at 23:37
• I said I went for intuition. The ideas carried over here are only the ideas that are behind the reason why the absolute value of the determinant is the factor that appears in the variable change. And yes, I just made a typo by writing things too quickly ; it is definitively required that $m=n$. I'll change that, thanks. – Patrick Da Silva Nov 10 '11 at 23:38