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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!

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    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.2011-11-10
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    Not necessarily rigorously? If you just want intuition, it's the same than the one from Riemann integration... What do you mean by that?2011-11-10
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    @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"?2011-11-10
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    Sorry, made it into an answer and forgot to delete this comment.2011-11-10
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    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.2011-11-10

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