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I am trying to work my way through the proof of the change of variables theorem for Lebesgue integrals. A key lemma in this context is as follows:

If $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ is a linear map and $A \subset \mathbb{R}^n$ is Lebesgue measurable then $\lambda(T(A)) = |\det T \ | \cdot \lambda (A)$, where $\lambda(X)$ denotes the Lebesgue measure of $X$.

Can anyone provide a reference for a proof of this lemma that clearly references the facts from linear algebra that are necessary to effect the proof? The source I have for this lemma refers to German texts that I am incapable of reading and I have been unsuccessful at finding an alternate proof.

Added For the Benefit of Future Readers: In addition to the excellent references I received in response to this question, I have managed to find an additional reference that also provides a good proof of this fact: Aliprantis and Burkinshaw's Principles of Real Analysis, Third Edition, Lemma 40.4 pp 389-390.

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    Jones' "Lebesgue Integration on Euclidean Spaces" contains a thorough proof of this fact (and the necessary linear algebra).2011-07-18

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Hint 1) Enough to show this in the case that $A$ is an $n$-dimensional parallelopiped (as John M pointed out).

Hint 2) Recall from linear algebra that any linear mapping can be written as a composition of elementary linear mappings of three types: (usually expressed in the language of matrices, so I will do the same here) A) swap two rows, B) multiply a row by a scalar, C) add a scalar multiple of one row to another.

Hint 3) Swapping two coordinates is geometrically a reflection with respect to a hyperplane, so type A is easy. Type B amounts to stretching one of the coordinates. Type C is geometrically a shearing, i.e. the type of mapping that turns a rectangle into a parallelogram with same base and height.

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    @3Sphere: Ok. Non-singular mappings use the identity matrix as sort of a base case in the sense that a sequence of elementary transformations turns any non-singular matrix into the identity matrix. You are correct in that with singular matrices this doesn't work. However, you can bring a singular matrix into its reduced row echelon form with a sequence of (invertible) elementary transformation. And that row echelon form will then have one or more rows all zeros, so checking what happens to the measure of a box is not difficult in that case either.2011-07-18
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I've taken a quick look at the proof given in the text that you reference. It largely follows Jyrki's approach, but with a small difference. The text (in part (v) of the proof) considers these type C shearing matrices, but with only a multiple of one, rather than a general multiple, and then refers to a rather specific theorem that allows for decomposition into elementary matrices, such that the elementary matrix with addition of one row to the next only requires a multiple of one. This theorem is stronger than the usual decomposition theorem, and I haven't been able to find a convenient reference for it.

Anyway, Jyrki's proof is nicer than your text's: There is no reason to restrict the multiple of your shearing matrix to one - the same argument goes through for any multiple. Once you allow for this general shearing matrix, you can then refer to the more standard proofs of decomposition into elementary matrices. I like Ch 1 of Artin's "Algebra" for this.

For another approach which might be quite illuminating, see Ch 5 of Lax's "Linear Algebra". He starts with the properties of what an operator for "signed volume" must look like, and then he deduces a formula which turns out to be the usual determinant.

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    @Jyrki Lahtonen - I think either usage is correct, since "shear" is can be used as a noun or a verb. "Shearing" then takes the verb "shear" and makes it a noun and an adjective.2011-07-18