Suppose we have $n$ linearly independent vectors $\mathbf{x}_1$, $\cdots$, $\mathbf{x}_n$ in $\mathbb{R}^n$. Let $\mathbf{X}$ be the $n \times n$ matrix with column $k$ given by $\mathbf{x}_k$, $k = 1, \cdots, n$. I know that the volume of the parallelepiped defined by the set $$ E = \{\mathbf{x} \in \mathbb{R}^n: \mathbf{x} = \mathbf{X}\theta \text{ for some $\theta \in [0,1]^n$}\} $$ is given by $|\det \mathbf{X}|$, the absolute value of the determinant of $\mathbf{X}$.

How can I show that the Lebesgue measure of $E$ is also equal to $|\det \mathbf{X}|$?

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    $\begingroup$ This isn't really an answer, per se, but the "volume" of a simple geometric feature coincides with its Lebesgue measure. $\endgroup$ – Math1000 Jun 8 '15 at 3:30
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    $\begingroup$ There is a full proof in chapter 3 of the book Lebesgue Integration on Euclidean Space by Frank Jones. Actually I think it is chapter 3. The idea is to decompose the linear map into maps for which the volume distortion is easier to see. $\endgroup$ – Tim kinsella Jun 8 '15 at 5:48
  • $\begingroup$ @Math1000 Yes I'm trying to prove that fact $\endgroup$ – David Jun 8 '15 at 15:44

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