When Lebesgue measure of $A$ is equal to Lebesgue measure of $TA$

Asked Aug 06 '17 at 23:22

Active Aug 06 '17 at 23:22

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Let $T: \mathbb{R}^{2} \rightarrow \mathbb{R}^2$ be a linear transformation.

Let $T A = \{T x\;|\;x \in A\}$ be the image of a measurable set $A$. I have the following question:

Under what conditions on $A$ and $T$ is it true that the Lebesgue measure $m(A)$ of $A$ is the same as the Lebesgue measure $m(T A)$ of $TA$.

asked Aug 06 '17 at 23:22

Tensor_Product

3

This should be helpful: https://math.stackexchange.com/q/52161/169852 – Aug 06 '17 at 23:27
1

If the determinant equals $1$. – gary Aug 06 '17 at 23:38
@gary Kindly mention some reference. – Tensor_Product Aug 06 '17 at 23:44
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@Tensor_Product: Please see link above my post, it describes the general result. Please let me know if this is not enough. – gary Aug 06 '17 at 23:48
Thank you both of you. Reference: Aliprantis and Burkinshaw's Principles of Real Analysis, Third Edition, Lemma 40.4 pp 389-390. – Tensor_Product Aug 06 '17 at 23:50
Tiny nit: If the absolute value of the determinant equals $1$. This is of course implied by the link above but contradicted by gary's comment. – kimchi lover Aug 07 '17 at 00:32
Note also that the result holds if $m(A) = 0$, regardless of the determinant of $T$. – Aug 07 '17 at 03:17

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