Help in understanding a property of Lebesgue measure and linear applications.

Question

I've been trying to understand why this relation is true and can't find the right way to attack it.

Let $\Phi$ be a linear map from $R^d$ to $R^d$ and $\lambda$ the Lebesgue measure in $R^d$, then for all E measurable set:

$$ \lambda( \Phi (E))= |(det(\Phi))|*\lambda(E)$$

According to my textbook, this should easily lead to the known formula of integration by variable change.

Démarche:
I consider the characteristc function of $\Phi(E)$:
if for $b$ its equal to 1 then $$\exists a\in E \Rightarrow \Phi(a)=b [1]$$ and the characteristic function of $E$ in $a$ is 1.
Since, $$\lambda(\Phi (E))=\int characteristicfunction(\Phi( E))(x)$$ $$\forall x \in R^d $$
Now, I know that [1] implies: $$det(b)=det(\Phi)*det(a)$$
Is this enough to prove that: $$\int characteristicfuntion(\Phi(E)) = \int characteristicfuntion(E)*det(\Phi)$$

Answer 1

For intuition, you can think of the SVD of $\Phi$, which is $\Phi = U \Lambda V^*$, where $U$ and $V$ are orthogonal and $\Lambda$ is diagonal with nonnegative diagonal entries. Note that $U$ and $V$ have no effect on Lebesgue measure, while $\Lambda$ simply performs a component-wise scaling, which has the effect of multiplying the measure by $|\text{det} \Lambda | = |\text{det} \Phi |$. A rigorous proof based on this intuition appears in the book Mathematical Analysis:An Introduction by Browder.