## Lebesgue Measure and Linear Transformations II

In the first installment on the pushforward of the Lebesgue measure by a linear transformation, we restricted our attention to linear transformations defined by an orthogonal matrix and more generally, isometric affine transformations of Euclidean space. Specifically, we showed that if $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is an isometry, then

$\lambda_{n}(T(A))=\lambda_{n}(A),\indent\forall A\in\mathcal{B}(\mathbb{R}^{n})$

where $\lambda_{n}$ denotes the $n$-dimensional Lebesgue measure and $\mathcal{B}(\mathbb{R}^{n})$ denotes the Borel $\sigma$-algebra on $\mathbb{R}^{n}$.

Today’s goal is to obtain a change-of-measure formula, when $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is some arbitrary affine transformation. Since the Lebesgue measure is translation-invariant, we may assume that $T$ is linear. Proving this formula is not difficult: we only need our previous orthogonal-invariance result and some linear algebra in the form of the singular value decomposition (SVD) and the determinant function. For a review of determinants using a coordinate-free approach, check out these notes.

Using the SVD, we may write $T$ as

$\displaystyle T=UDV:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$,

where $U,V:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ are unitary operators and $D:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ is diagonalizable with respect to the standard basis. Since the determinant is a multiplicative homomorphism and unitary operators have unit determinant, we see that

$\displaystyle\left|\det T\right|=\left|\det U\right|\cdot\left|\det D\right|\cdot\left|\det V\right|=\left|d_{1}\cdots d_{n}\right|$,

where $d_{1},\ldots,d_{n}$ are the diagonal entries of $D$. If $T$ is non-invertible, then the image $T(\mathbb{R}^{n})$ is a proper subspace of $\mathbb{R}^{n}$ and therefore of measure zero. Moreover, $\det T=0$, so we see that the formula holds for the singular case. Now suppose that $T$ is invertible. We know that, for any Borel set $E$,

$\displaystyle\lambda_{n}(UDV(E))=\lambda_{n}(DV(E))$

Since $V(E)$ is again a Borel set (by continuity) and $\lambda_{n}(V(E))=\lambda_{n}(E)$, we may assume by an approximation argument that $V(E)$ is a rectangle $R$ of the form

$R=[a_{1},b_{1}]\times\cdots\times[a_{n},b_{n}]$

Then the measure of $R$ under the map $D$ is

$\begin{array}{lcl}\displaystyle\lambda_{n}(D(R))=\left|(d_{1}b_{1}-d_{1}a_{1})\cdots(d_{n}b_{n}-d_{n}a_{n})\right|&=&\displaystyle\left|d_{1}\cdots d_{n}\right|\cdot\lambda_{n}(R)\\[.7 em]&=&\displaystyle\left|\det D\right|\cdot\lambda_{n}(R),\end{array}$

where we use the determinant formula for a diagonal matrix to obtain the ultimate equality.