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 is an isometry, then

where denotes the -dimensional Lebesgue measure and denotes the Borel -algebra on .

Today’s goal is to obtain a change-of-measure formula, when is some arbitrary affine transformation. Since the Lebesgue measure is translation-invariant, we may assume that 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 as

,

where are unitary operators and is diagonalizable with respect to the standard basis. Since the determinant is a multiplicative homomorphism and unitary operators have unit determinant, we see that

,

where are the diagonal entries of . If is non-invertible, then the image is a proper subspace of and therefore of measure zero. Moreover, , so we see that the formula holds for the singular case. Now suppose that is invertible. We know that, for any Borel set ,

Since is again a Borel set (by continuity) and , we may assume by an approximation argument that is a rectangle of the form

Then the measure of under the map is

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

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