Lemma 1. Let be open. If is locally Lipschitz on , then the set is measurable.
Proof. Let be measurable. Replacing by , we may assume that is bounded. Observe that by the inner regularity of the Lebsgue measure, we may write
where the sets are compact and is a set of measure zero. Since the set is a Borel set, being the union of compact sets, it suffices to show that is measurable. Fix . Since has measure zero, we can cover by a sequence of cubes with edge lengths and with sum of measures less than . Since is locally Lipschitz, it has Lipschitz constant on the set . Hence, is contained in a ball of radius . Hence, is contained a cube with edge length . Thus,
Since was arbitrary, we conclude that has measure zero, and therefore necessarily measurable.
Recall that we say that a real matrix is orthogonal if is invertible and . There are a number of properties which equivalently characterize an orthogonal matrix. We summarize them with the following proposition.
Proposition 2. The following are equivalent:
- The matrix is orthogonal.
- The columns of form an orthonormal basis.
- , for all vectors .
- The linear transformation defined by is an isometry.
Proof. In what follows, denotes the standard basis of , and denotes the column vectors of the matrix . Suppose (1) holds. Then, for ,
Now suppose that (2) holds. By hypothesis, for any vectors , we can write
Using the bilinearity of the inner product, we obtain that
(3)(4) is obvious. To see that (4)(1), we will prove the more general result that any isometry of is an affine transformation. Recall that an isometry of Euclidean space is a function which preserves Euclidean distance: , where denotes the Euclidean norm. It turns out that every isometry of is an affine map such that
and is orthogonal. To see this, note that by the polarization identity, the map peserves the Euclidean inner product. Hence, for any and , we see that
So, is linear, equivalently is affine. That is orthogonal is an immediate consequence of the preservation of the inner product under .
Although every orthogonal matrix necessarily has determinant satisfying , this condition is not sufficient for a matrix to be orthogonal. Indeed, consider the matrix
which has both determinant and orthogonal columns.
If is a linear transformation given by an orthogonal matrix, then is tautologically invertible and moreover, its inverse is given by an orthogonal matrix. Consider the pushforward of the Lebesgue measure, which we denote by . If we can show that is a translation-invariant measure on the measure space , and , then the uniqueness of the Lebesgue measure (see Proposition 5 here) proves that . Replacing by , we obtain the desired invariance property of Lebesgue measure.
For any Borel set and , we have that
where we use the translation-invariance of the Lebesgue measure in the penultimate equality. By Theorem 6 here, we conclude that , for some nonzero real constant . To see that , observe that since is orthogonal,
If is the open unit ball in , then . We conclude that
and therefore .