Lebesgue measure is invariant under isometries
A friend of mine and I had a conversation once about simple results in mathematics with cumbersome proofs. He suggested as an example the fact that Lebesgue measure on \(\mathbb{R}^n\) is invariant under isometries. At first I agreed with him: it seemed to us that the only reasonable way to do it was to show that the Lebesgue measure of a rotated box is just the product of the length of its edges (once you’ve shown that, the result is clear since the pushforward of Lebesgue measure under a rotation is just “Lebesgue measure defined in terms of rotated boxes”). If you have to do it that way, it is annoying: to compute the measure of the rotated box you have to “pixelate” it, i.e., approximate it with boxes with edges parallel to the coordinates axes.
But after thinking about it a little bit I realized it is much easier than that. This proof is probably well known but the first book I thought to check didn’t have it (I forget which book that was), and I was too lazy to check any others.
Lebesgue measure is the only measure on \(\mathbb{R}^n\) defined on all Borel subsets, invariant under translations and such that the measure of \([0,1)^n\) is 1.
Let \(\mu\) be a measure satisfying those conditions. Using translation invariance and additivity we get succesively that
- \(\mu(\prod [0,m_i)) = \prod m_i\) for positive integers \(m_i\),
- \((\prod n_i) \cdot \mu(\prod [0,m_i/n_i)) = \prod m_i\), for positive integers \(m_i\) and \(n_i\),
- \(\mu\) coincides with Lebesgue measure on any product of half-open intervals with rational endpoints.1
Since these products generate the \(\sigma-\)algebra of Borel sets, \(\mu\) must agree with Lebesgue measure.
Lebesgue measure is invariant under isometries.
It certainly is invariant under translations. Let \(S\) be an isometry fixing the origin and consider \(S_* \lambda := \lambda \circ S^{-1}\), the pushforward of Lebesgue measure. If \(T_v\) denotes the translation by \(v\), we have that \((T_v)_* S_* \lambda = (T_v \circ S)_* \lambda = (S \circ T_{S^{-1} v})_* \lambda = S_* (T_{S^{-1}v})_* \lambda = S_* \lambda\), so \(S_* \lambda\) is invariant under translations. By the lemma, this means that \(S_* \lambda = c \lambda\) for some constant2 \(c\). Applying both sides to a ball \(B\) around the origin we get \(\lambda(B) = \lambda(S^{-1} B) = c \lambda(B)\), from which \(c=1\) and \(S_* \lambda = \lambda\).