Coordinate Permutations and Reflections

Coordinate Changes That Preserve Volume

A coordinate permutation changes the order in which coordinates are listed, while a coordinate reflection changes the sign of one or more coordinates. Neither operation stretches or compresses any coordinate interval, so the volume of a box should not change. To extend that observation from boxes to all Lebesgue measurable sets, we will first verify preservation of outer measure and then apply the Carathéodory criterion for measurability.

Throughout, \(n\ge1\), and we work in \((\mathbb{R}^n,\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This measure space is complete and sigma-finite, but not finite when \(n\ge1\). Sets considered in the main measure-preservation theorem are Lebesgue measurable; the outer-measure statement applies to every subset of \(\mathbb{R}^n\). We use the earlier construction of Lebesgue outer measure by countable covers with bounded half-open boxes. The volume of \(B=\prod_{i=1}^n[a_i,b_i)\), where each \(a_i<b_i\) is finite, is \(\prod_{i=1}^n(b_i-a_i)\).

Permutations and reflections can be combined in one map \(T\). The crucial fact about its action on a box is that every side length is preserved, although the endpoint inclusion can change. For example, reflection sends \([a,b)\) to \((-b,-a]\), not to \([-b,-a)\). Since the outer-measure construction uses half-open boxes, we account for this endpoint change in the proof below.

Outer Measure and Measurability

Proof. The map \(T\) is a bijection and sends a half-open box to a rectangle whose coordinate intervals have the same lengths as the original intervals, but may have different endpoint inclusions. Let \(B=\prod_{i=1}^n[a_i,b_i)\) be a box in a countable cover of \(A\). For any \(\delta>0\), each coordinate interval of \(T(B)\) is contained in a half-open interval of the form \([c_i-\delta,d_i+\delta)\), where \(d_i-c_i=b_{\pi(i)}-a_{\pi(i)}\). Thus \(T(B)\) is contained in a half-open box with volume \(\prod_{i=1}^n(b_{\pi(i)}-a_{\pi(i)}+2\delta)\). As \(\delta\) decreases to zero, this volume tends to \(\prod_{i=1}^n(b_i-a_i)\). Consequently, for every \(\eta>0\), \(T(B)\) can be covered by a half-open box whose volume is less than \(m^*(B)+\eta\), where \(m^*(B)=\prod_i(b_i-a_i)\).

Now let \(B_1,B_2,\ldots\) be any countable half-open box cover of \(A\) with finite total volume. For each \(j\ge1\), cover \(T(B_j)\) by a half-open box of volume less than \(m^*(B_j)+\eta 2^{-j}\). These boxes cover \(T(A)\), so the definition of outer measure gives \(m^*(T(A))\le\sum_{j=1}^\infty m^*(B_j)+\eta\). Taking the infimum over covers and then letting \(\eta\downarrow0\) gives \(m^*(T(A))\le m^*(A)\). If \(m^*(A)=\infty\), this inequality is automatic. The inverse \(T^{-1}\) is also a coordinate permutation and reflection map, so the same argument applied to \(T^{-1}\) gives \(m^*(A)\le m^*(T(A))\). Therefore the two outer measures are equal. \(\square\)

Proof. Let \(E\in\mathcal{L}\). By the Carathéodory criterion, for every set \(A\subseteq\mathbb{R}^n\), \(m^*(A)=m^*(A\cap E)+m^*(A\setminus E)\). To check the criterion for \(T(E)\), take any \(C\subseteq\mathbb{R}^n\) and put \(A=T^{-1}(C)\). Bijection gives \(T^{-1}(C\cap T(E))=A\cap E\) and \(T^{-1}(C\setminus T(E))=A\setminus E\). Outer-measure preservation, already proved for every subset, therefore gives \(m^*(C)=m^*(A)=m^*(A\cap E)+m^*(A\setminus E)=m^*(C\cap T(E))+m^*(C\setminus T(E))\). The Carathéodory criterion implies \(T(E)\in\mathcal{L}\). Applying outer-measure preservation to \(E\), and using that \(m=m^*\) on Lebesgue measurable sets, gives \(m(T(E))=m(E)\). \(\square\)

This theorem applies to sets of finite or infinite measure. In particular, it does not require a bounded set or a finite measure space. It also says that images and inverse images of Lebesgue measurable sets are measurable, since \(T^{-1}\) is another map of the same type.

Worked Images and Measures

Endpoint Pitfall and Scope

A common mistake is to reflect \([a,b)\) and write \([-b,-a)\). This is not the image: it gets the endpoint inclusions wrong. For example, reflecting \([1,3)\) gives \((-3,-1]\), because \(1\) is included and maps to \(-1\), while \(3\) is excluded and maps to \(-3\). The incorrect interval \([-3,-1)\) includes a point that is not in the image and omits a point that is. The two intervals nevertheless have the same Lebesgue measure, since changing endpoints changes only a finite set of points of measure zero. Equality of measures does not make the sets themselves equal.

The preservation result concerns only coordinate permutations and reflections. It establishes their volume factor as \(1\), using box covers and the Carathéodory criterion. It does not establish a volume formula for coordinate scalings or for general linear maps; those transformations require additional arguments.