Translation Invariance in Rn

Moving a Set Without Changing Its Measure

A translation changes where a set lies, but not its size as measured by Lebesgue measure. The key is to compare covers before and after moving the set: every box in a countable cover can be translated by the same vector, with all its side lengths unchanged. This observation proves invariance first for Lebesgue outer measure, which is defined for every subset of \(\mathbb{R}^n\). The Carathéodory criterion then shows that the translation of a Lebesgue-measurable set is measurable as well.

Throughout, the ambient measure space is \((\mathbb{R}^n,\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra and \(m\) is Lebesgue measure. This space is sigma-finite but not finite: \(\mathbb{R}^n\) has infinite measure. We also use Lebesgue outer measure \(m^*\), defined on all subsets of \(\mathbb{R}^n\). For \(a=(a_1,\ldots,a_n)\in\mathbb{R}^n\) and \(E\subseteq\mathbb{R}^n\), write \(E+a=\{x+a:x\in E\}\).

A bounded half-open box is a set \(Q=\prod_{i=1}^n[u_i,v_i)\), where \(u_i<v_i\) for each coordinate. Its coordinate volume is \(\operatorname{vol}(Q)=\prod_{i=1}^n(v_i-u_i)\). In the box-cover definition of \(m^*\), one takes the infimum of the sums of coordinate volumes over countable covers by such boxes. The established exact outer measure of rectangles says that a box is Lebesgue measurable and its measure is its coordinate volume.

Translating Box Covers

The box-cover argument works for arbitrary sets, whether or not they are measurable. If \(Q=\prod_{i=1}^n[u_i,v_i)\), then

The half-open endpoint convention is preserved: \(u_i\le x_i<v_i\) holds exactly when \(u_i+a_i\le x_i+a_i<v_i+a_i\). Translation is also a bijection, with inverse translation by \(-a\). Thus a countable box cover of \(S\) translates to a countable box cover of \(S+a\) with exactly the same sum of volumes.

Proof. Fix any \(S\subseteq\mathbb{R}^n\) and \(a\in\mathbb{R}^n\). First suppose \(m^*(S)<\infty\), and let \(\varepsilon>0\). By the definition of outer measure as an infimum of covering sums, there is a countable box cover \((Q_j)_{j\ge1}\) of \(S\) such that \(\sum_{j\ge1}\operatorname{vol}(Q_j)<m^*(S)+\varepsilon\). The boxes \(Q_j+a\) cover \(S+a\), and their volume sum is the same. Therefore \(m^*(S+a)\le m^*(S)+\varepsilon\) for every \(\varepsilon>0\), so \(m^*(S+a)\le m^*(S)\). If \(m^*(S)=\infty\), this inequality holds in the extended real order without choosing a cover. Apply the resulting inequality to \(S+a\) and the vector \(-a\). Since \((S+a)+(-a)=S\), it gives \(m^*(S)\le m^*(S+a)\). Hence \(m^*(S+a)=m^*(S)\), including the case where either value is infinite.

It remains to check measurability, since equality of outer measures alone does not, by itself, establish that a set is Lebesgue measurable. The Carathéodory criterion says that \(E\subseteq\mathbb{R}^n\) is Lebesgue measurable exactly when, for every \(A\subseteq\mathbb{R}^n\), \(m^*(A)=m^*(A\cap E)+m^*(A\setminus E)\). This criterion is applied to outer measure, so the test set \(A\) need not be measurable.

Now let \(E\in\mathcal{L}\) and fix \(a\in\mathbb{R}^n\). For an arbitrary \(A\subseteq\mathbb{R}^n\), set \(A'=A-a\). Because translation is a bijection, \(A\cap(E+a)=(A'\cap E)+a\) and \(A\setminus(E+a)=(A'\setminus E)+a\). The outer-measure invariance just proved, followed by the Carathéodory criterion for \(E\), gives

This holds for every \(A\), so the criterion proves \(E+a\in\mathcal{L}\). Its measure is the restriction of outer measure to measurable sets; consequently \(m(E+a)=m^*(E+a)=m^*(E)=m(E)\). This argument also covers \(m(E)=\infty\). Proof. \(\square\)

Worked Calculations

Common Pitfall: Using Endpoint Locations as Side Lengths

Translation invariance does not mean that the endpoints of a box are irrelevant to describing the box. It means that the measure depends on its side lengths, which are differences of endpoints, not on the endpoints themselves. For example, translating \([2,5)\times[-1,1)\) by \((10,-4)\) gives \([12,15)\times[-5,-3)\). A wrong calculation might multiply the translated upper endpoints, obtaining \(15\cdot(-3)=-45\), and treat that as an area. This is incorrect: a negative “area” cannot be the measure of a rectangle, and products of endpoints do not give its coordinate volume.

The correction is to subtract the lower endpoint from the upper endpoint in each coordinate. The translated widths are \(15-12=3\) and \(-3-(-5)=2\), so the translated box has measure \(3\cdot2=6\). For the original box, the widths are \(5-2=3\) and \(1-(-1)=2\), giving \(3\cdot2=6\) again. In particular, a box may lie partly or wholly at negative coordinates and still have nonnegative measure.

What the Invariance Result Gives

The box calculation is only the starting point. The outer-measure proof shows why the same principle applies even when a measurable set has no finite box decomposition: a countable cover can be shifted box by box, preserving its total cost. The Carathéodory step is what extends this outer-measure fact to the Lebesgue-measurable sets on which \(m\) is defined. For the empty set, the equality also holds: its translate is empty and both measures are zero. For sets of infinite measure, the theorem asserts equality in the extended real numbers, not a finite numerical value.

When a finite union of boxes overlaps, do not add the individual box measures without accounting for the overlap. One may partition the union into disjoint measurable pieces or subtract overlaps as in the second example. Translation preserves the measure of the union itself; the direct calculations confirm that side lengths and intersection volumes are unchanged. These calculations concern translations only; other coordinate transformations require separate arguments.