Translation Invariance from Translated Covers

Why Translate the Cover?

Lebesgue outer measure on \(\mathbb{R}\) is defined through the costs of interval covers. A translation moves every point by the same real number, so it also moves every interval in a cover without changing that interval’s length. This observation gives a direct proof that outer measure is unchanged by translation; the reverse inequality comes from translating back.

Throughout, \(m^*\) denotes Lebesgue outer measure on all subsets of \(\mathbb{R}\), not a measure restricted to measurable sets. For \(A\subseteq\mathbb{R}\), a countable open-interval cover is a finite or countable family \((I_n)\) of bounded open intervals whose union contains \(A\). If \(I_n=(a_n,b_n)\), its length is \(\ell(I_n)=b_n-a_n\). The cost of the cover is \(\sum_n\ell(I_n)\), with an infinite sum allowed. The empty family covers the empty set. The definition is

Every subset of \(\mathbb{R}\) has at least one such cover: the intervals \((-n,n)\), \(n\ge1\), cover \(\mathbb{R}\). Thus the infimum is defined in \([0,\infty]\). It need not be attained by any cover. For \(t\in\mathbb{R}\), write \(A+t=\{x+t:x\in A\}\) for the translate of \(A\).

Translating Intervals and Their Covers

Proof. A point \(y\) belongs to \(I+t\) exactly when \(y=x+t\) for some \(x\in(a,b)\). This is equivalent to \(a+t<y<b+t\), so \(I+t=(a+t,b+t)\). Its length is \((b+t)-(a+t)=b-a=\ell(I)\). For a countable family \((I_n)\), this equality holds for each \(n\), so the sums of the nonnegative lengths agree term by term, including when the sum is infinite. \(\square\)

If \(A\subseteq\bigcup_n I_n\), then \(A+t\subseteq\bigcup_n(I_n+t)\): for each \(x\in A\), some \(I_n\) contains \(x\), and therefore \(I_n+t\) contains \(x+t\). The lemma shows that the new cover has the same cost as the original one. This proves a one-sided bound on the infima. To obtain equality, the same reasoning must also be applied to the inverse translation by \(-t\).

Translation Invariance

Proof. First we show \(m^*(A+t)\le m^*(A)\). If \(m^*(A)=\infty\), this inequality holds in the extended order because every value in \([0,\infty]\) is at most \(+\infty\). Suppose instead that \(m^*(A)<\infty\). For any \(\varepsilon>0\), the definition of the infimum gives a countable interval cover \((I_n)\) of \(A\) whose cost is less than \(m^*(A)+\varepsilon\). Translating every interval by \(t\) covers \(A+t\) and has exactly the same cost. Hence \(m^*(A+t)<m^*(A)+\varepsilon\) for every \(\varepsilon>0\). If \(m^*(A+t)>m^*(A)\), choosing \(\varepsilon\) smaller than their positive difference would contradict this inequality. Thus \(m^*(A+t)\le m^*(A)\).

Apply the inequality just proved to the set \(A+t\) and translation \(-t\). Since \((A+t)+(-t)=A\), it gives \(m^*(A)\le m^*(A+t)\). Combining the two inequalities proves equality. This also covers the empty set, whose translate is empty, and the case where either outer measure is infinite. \(\square\)

Proof. Translating each interval in a cover of \(A\) gives a cover of \(A+t\) with the same cost, by the lemma. Translating a cover of \(A+t\) by \(-t\) gives a cover of \(A\), again with the same cost. These operations are inverse to one another, since translating by \(t\) and then by \(-t\) returns each interval to its starting position. Therefore the two collections of cover costs coincide. \(\square\)

Worked Cover Calculations

Common Mistakes and Scope

Mistake: changing only one endpoint. Translating an interval means adding \(t\) to both endpoints. For example, translating \((-1,2)\) by \(4\) gives \((-1+4,2+4)=(3,6)\), of length \(6-3=3\). The incorrect interval \((-1,6)\) has length \(6-(-1)=7\); it changes only the right endpoint and is not the translated interval.

Mistake: treating one inequality as equality. The fact that a translated cover has the same cost proves \(m^*(A+t)\le m^*(A)\), but that one inequality alone does not establish equality. The reverse inequality requires a cover of \(A+t\) translated by \(-t\). The proof uses the same interval-length calculation in each direction.

These arguments concern outer measure, defined for every subset of \(\mathbb{R}\) by an infimum over covers. They do not require that a set be Borel or Lebesgue measurable, and they do not assert that an outer-measure infimum is attained. Translation invariance is a structural property of the covering definition itself; later, when measure is restricted to measurable sets, this property remains available there as well.