Translating Explicit Interval Covers

Translating a Cover Without Changing Its Cost

A cover of a set is a collection of intervals whose union contains that set. If every interval is shifted by the same amount, the covered set shifts by that amount as well. The key calculation is that shifting both endpoints of an interval does not change the difference between them. This simple fact applies to finite covers and countable covers, and it allows us to compare the outer measures of a set and any translate of it.

Here \(E\) denotes an arbitrary subset of \(\mathbb{R}\), not necessarily a measurable set, and \(t\in\mathbb{R}\). Write \(E+t=\{x+t:x\in E\}\). We use Lebesgue outer measure \(m^*\) on all subsets of \(\mathbb{R}\), defined by taking the infimum of the total lengths of countable covers by bounded open intervals. Thus, if \(I_k=(a_k,b_k)\), its length is \(|I_k|=b_k-a_k\), and a cover \(E\subseteq\bigcup_{k=1}^{\infty} I_k\) has total length \(\sum_{k=1}^{\infty}(b_k-a_k)\). The sum and the outer measure may be \(+\infty\). The empty set has outer measure zero.

If \(E\subseteq\bigcup_k I_k\), then \(E+t\subseteq\bigcup_k(I_k+t)\): for \(x\in E\), choose an interval \(I_k\) containing \(x\); then \(x+t\in I_k+t\). This inclusion and the endpoint calculation are the two ingredients needed below. They do not require that the intervals be disjoint, that the cover be optimal, or that \(E\) be measurable.

Finite Covers: Track Each Interval

For a finite family \(I_1,\ldots,I_N\), the verification is term by term: \(|I_k+t|=|I_k|\) for every \(k\), so \(\sum_{k=1}^N|I_k+t|=\sum_{k=1}^N|I_k|\). The same conclusion holds for a countable family. Indeed, each finite partial sum of the translated lengths equals the corresponding original partial sum; taking the limit of these nondecreasing partial sums gives equal total sums, including when the value is infinite.

Countable Covers: The Same Calculation at Every Index

It is important not to confuse the cost of a displayed cover with the outer measure. A cover can overlap, include space outside the set, or simply fail to be close to optimal. Translation preserves the cost of each chosen cover; the outer measure is obtained only after taking the infimum over all covers.

Proof. Write \(I_k=(a_k,b_k)\). For each index \(k\), \(|I_k+t|=(b_k+t)-(a_k+t)=b_k-a_k=|I_k|\). For a finite family, summing these equalities gives equality of total lengths. For a countable family, the sums are the limits of their nondecreasing finite partial sums; corresponding partial sums are equal, so their limits are equal, whether finite or infinite. If \(x\in E\), the cover assumption gives some index \(k\) with \(x\in I_k\). Adding \(t\) yields \(x+t\in I_k+t\), and therefore \(E+t\subseteq\bigcup_k(I_k+t)\). \(\square\)

Translation Invariance of Lebesgue Outer Measure

Proof. First suppose \(E=\varnothing\). Then \(E+t=\varnothing\), so both outer measures are zero. Now suppose \(E\ne\varnothing\) and \(m^*(E)<\infty\). For every \(\varepsilon>0\), the definition of an infimum gives a countable interval cover \((I_k)\) of \(E\) whose total length is less than \(m^*(E)+\varepsilon\). By the proposition, \((I_k+t)\) covers \(E+t\) and has the same total length. Therefore \(m^*(E+t)\le m^*(E)+\varepsilon\) for every \(\varepsilon>0\), which implies \(m^*(E+t)\le m^*(E)\).

If \(m^*(E)=+\infty\), the inequality \(m^*(E+t)\le m^*(E)\) holds in the extended order, since every outer measure is at most \(+\infty\). Thus this inequality holds in all cases. Apply the same argument to \(E+t\) with translation \(-t\). Since \((E+t)-t=E\), it gives \(m^*(E)\le m^*(E+t)\). The two inequalities yield \(m^*(E+t)=m^*(E)\), including the case where the common value is infinite. \(\square\)

The infimum in the definition need not be attained. The proof therefore does not assume that a shortest cover exists: when the outer measure is finite, it uses a cover within an arbitrary \(\varepsilon\) of the infimum, and then lets \(\varepsilon\) decrease to zero. The reverse inequality is essential; a translated cover alone gives only one direction.

A Common Endpoint-Translation Mistake

A frequent error is to add \(t\) to only one endpoint when calculating a translated interval’s length. For example, \(I=(1/2,5/2)\) has length \(5/2-1/2=2\). Translating by \(3\) gives \(I+3=(7/2,11/2)\). The incorrect calculation \((11/2)-(1/2)=5\) leaves the original left endpoint in place. The correct calculation uses both translated endpoints: \((11/2)-(7/2)=4/2=2\). The interval has moved, but its length has not changed.

A second mistake is to infer equality of outer measures from just one cover. A cover of \(E\) gives an upper bound for \(m^*(E)\), and its translate gives the same upper bound for \(m^*(E+t)\). To prove equality, one must compare the infima in both directions. Translating covers by \(t\) gives one inequality; translating covers of \(E+t\) by \(-t\) gives the other. This argument works for arbitrary sets and does not claim that any particular cover realizes the infimum.