Near-Optimal Open Covers for Intervals and Null Sets

Open Covers with Controlled Length

An open cover can do more than contain a set: its intervals can be chosen so that their total length is as close as desired to the length of a bounded interval, or as small as desired for a countable set. The essential step is to allocate a positive length budget among the intervals, and then check both that every required point is covered and that the budget sums to the stated bound.

All statements below concern Lebesgue measure on \((\mathbb{R},\mathcal{L},m)\), which is complete and sigma-finite, but not finite. We use the established facts that every interval is Lebesgue measurable and has measure equal to its length, and that \(m\) is monotone and countably subadditive. For an interval with finite endpoints \(a\le b\), write its length as \(b-a\), regardless of whether its endpoints are included. The empty interval has length zero. In particular, a singleton has measure zero.

Proof. First suppose \(I\) is nonempty and has endpoints \(a<b\), so \(I\) is one of the intervals with endpoints \(a,b\), with any allowed choice of including or excluding those endpoints. Set \(J=(a-\varepsilon/2,b+\varepsilon/2)\). Since \(a-\varepsilon/2<a\le x\le b<b+\varepsilon/2\) for every \(x\in I\), we have \(I\subset J\), and \(J\) is open. Its measure is its length: \(m(J)=b+\varepsilon/2-(a-\varepsilon/2)=b-a+\varepsilon=m(I)+\varepsilon\). If \(I=\{a\}\), take \(J=(a-\varepsilon/2,a+\varepsilon/2)\). It contains \(I\), and \(m(J)=\varepsilon=m(I)+\varepsilon\). If \(I=\varnothing\), take \(J=\varnothing\), which is open and has measure zero. These cases prove the claim. \(\square\)

The equality in the nondegenerate case is a convenient explicit bound. The cover is also optimal up to the prescribed tolerance: if \(O\) is any measurable set containing \(I\), monotonicity gives \(m(O)\ge m(I)\). Thus no cover can have measure below the interval’s measure, while the construction adds only the chosen \(\varepsilon\). The theorem controls the measure of one open covering interval; here its measure and its length are exactly the same.

Covering Countable Sets by a Length Budget

For a countable set, there is no single finite interval that need contain the entire set. Instead, assign a small open interval to each point. The lengths must form a convergent series: giving every point the same positive length would give an infinite total for an infinite set. A geometric allocation provides a direct construction.

Proof. If \(E\) is empty, take \(U=\varnothing\). If \(E\) is nonempty and finite or countably infinite, list its distinct points as \(x_1,\ldots,x_N\) in the finite case, or as \(x_1,x_2,\ldots\) in the infinite case. For each listed point define \(J_n=(x_n-\varepsilon/2^{n+1},x_n+\varepsilon/2^{n+1})\). Then \(x_n\in J_n\), and the length of \(J_n\) is \(2(\varepsilon/2^{n+1})=\varepsilon/2^n\). Set \(U=\bigcup_nJ_n\). This is open, since it is a union of open intervals, and it contains \(E\). In the infinite case the sum of the interval lengths is \(\sum_{n=1}^{\infty}\varepsilon/2^n=\varepsilon\); in the finite case the sum is at most \(\varepsilon\), because it is a finite partial sum of positive terms of that series. Countable subadditivity gives \(m(U)\le\sum_n m(J_n)=\sum_n\varepsilon/2^n\le\varepsilon\). Since \(E\subset U\), monotonicity yields \(m^*(E)\le\varepsilon\) for every \(\varepsilon>0\), where \(m^*\) is Lebesgue outer measure. As outer measure is nonnegative, this implies \(m^*(E)=0\). A set of outer measure zero is Lebesgue measurable by the Caratheodory criterion; its Lebesgue measure equals its outer measure, so \(m(E)=0\). \(\square\)

The lengths in this construction are budgets for the covering intervals, not claims that the intervals are disjoint. If intervals overlap, the measure of their union can be strictly smaller than the sum of their lengths. Countable subadditivity gives the needed upper bound in either case; equality is not required.

Common Pitfalls and the Meaning of the Bound

Check the center, not just the radius. A proposed interval can have the intended small length and still fail to cover its assigned point if its center is wrong. For instance, \((\tfrac12-\tfrac1{32},\tfrac12+\tfrac1{32})=(\tfrac{15}{32},\tfrac{17}{32})\) has midpoint \((15/32+17/32)/2=1/2\), not \(1\). It does not contain \(1\), since \(17/32<1\). Likewise, \((\tfrac13-\tfrac1{64},\tfrac13+\tfrac1{64})=(\tfrac{61}{192},\tfrac{67}{192})\) has midpoint \(1/3\), not \(1/2\). If the intended first centers are \(1\) and \(1/2\), the correct intervals are \((1-\tfrac1{32},1+\tfrac1{32})=(\tfrac{31}{32},\tfrac{33}{32})\) and \((\tfrac12-\tfrac1{64},\tfrac12+\tfrac1{64})=(\tfrac{31}{64},\tfrac{33}{64})\). Each point is then the midpoint and lies strictly inside its open interval.

Do not confuse a sum of lengths with the measure of a union. For intervals \(J_n\), countable subadditivity says \(m(\bigcup_nJ_n)\le\sum_nm(J_n)\); it does not say equality. Overlaps can make the union smaller. The constructions rely only on the upper bound. They also require a positive, summable allocation: assigning an interval of fixed length \(c>0\) to every point of an infinite set gives total length \(\sum_{n=1}^{\infty}c=\infty\), not an arbitrarily small cover.

For a bounded interval, one enlarged interval gives a measure within the prescribed excess of the interval’s measure, and monotonicity explains why its measure cannot be smaller than that of the interval it contains. For a countable set, the geometric allocation works point by point and proves nullity through covers of arbitrarily small total length. These are distinct constructions: the interval estimate uses its two endpoints, while the countable-set estimate depends on distributing a finite length budget across a list.