The Length Lower Bound for Interval Covers

Why a Countable Cover Has a Length Cost

An interval may be covered by many smaller intervals, with overlaps and with endpoints included or excluded in different ways. The basic constraint is that the sum of their lengths cannot be smaller than the length of the interval being covered. The finite version follows from an elementary estimate for open intervals. Compactness then lets us use that estimate even when the original cover has infinitely many members.

All intervals in this discussion are subsets of \(\mathbb{R}\). For a nonempty bounded interval \(I\), define its length by \(|I|=\sup I-\inf I\). This definition does not depend on whether either endpoint belongs to \(I\); in particular, a singleton has length zero. An empty interval is assigned length zero, and an unbounded interval has length \(+\infty\). A countable cover \((I_n)_{n\ge1}\) has total length \(\sum_{n=1}^{\infty}|I_n|\), where the sum is allowed to be infinite.

The proof has two parts. First, a finite open cover of a closed segment must have enough total length to reach from one endpoint to the other. Second, intervals in a countable cover can be enlarged by a total length as small as desired. After enlargement, compactness supplies a finite subcover.

A Finite Estimate for Open Covers

Proof. Write each \(U_j=(u_j,v_j)\). Choose one of these intervals containing \(c\), and call it the first selected interval. Its right endpoint \(r_1\) satisfies \(r_1>c\), and its length satisfies \(|U_{j_1}|=v_{j_1}-u_{j_1}>r_1-c\), since \(u_{j_1}<c\). If \(r_1\ge d\), this one interval already has length greater than or equal to \(d-c\).

Otherwise, suppose the selected intervals so far have right endpoints with maximum \(r_k<d\). The point \(r_k\) lies in \([c,d]\), so it is covered by some \(U_j=(u_j,v_j)\). Since this interval is open and contains \(r_k\), \(u_j<r_k<v_j\). It cannot be one of the intervals already selected, because all their right endpoints are at most \(r_k\). Select it next and put \(r_{k+1}=v_j\); this is a strict extension of the covered reach.

Because there are only finitely many intervals, this process reaches a selected right endpoint \(r_K\ge d\). For each new interval after the first, its length satisfies \(|U_{j_k}|=v_{j_k}-u_{j_k}>r_k-r_{k-1}\), since \(u_{j_k}<r_{k-1}\) and \(v_{j_k}=r_k\). Adding these inequalities, including the first interval’s estimate, gives \(\sum_{k=1}^{K}|U_{j_k}|>r_K-c\ge d-c\). The selected intervals are distinct, and all lengths are nonnegative, so their sum is at most \(\sum_{j=1}^{N}|U_j|\). Therefore \(\sum_{j=1}^{N}|U_j|\ge d-c\). \(\square\)

From Finite Covers to Countable Covers

We first prove the countable-cover estimate for a closed segment \([c,d]\), with \(c<d\). If any covering interval is unbounded, the total length is infinite and the estimate holds. Empty intervals can be discarded. Thus we may assume the intervals are bounded and nonempty; write \(p_n=\inf I_n\) and \(q_n=\sup I_n\). If the total length is infinite, the estimate again holds, so suppose \(S=\sum_{n\ge1}(q_n-p_n)<\infty\).

Fix \(\varepsilon>0\). Enlarge each interval to the open interval \(U_n=(p_n-\delta_n,q_n+\delta_n)\), where \(\delta_n=\varepsilon/2^{n+1}\). Every point of \(I_n\) belongs to \(U_n\), regardless of the endpoint convention for \(I_n\). The enlarged intervals cover \([c,d]\), and their total length is controlled by \(\sum_{n\ge1}|U_n|=S+2\sum_{n\ge1}\delta_n=S+\varepsilon\), because \(\sum_{n\ge1}2^{-(n+1)}=1/2\).

The closed bounded interval \([c,d]\) is compact. By the Heine–Borel theorem, every open cover of it has a finite subcover. Apply this to \((U_n)\) and use the finite-cover lemma. The sum of the lengths in the finite subcover is at least \(d-c\), and it is at most the sum of the lengths of all the enlarged intervals. Consequently \(d-c\le S+\varepsilon\). This is true for every \(\varepsilon>0\), so \(d-c\le S\): if \(d-c>S\), choosing \(\varepsilon=(d-c-S)/2\) would contradict \(d-c\le S+\varepsilon\).

Now let the target \(I\) be any bounded interval with endpoints \(\alpha=\inf I\) and \(\beta=\sup I\), where \(\alpha<\beta\). Every point strictly between \(\alpha\) and \(\beta\) belongs to \(I\): there are points of \(I\) on either side, and \(I\) contains every point between any two of its points. Thus, for \(0<\eta<(\beta-\alpha)/2\), the closed segment \([\alpha+\eta,\beta-\eta]\) is contained in \(I\). Applying the closed-segment result gives \(\beta-\alpha-2\eta\le\sum_n|I_n|\). Letting \(\eta\) decrease to zero proves \(|I|=\beta-\alpha\le\sum_n|I_n|\). If \(I\) is empty or a singleton, its length is zero and the result follows from nonnegativity of the sum. This completes the proof for all bounded intervals. \(\square\)

What the Estimate Does—and Does Not—Say

The lower bound is a statement about every particular cover, not a claim that a cover of minimum total length must exist. Its usefulness is that it rules out covers that are too cheap. For example, the definition of Lebesgue outer measure \(m^*\) takes an infimum of total lengths over interval covers. The theorem supplies a lower bound for each such total; identifying the exact outer measure of particular intervals requires a matching upper bound as well.

Common pitfall: checking only the interior of the target. The two open intervals \((0,1/2)\) and \((1/2,1)\) have total length \(1\), but they do not cover \([0,1]\): they omit \(0\), \(1/2\), and \(1\). It is therefore invalid to apply the finite-cover lemma to that pair and claim it covers the closed segment. A valid open cover is \((-1/10,3/5)\) and \((2/5,11/10)\). The first contains \([0,1/2]\), and the second contains \([1/2,1]\); their total length is \(7/10+7/10=14/10\ge1\). In the countable-cover proof, the slight enlargement ensures that the sets really are an open cover before compactness is used.