Comparing Covers of the Same Interval

What Does a Cover’s Total Length Tell Us?

A single interval can cover a fixed interval with little bookkeeping, but competing covers may overlap, split the target into pieces, or use infinitely many intervals. Comparing them requires two separate checks: first verify that every point of the target is covered; then add the lengths of the covering intervals. The sum is an upper bound for outer measure only when the family really is a cover. Even for a valid cover, its total length need not be the best possible estimate.

We work on \(\mathbb{R}\) with Lebesgue outer measure \(m^*\), not with a measure on a sigma-algebra. For a bounded interval \(J\), its length \(\ell(J)\) is its supremum minus its infimum; endpoint inclusion does not change this length. A singleton has length zero, and an unbounded interval has infinite length. If \(J_1,\ldots,J_N\) is a finite cover of a set \(E\), or if \((J_n)_{n\ge1}\) is a countable cover, its total length is the corresponding sum of nonnegative lengths.

We use the length lower bound for interval covers, established earlier in the material on Lebesgue outer measure: if intervals \(J_j\) cover a bounded interval \([a,b]\), then \(b-a\le\sum_j\ell(J_j)\). It applies to finite and countable covers, and to intervals with any endpoint conventions. In particular, no valid cover of \([0,1]\) can have total length less than \(1\). The bound does not say that every cover has total length \(1\); overlaps and expansions can make the sum larger.

Finite Covers: Overlap Creates Excess

The comparison illustrates why the number of intervals does not determine the quality of a cover. Two intervals can have a larger total length than one interval if they overlap, and a split into pieces can preserve the total exactly. The choice between \([0,1/3]\) and \((1/3,1]\) matters for coverage at the shared endpoint: replacing both by intervals that exclude \(1/3\) would leave that point uncovered.

A Countable Cover with No Excess

A countable cover can have total length exactly equal to the length of the target interval. The following construction divides \([0,1)\) into consecutive pieces whose lengths form a geometric series, and adds the endpoint \(1\) as a zero-length interval. Endpoint conventions ensure that no point is lost between consecutive pieces.

Proof. Define \(P_1=[0,1/2]\), and for each integer \(n\ge2\) define \(P_n=(1-2^{-(n-1)},1-2^{-n}]\). Add \(P_*=\{1\}\). The intervals \(P_1,P_2,\ldots\) cover \([0,1)\): \(P_1\) covers \([0,1/2]\), and each \(P_n\), for \(n\ge2\), begins just above the right endpoint of \(P_{n-1}\) and includes its own right endpoint. More explicitly, the right endpoint \(1-2^{-(n-1)}\) of \(P_{n-1}\) is included in \(P_{n-1}\), and points strictly between it and \(1-2^{-n}\), together with \(1-2^{-n}\), belong to \(P_n\). The right endpoints \(1-2^{-n}\) increase to \(1\). The additional interval \(P_*\) covers \(1\), so the family covers all of \([0,1]\). The length of \(P_1\) is \(1/2=2^{-1}\), and for \(n\ge2\) the length of \(P_n\) is \( (1-2^{-n})-(1-2^{-(n-1)})=2^{-n}\). Also \(\ell(P_*)=0\). Therefore the total length is \(\sum_{n=1}^{\infty}2^{-n}+0=1\), by the geometric-series formula. The cover has total length \(1\), as claimed. \(\square\)

This example also clarifies the difference between a countable cover and a disjoint decomposition. Here the pieces do not overlap, and each boundary point below \(1\) belongs to the piece on its left. The singleton \(\{1\}\) contributes nothing to the sum of lengths, but it is necessary: none of the \(P_n\) contains \(1\).

Open Covers Can Be Arbitrarily Close to Sharp

Sometimes a desired cover is required to use open intervals. A half-open partition is then not available as written. Instead, expand finitely many consecutive pieces slightly past their endpoints. The added lengths can be controlled, and the total can be made as close to \(1\) as desired while remaining strictly greater than \(1\).

Proof. Fix \(\varepsilon>0\) and a positive integer \(N\), and set \(\delta=\varepsilon/(2N)\). For \(k=1,\ldots,N\), define the open interval \(J_k=((k-1)/N-\delta,\ k/N+\delta)\). Every point of the closed cell \([(k-1)/N,k/N]\) lies strictly between the endpoints of \(J_k\), because \(\delta>0\). These \(N\) closed cells together cover \([0,1]\), so the open intervals \(J_k\) cover \([0,1]\). Each has length \(1/N+2\delta=1/N+\varepsilon/N\); hence the total length is \(N(1/N+\varepsilon/N)=1+\varepsilon\). The length lower bound for interval covers gives a total of at least \(1\) for every open cover of \([0,1]\). Since for every \(\varepsilon>0\) the construction gives a total \(1+\varepsilon\), the infimum is both at least \(1\) and at most \(1+\varepsilon\) for every positive \(\varepsilon\), and therefore equals \(1\). \(\square\)

For a smaller excess, the same construction uses a smaller \(\varepsilon\). The lower bound explains why this improvement has a limit: no valid family, open or otherwise, can have total length strictly below \(1\). The construction establishes an infimum, not necessarily an open cover whose total equals that infimum.

A Coverage Error to Avoid

A common mistake is to add interval lengths before checking that the intervals cover the target. For example, \([0,2/5]\) and \([3/5,1]\) have lengths \(2/5\) and \(2/5\), so their total is \(4/5\). But their union omits every point in \((2/5,3/5)\), and therefore does not cover \([0,1]\). The value \(4/5\) is not an upper bound obtained from this family for \(m^*([0,1])\). Correct the family by adding \([2/5,3/5]\). The three intervals then cover \([0,1]\), including both gap endpoints, and the total is \(2/5+1/5+2/5=1\). This agrees with the lower bound.

When comparing covers, keep the following questions separate:

• Coverage: Is every point of the target included in at least one interval? Check boundary points as well as interior points.
• Length: What is each interval’s length, and what is the sum? Overlap does not reduce the sum of the individual lengths.
• Sharpness: Does the sum equal the target interval’s length, or is it merely an upper estimate with excess?