Exact Outer Measure of Closed Intervals

From Endpoint Extensions to Exact Outer Measure

For an open interval, the interval itself provides a cover whose length equals the distance between its endpoints. A closed interval presents a small but important difference: an open interval with the same endpoints omits those endpoints. The remedy is to extend the covering interval slightly beyond each endpoint. The added length can be made arbitrarily small, while the lower bound for interval covers prevents any cover from having total length less than the interval's length.

Throughout, \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\), defined for every set \(E\subseteq\mathbb{R}\), whether or not \(E\) is measurable. A countable open-interval cover of \(E\) is a finite or countable collection of open intervals \(I_k=(u_k,v_k)\) whose union contains \(E\). Its total length is \(\sum_k(v_k-u_k)\). The outer measure is the infimum of these total lengths over all such covers. The infimum need not be achieved by a particular cover.

The key lower-bound fact, established in the earlier material on interval-cover length lower bounds, is this: if a countable collection of open intervals covers \([a,b]\), then the sum of their lengths is at least \(b-a\). This applies to every admissible cover, not only to a particular choice of endpoint extensions. For the upper bound, it is enough to give covers whose lengths approach \(b-a\) from above.

Proof by Two Bounds

Proof. First suppose \(a<b\). Let \(\varepsilon>0\). The open interval \[ I_\varepsilon=\left(a-\frac{\varepsilon}{3},\,b+\frac{\varepsilon}{3}\right) \] contains every point of \([a,b]\): its left endpoint is strictly less than \(a\), and its right endpoint is strictly greater than \(b\). Its length is \[ \left(b+\frac{\varepsilon}{3}\right)-\left(a-\frac{\varepsilon}{3}\right) =b-a+\frac{2\varepsilon}{3} <b-a+\varepsilon. \] Since this is a valid cover, the definition of outer measure gives \(m^*([a,b])\le b-a+\varepsilon\) for every \(\varepsilon>0\). If \(m^*([a,b])>b-a\), choosing \(\varepsilon\) smaller than \(m^*([a,b])-(b-a)\) would contradict that inequality. Hence \(m^*([a,b])\le b-a\).

For the reverse inequality, take any countable open-interval cover \((I_k)\) of \([a,b]\). By the interval-cover length lower bound, \(\sum_k |I_k|\ge b-a\). Taking the infimum over all such covers gives \(m^*([a,b])\ge b-a\). Combining the two bounds proves equality when \(a<b\). If \(a=b\), then \([a,b]=\{a\}\). For any \(\varepsilon>0\), the interval \((a-\varepsilon/2,a+\varepsilon/2)\) covers \(\{a\}\) and has length \(\varepsilon\), so \(0\le m^*(\{a\})\le\varepsilon\) for every \(\varepsilon>0\), which implies \(m^*(\{a\})=0=b-a\). Thus the formula holds for all \(a\le b\). \(\square\)

Proof. If \(a<b\), the single interval \((a-\eta/3,b+\eta/3)\) covers \([a,b]\) and has length \(b-a+2\eta/3<b-a+\eta\). If \(a=b\), the interval \((a-\eta/3,a+\eta/3)\) has length \(2\eta/3<\eta=b-a+\eta\) and covers the singleton. In both cases, the lower bound for every cover follows from the interval-cover length lower bound when \(a<b\); when \(a=b\), it follows because lengths are nonnegative. \(\square\)

Worked Endpoint Extensions

Why the Lower Bound Matters

The endpoint extension alone proves only that outer measure is no larger than the interval length: it constructs efficient covers. It cannot rule out the possibility of a very different countable cover with even smaller total length. The lower bound is what rules that out. It applies to every admissible cover, including covers that use many intervals and do not resemble the single extended interval used for the upper bound.

A common mistake is to use \((a,b)\) as a cover of \([a,b]\) and record its length as \(b-a\). The length calculation is correct, but the cover is not: \(a\notin(a,b)\) and \(b\notin(a,b)\), so it fails the requirement \([a,b]\subseteq\bigcup_k I_k\). For example, \((2,5)\) has length \(5-2=3\), but neither 2 nor 5 belongs to it. Replacing it with \((2-\delta,5+\delta)\), where \(\delta>0\), makes it a valid cover, with length \(3+2\delta\). As \(\delta\) decreases, these valid lengths approach 3; the infimum is 3 even though no zero extension is used.

The calculation concerns outer measure, which is defined through covers and applies to all subsets of \(\mathbb{R}\). No measurability assumption is needed to define \(m^*([a,b])\). The proof also illustrates the role of the infimum: one need not find a cover of length exactly \(b-a\). It suffices to show that every cover has length at least \(b-a\), and that for each positive tolerance there is a cover whose length is less than \(b-a\) plus that tolerance.