Why Covering Infima Need Not Be Attained

Infimum Does Not Mean Minimum

In the interval-cover definition of Lebesgue outer measure, we take the infimum of the total lengths of covers. An infimum is a lower bound that can be approached as closely as desired; it need not be the length of any single cover. The distinction is especially visible for a closed interval: open intervals can cover it with total lengths approaching its length, but no countable open-interval cover has total length exactly that length.

Throughout this tutorial, \(m^*\) denotes Lebesgue outer measure on \(\mathbb{R}\), defined for every set \(E\subseteq\mathbb{R}\); no measurability assumption on \(E\) is needed. A cover here means a finite or countable family of open intervals whose union contains \(E\). The length of a bounded open interval \((u,v)\), where \(u<v\), is \(v-u\); an unbounded interval has infinite length. The total length of a cover is the sum of its interval lengths, with \(+\infty\) allowed. The defining infimum is

For an interval \(I\) of finite length, the length lower bound for interval covers says that any countable interval cover of \(I\) has total length at least the length of \(I\). In particular, this gives a lower bound for every cover used in the definition above. The question in this tutorial is whether a cover with exactly that lower-bound length must exist. For closed intervals, it does not.

Explicit Covers Approaching the Length

This calculation determines the infimum, but the fact that the cover lengths converge to \(3\) does not produce a cover of length \(3\). The intervals \(J_n\) are covers at finite \(n\); their endpoint limits give \((2,5)\), which omits the endpoints of \([2,5]\). The next result rules out not just this particular limiting interval, but every possible countable open-interval cover of the closed interval.

Why No Open Cover Attains the Infimum

Proof. The length lower bound gives \(m^*([a,b])\ge b-a\). The single open interval \((a-\varepsilon/2,b+\varepsilon/2)\), for any \(\varepsilon>0\), covers \([a,b]\) and has length \(b-a+\varepsilon\). Since this holds for every \(\varepsilon>0\), the infimum is at most \(b-a\). Hence \(m^*([a,b])=b-a\).

It remains to prove that no cover has total length exactly \(b-a\). Let \(\{I_k\}\) be any finite or countable open-interval cover of \([a,b]\). The closed bounded interval \([a,b]\) is compact, so the open cover has a finite subcover \(I_1,\ldots,I_N\). Let \(U=I_1\cup\cdots\cup I_N\). Since \(a\in U\) and \(U\) is open, there is \(r_a>0\) such that \((a-r_a,a+r_a)\subseteq U\). Since \(b\in U\) and \(U\) is open, there is \(r_b>0\) such that \((b-r_b,b+r_b)\subseteq U\). Choose \(\delta=\min(r_a,r_b)/2>0\). The intervals just found cover the portions of \((a-\delta,b+\delta)\) outside \([a,b]\), and \(U\) covers all of \([a,b]\). Thus \(I_1,\ldots,I_N\) cover \((a-\delta,b+\delta)\).

By the length lower bound, the sum of the lengths of these finitely many intervals is at least the length of the interval they cover, namely \(b-a+2\delta\). The total length of the original cover is at least the sum for this finite subcover, because all interval lengths are nonnegative. Therefore its total length is at least \(b-a+2\delta>b-a\). This proves strict inequality for every cover. As the infimum is \(b-a\), no cover attains it. \(\square\)

A Common Mistake and Its Correction

A tempting but incorrect calculation is: “The interval \((2,5)\) has length \(3\), so it is a cover of \([2,5]\) with the minimum possible length.” Its length calculation is correct, but its membership claim is not: \(2\notin(2,5)\) and \(5\notin(2,5)\). The interval is a cover of the open interval \((2,5)\), not of the closed interval \([2,5]\).

Expanding the endpoints repairs the cover: \((2-\varepsilon/2,5+\varepsilon/2)\) contains \([2,5]\) and has length \(3+\varepsilon\). This shows that lengths can be made arbitrarily close to \(3\) from above. It does not show that one can set \(\varepsilon=0\), because that would return to \((2,5)\), which misses the endpoints. The theorem proves that every open-interval cover of \([2,5]\), not only this family of expanded intervals, has length strictly greater than \(3\).

The same distinction also handles the degenerate closed interval \([a,a]=\{a\}\). The intervals \((a-1/n,a+1/n)\) cover it and have lengths \(2/n\), so \(m^*(\{a\})=0\). But any open interval containing \(a\) has positive length, and any cover of \(\{a\}\) contains at least one such interval. Thus no cover has total length \(0\). The infimum is still not attained, even though the interval has length zero.

The practical point is that an infimum describes the best lower bound for all admissible covers; it does not promise an optimal cover. To prove that a particular cover achieves an infimum, one must exhibit a cover with that total length and separately show that no cover can do better. For closed intervals, the lower bound is exact as an infimum, while the open neighborhoods forced at the endpoints prevent equality for any individual cover.