Epsilon-Optimal Covers and Infima

From an Infimum to an Almost-Minimal Cover

Lebesgue outer measure is defined by taking an infimum over the total lengths of countable interval covers. An infimum need not itself be the cost of any cover: it is a greatest lower bound, not necessarily a minimum. The useful consequence of the definition is instead an approximation statement. If a set has finite outer measure, then for every prescribed positive error there is a cover whose total length is less than the outer measure plus that error.

We work on \(\mathbb{R}\). An interval cover of \(E\subseteq\mathbb{R}\) is a finite or countable collection of intervals \(I_k\) whose union contains \(E\), using the interval-cover conventions and length \(|I_k|\) established for Lebesgue outer measure. Its cost is the extended nonnegative sum \(\sum_k |I_k|\). The empty collection is allowed as a cover of the empty set and has cost zero. Unbounded intervals have infinite length. The Lebesgue outer measure \(m^*\), defined on every subset of \(\mathbb{R}\), is

The expression inside the infimum is a cover cost. Every set \(E\) has at least one interval cover, since \(\mathbb{R}\) covers \(E\); thus the family of costs is nonempty. The infimum is taken in \([0,\infty]\). This tutorial concerns the approximation property of that infimum, not whether outer measure is a measure or how to compute its value for particular sets.

The Epsilon-Approximation Property

Proof. Let \(S_E\) be the nonempty set of costs of interval covers of \(E\), so \(m^*(E)=\inf S_E\). Put \(\alpha=m^*(E)\), which is finite by hypothesis. Suppose, contrary to the claim, that no cover has cost less than \(\alpha+\varepsilon\). Then every \(s\in S_E\) satisfies \(s\ge\alpha+\varepsilon\). Consequently \(\alpha+\varepsilon\) is a lower bound for \(S_E\). By the definition of infimum, \(\alpha\) is the greatest lower bound, so every lower bound of \(S_E\) is at most \(\alpha\). But \(\alpha+\varepsilon>\alpha\), a contradiction. Therefore some cover has cost less than \(\alpha+\varepsilon\). \(\square\)

The proof uses only the defining property of an infimum. It does not produce a formula for the intervals: when the cover costs are not explicitly known, the existence conclusion can be nonconstructive. It does, however, give control at any threshold chosen in advance. For example, if an error bound of \(10^{-3}\) is required, set \(\varepsilon=10^{-3}\) in the theorem; a cover exists with cost strictly below \(m^*(E)+10^{-3}\).

The finite hypothesis is essential to this formulation. If \(m^*(E)=\infty\), then every cover cost is infinite: a finite-cost cover would put a finite upper bound on the infimum. In that case there is no finite-cost cover to approximate the outer measure. An inequality involving \(\infty+\varepsilon\) does not express finite accuracy, so the theorem makes no such claim.

Choosing a Sequence of Covers

Applying the theorem repeatedly gives a sequence of covers that approaches the infimum in cost. The conclusion is stronger than choosing one cover for one tolerance, but it still does not assert that any cover attains the limiting cost.

Proof. Write \(\alpha=m^*(E)<\infty\). For each \(n\ge1\), apply the theorem with \(\varepsilon=1/n>0\). This gives a cover with cost \(c_n<\alpha+1/n\). Since \(\alpha\) is the infimum of all cover costs, it is a lower bound for each such cost, so \(\alpha\le c_n\). Thus \(0\le c_n-\alpha<1/n\). Given any \(\delta>0\), choose an integer \(N>1/\delta\). For every \(n\ge N\), \(1/n\le1/N<\delta\), and therefore \(0\le c_n-\alpha<\delta\). This is the definition of \(c_n\to\alpha\). The covers may be chosen separately for each \(n\); no nesting or monotonicity of the covers is claimed. \(\square\)

What the Infimum Does—and Does Not—Guarantee

For a finite infimum, two statements are immediate and useful. First, every cover has cost at least \(m^*(E)\), because an infimum is a lower bound. Second, for every positive tolerance there is a cover whose cost is less than \(m^*(E)\) plus that tolerance. Together, these facts pin down the best possible cost to arbitrary precision. They do not say that one cover has cost equal to the infimum.

Common mistake: setting the tolerance to zero. In the numerical example above, one might incorrectly take the epsilon-approximation statement with \(\varepsilon=0\) and conclude that some cost is less than \(2\). This is wrong twice: the statement requires \(\varepsilon>0\), and every member \(2+1/n\) is greater than \(2\). For a positive tolerance, such as \(1/5\), the calculation \(13/6<11/5\) verifies the correct conclusion. The error is confusing arbitrary closeness with attainment.

There is also a difference between an upper bound from one known cover and an epsilon-optimal cover. A cover of cost \(L\) shows that \(m^*(E)\le L\), but it need not have cost less than \(m^*(E)+\varepsilon\) for a small prescribed \(\varepsilon\). The infimum property supplies a possibly different cover for that tolerance. No assumption that covers are nested, or that their costs decrease, appears in either the theorem or its corollary.

This distinction is useful whenever an object is defined as an infimum of admissible costs. In the outer-measure construction, one can work with covers whose excess over the infimum is as small as needed, while avoiding an unjustified assumption that an optimal cover exists. The epsilon is chosen before the cover; the definition then guarantees a cover meeting that particular strict bound.