Outer Regularity for Finite-Measure Sets

From Interval Covers to Open Supersets

Lebesgue outer measure is built by measuring sets through interval covers. For a Lebesgue measurable set of finite measure, this gives more than an upper bound: it lets us place the set inside an open set whose measure is as close as desired to the measure of the set itself. The set need not be bounded, and the open superset need not be unique or nested with the ones chosen for other error tolerances.

Throughout, the ambient measure space is \((\mathbb{R},\mathcal{L},m)\), Lebesgue measure on the real line. This space is complete and sigma-finite, but is not a finite measure space. We consider a Lebesgue measurable set \(E\in\mathcal{L}\) with \(m(E)<\infty\). The Lebesgue outer measure \(m^*(A)\) of any \(A\subseteq\mathbb{R}\) is the infimum of the sums of lengths of countable interval covers of \(A\). Since \(E\) is measurable, \(m(E)=m^*(E)\). We use the standard facts that open subsets of \(\mathbb{R}\) are Lebesgue measurable, that an interval’s Lebesgue measure is its length, and that \(m\) is countably subadditive.

Proof: Spending a Length Budget

The interval-cover definition gives a cover whose total length is close to \(m(E)\), but the covering intervals need not themselves be open. We enlarge each interval slightly. The enlargements have to be chosen so that their total added length fits inside the remaining error allowance.

Proof. Fix \(\varepsilon>0\). If \(E=\varnothing\), take \(O=\varnothing\); then \(m(O)=m(E)=0\), so the claim holds. Now suppose \(E\ne\varnothing\). Because \(m^*(E)=m(E)<\infty\), the definition of the infimum gives a finite or countable interval cover \((I_j)\) of \(E\) such that

Here \(|I_j|\) denotes the length of \(I_j\). Every nonempty \(I_j\) in this cover has finite length: an interval of infinite length would make the sum infinite. Write \(a_j=\inf I_j\) and \(b_j=\sup I_j\). Choose \(\delta_j=\varepsilon/2^{j+2}\), and replace \(I_j\) by the open interval \(J_j=(a_j-\delta_j,b_j+\delta_j)\). This contains \(I_j\), including its endpoints when they belong to \(I_j\), and its length is \(|J_j|=|I_j|+2\delta_j\). Empty intervals in a cover can be discarded.

For a finite or countable family indexed by positive integers, the geometric-series bound gives \(\sum_j\delta_j\le\sum_{j=1}^{\infty}\varepsilon/2^{j+2}=\varepsilon/4\). Consequently, the total length of the enlarged intervals satisfies

Let \(O=\bigcup_j J_j\). This is open and contains \(E\). By countable subadditivity and the interval-length formula, \(m(O)\le\sum_j|J_j|<m(E)+\varepsilon\); in particular, \(m(O)\) is finite. Since \(E\) and \(O\) are measurable and \(E\subseteq O\), countable additivity applied to the disjoint union \(O=E\cup(O\setminus E)\) gives \(m(O)=m(E)+m(O\setminus E)\ge m(E)\). This proves the first assertion, including the claimed upper and lower bounds.

For the sequence assertion, apply the first assertion with \(\varepsilon=1/k\) for each positive integer \(k\), obtaining open \(O_k\supseteq E\) with \(m(E)\le m(O_k)<m(E)+1/k\). The inequalities imply \(0\le m(O_k)-m(E)<1/k\), and \(1/k\to0\); hence \(m(O_k)\to m(E)\). The sets \(O_k\) need not be nested. \(\square\)

Worked Applications

What the Result Does and Does Not Say

The proof works for any finite-measure Lebesgue measurable \(E\subseteq\mathbb{R}\), including unbounded sets. Boundedness was never assumed: the crucial input was \(m^*(E)=m(E)<\infty\), which allowed an interval cover with finite total length and then a controlled enlargement. The conclusion is an approximation from outside. It does not assert that \(E\) is open, that one open superset has measure exactly \(m(E)\), or that the approximating open sets can be chosen nested.

Proof. Every open \(O\supseteq E\) is measurable, and additivity gives \(m(O)\ge m(E)\), so the infimum is at least \(m(E)\). Conversely, for each \(\varepsilon>0\), outer regularity supplies an open \(O\supseteq E\) with \(m(O)<m(E)+\varepsilon\). Therefore the infimum is at most \(m(E)+\varepsilon\) for every \(\varepsilon>0\), and hence at most \(m(E)\). The two bounds prove equality. \(\square\)

A common error is to replace the measure of a union by the sum of the lengths of its covering intervals. For example, the two open intervals \((-1/10,3/5)\) and \((2/5,11/10)\) each have length \(7/10\), so their lengths sum to \(14/10\). But their union is \((-1/10,11/10)\), whose measure is \(12/10\), not \(14/10\). The correct general statement is \(m(\bigcup_jJ_j)\le\sum_jm(J_j)\), with equality guaranteed when the sets are pairwise disjoint. The proof of outer regularity needs only the inequality; in examples where we calculate the union’s measure by adding lengths, we verify disjointness.