Restricting Outer Measure
Lebesgue outer measure \(m^*\) assigns a size to every subset of \(\mathbb{R}\), but a measure must be countably additive on its domain. The Carathéodory measurability criterion supplies that domain: its measurable sets form a sigma-algebra, and outer measure is countably additive there. We can now make the construction explicit by restricting \(m^*\) to those sets.
This is a restriction, not a new formula for the size of a set. The outer measure \(m^*\) is still defined for every subset of \(\mathbb{R}\), while \(\lambda\) is defined as a measure on \(\mathcal{L}\). By the Countable Additivity on the Carathéodory Measurable Sets Theorem from the previous tutorial, \(\lambda\) is countably additive on \(\mathcal{L}\). Also \(m^*(\varnothing)=0\), so \(\lambda\) satisfies the measure axioms.
In particular, if \(E_1,E_2,\ldots\) are pairwise disjoint Lebesgue measurable sets, then
The equality is understood in the extended nonnegative real numbers, so the measure and the sum are allowed to be infinite. This construction gives a measure while preserving the outer-measure values already obtained for sets such as intervals.
Interval Length and Basic Examples
The previous tutorial established that every bounded interval has outer measure equal to its length, regardless of whether either endpoint is included. Each such interval is measurable: open intervals were proved measurable there, and the sigma-algebra operations give the other endpoint conventions. Consequently, the restriction defining \(\lambda\) preserves the familiar length calculation.
For a nondegenerate interval, measurability follows from the measurability of open intervals and closure under complements and countable unions; for example, a closed interval can be obtained by intersecting closed half-lines. A half-open interval can also be formed from an open interval by adding or removing an endpoint, and singletons are measurable because they have outer measure zero. The stated values then follow from the bounded-interval outer-measure calculation and \(\lambda(E)=m^*(E)\). When \(a=b\), the only nonempty interval with those endpoints is the singleton \(\{a\}\), whose measure is zero.
Worked Example: A Half-Open Interval
Consider \(E=[2,7)\). This is a bounded interval, so it is Lebesgue measurable, and its measure is its length:
Including the endpoint \(7\) does not change the answer: \([2,7]=[2,7)\cup\{7\}\), where the union is disjoint and \(\lambda(\{7\})=0\). Thus countable additivity gives \(\lambda([2,7])=5+0=5\).
Worked Example: A Countable Union of Disjoint Intervals
For each positive integer \(n\), let \(I_n=(4n,4n+2^{-n})\). These intervals are pairwise disjoint: the right endpoint of \(I_n\) is less than \(4n+1\), while the left endpoint of \(I_{n+1}\) is \(4n+4\). Each interval is measurable and has measure \(2^{-n}\). Therefore their union \(U\) is measurable, and countable additivity gives
The series has this value because its first \(N\) terms sum to \(1-2^{-N}\), which tends to \(1\). The union consists of infinitely many separated intervals, yet its total measure is finite.
Worked Example: An Unbounded Interval
The ray \([3,\infty)\) is measurable: it is the complement of the open interval \((-\infty,3)\), which is a countable union of open intervals and hence measurable. For every positive integer \(N\), it contains \([3,3+N]\). By monotonicity of measure and the bounded-interval formula,
No finite real number is at least every positive integer, so \(\lambda([3,\infty))=+\infty\). This calculation uses finite subintervals to establish the measure of an unbounded set; it does not assign an infinite interval a finite length.
Borel Sets Are Lebesgue Measurable
The Borel sigma-algebra \(\mathcal{B}(\mathbb{R})\) is generated by the open subsets of the real line. Every open interval is Carathéodory measurable, and the collection \(\mathcal{L}\) is a sigma-algebra. It follows that every open set, and then every Borel set, is Lebesgue measurable.
Proof. Let \(O\subseteq\mathbb{R}\) be open. Consider all open intervals \((p,q)\) with rational endpoints such that \((p,q)\subseteq O\). There are at most countably many such intervals because the rational pairs \((p,q)\) form a countable set. Their union is \(O\): if \(x\in O\), openness gives some interval around \(x\) contained in \(O\); density of the rationals provides rational \(p<x<q\) with \((p,q)\) contained in that interval. Thus \(O\) is a countable union of open intervals. Each is Carathéodory measurable, so the sigma-algebra property implies \(O\in\mathcal{L}\). Since \(\mathcal{L}\) is a sigma-algebra containing every open set, it contains the sigma-algebra generated by the open sets, namely \(\mathcal{B}(\mathbb{R})\). \(\square\)
This inclusion says that Borel sets are measurable, not that all Lebesgue measurable sets must be Borel. In fact, Lebesgue measure has the additional completeness property proved below: every subset of a measurable null set is measurable. This property is one reason the Lebesgue sigma-algebra is larger than the Borel sigma-algebra.
Completeness of Lebesgue Measure
A measure space is called complete if every subset of a measurable set of measure zero is measurable. For Lebesgue measure, this follows directly from the outer-measure-zero result established in the previous tutorial. The argument uses no assumption that the subset itself is Borel.
Proof. Since \(N\in\mathcal{L}\) and \(\lambda(N)=0\), the definition of \(\lambda\) gives \(m^*(N)=0\). By monotonicity of outer measure and \(S\subseteq N\),
Hence \(m^*(S)=0\). The Every Outer-Measure-Zero Set Is Carathéodory Measurable Theorem from the previous tutorial now shows that \(S\in\mathcal{L}\). Applying the definition of \(\lambda\) once more gives \(\lambda(S)=m^*(S)=0\). This proves completeness. \(\square\)
Completeness matters when a set is changed on a negligible subset. A set contained in a null set need not be presented as a Borel set, but the theorem guarantees it is Lebesgue measurable and has measure zero. In particular, the measure space does not omit subsets merely because they are irregular as topological sets.
Translation Invariance
The interval-cover definition of Lebesgue outer measure treats translated intervals identically: translation changes an interval's location but not its length. This implies that Lebesgue measure is translation invariant, both in its values and in its collection of measurable sets.
Proof. First, translation preserves outer measure for every subset \(F\subseteq\mathbb{R}\). Translating a countable cover of \(F\) by open intervals \(I_j\) gives a cover of \(F+t\) by intervals \(I_j+t\), and \(|I_j+t|=|I_j|\). Taking infima over covers gives \(m^*(F+t)\leq m^*(F)\). Applying this inequality to \(F+t\) translated by \(-t\) gives \(m^*(F)\leq m^*(F+t)\). Therefore
Now let \(E\in\mathcal{L}\), and take an arbitrary test set \(A\subseteq\mathbb{R}\). Since \(E\) is Carathéodory measurable, applying its defining equality to \(A-t\) gives
Translate each set in this equality by \(t\). Translation preserves outer measure, and the translated pieces are \((A-t)\cap E+t=A\cap(E+t)\) and \(((A-t)\setminus E)+t=A\setminus(E+t)\). Thus
Because this holds for every \(A\), the Carathéodory criterion shows that \(E+t\in\mathcal{L}\). Finally, the outer-measure translation equality and the definition of \(\lambda\) imply \(\lambda(E+t)=m^*(E+t)=m^*(E)=\lambda(E)\). \(\square\)
What the Construction Does—and Does Not—Claim
Lebesgue measure is obtained by keeping the outer-measure values and restricting their domain to sets that pass the Carathéodory splitting test. On this domain, countable additivity is available, intervals have their usual lengths, Borel sets are included, null sets contain only measurable subsets, and translation does not change measure. These facts are foundational for integration and for measuring limits of sequences of sets and functions.
It is important not to confuse the domain of \(m^*\) with the domain of \(\lambda\). Outer measure can be evaluated on every subset of \(\mathbb{R}\), but the construction defines Lebesgue measure as a measure only on \(\mathcal{L}\). One also should not infer that an arbitrary union has measure equal to the sum of the measures: countable additivity applies to pairwise disjoint measurable sets, while general unions are controlled by subadditivity. For example, overlapping intervals must not have their lengths added without first accounting for the overlap.
Check Your Understanding
Use the construction and results in this tutorial to answer the following questions.
- What is the domain of Lebesgue measure, and how does it differ from the domain of Lebesgue outer measure?
- Why does \([2,7]\) have the same Lebesgue measure as \([2,7)\)?
- How can bounded intervals be used to show that \([3,\infty)\) has infinite measure?
- Why does every Borel set belong to the Lebesgue sigma-algebra?
- Which result lets us conclude that every subset of a Lebesgue null set is measurable?
- How does translating interval covers prove translation invariance of Lebesgue measure?