Lebesgue measure is built by first assigning an outer size to every set, then selecting the sets whose interaction with outer size is consistent.
Outer measure
Outer measures are defined on all subsets, but only subadditively.
Definition. An outer measure on \(X\) is a function \(\mu^*:2^X\to[0,\infty]\) with \(\mu^*(\varnothing)=0\), monotonicity, and countable subadditivity.
Proposition. If \(\ell(I)\) is interval length, then \[ m^*(E)=\inf\left\{\sum_n\ell(I_n):E\subset\bigcup_n I_n\right\} \] defines an outer measure on \(\mathbb R\).
Proof. The empty set is covered with total length zero. Monotonicity follows because every cover of a larger set covers a smaller one. For subadditivity, cover each \(E_k\) within \(\varepsilon 2^{-k}\) of its outer measure; the union of these interval families covers \(\bigcup_kE_k\), and summing gives the inequality after \(\varepsilon\downarrow0\).
Carath?odory measurability
Measurable sets are those that split every set additively.
Definition. A set \(A\subset X\) is \(\mu^*\)-measurable if for every \(E\subset X\), \[ \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A). \]
Theorem. The \(\mu^*\)-measurable sets form a sigma-algebra, and \(\mu^*\) restricted to it is a complete measure.
Proof. The empty set is measurable, and the criterion is symmetric in \(A\) and \(A^c\). If \(A,B\) are measurable, split \(E\) first by \(A\) and then split \(E\cap A^c\) by \(B\): \[ \mu^*(E) =\mu^*(E\cap A)+\mu^*(E\cap A^c\cap B) +\mu^*(E\cap A^c\cap B^c). \] The first two pieces lie in \(E\cap(A\cup B)\), and subadditivity gives measurability of \(A\cup B\). Thus measurable sets form an algebra.
For disjoint measurable \(A_n\), repeated splitting gives \[ \mu^*(E)\ge \sum_{n=1}^N\mu^*(E\cap A_n) +\mu^*\!\left(E\setminus\bigcup_{n=1}^NA_n\right). \] Let \(N\to\infty\) and use subadditivity for the reverse direction to show that \(\bigcup_nA_n\) is measurable and \[ \mu^*\!\left(\bigcup_nA_n\right)=\sum_n\mu^*(A_n). \] Arbitrary countable unions are reduced to disjoint ones. If \(\mu^*(N)=0\) and \(A\subset N\), then both \(\mu^*(E\cap A)\) and the possible loss between \(E\) and \(E\setminus A\) are zero; hence \(A\) satisfies the criterion. This also proves completeness.
Theorem (metric outer measures). If an outer measure on a metric space satisfies \[ d(A,B)>0\quad\Longrightarrow\quad \mu^*(A\cup B)=\mu^*(A)+\mu^*(B), \] then every closed set, hence every Borel set, is Carath?odory measurable.
Proof. Let \(F\) be closed and \(E\subset X\). For \(n\ge1\), put \[ E_n=\{x\in E:d(x,F)\ge1/n\}. \] The sets \(E\cap F\) and \(E_n\) are positively separated, so \[ \mu^*(E)\ge\mu^*(E\cap F)+\mu^*(E_n). \] As \(E_n\uparrow E\setminus F\), group alternating annular differences to use metric additivity and obtain \(\mu^*(E_n)\uparrow\mu^*(E\setminus F)\). Thus \(F\) satisfies the Carath?odory criterion. Closed sets generate the Borel sigma-algebra.
Lebesgue measurable sets
Lebesgue measure is the Carath?odory measure induced by interval covers.
Theorem. All Borel subsets of \(\mathbb R\) are Lebesgue measurable, and the resulting measure agrees with interval length on intervals.
Proof. The interval-cover outer measure is a metric outer measure: positively separated sets can be covered, after subdividing each interval cover into sufficiently short intervals, without one interval meeting both sets. Hence all Borel sets are measurable by the preceding theorem.
For an interval \(I\), the cover \(I\) itself gives \(m^*(I)\le\ell(I)\). For the reverse inequality on a compact interval \([a,b]\), enlarge every interval in an arbitrary countable cover by a total amount \(<\varepsilon\). Compactness extracts a finite subcover. Ordering its endpoints and deleting overlaps shows that the sum of its lengths is at least \(b-a\). Therefore every original cover has total length at least \(b-a-\varepsilon\), and \(\varepsilon\downarrow0\) gives \(m^*([a,b])=b-a\). Open, closed, and half-open intervals follow by monotonicity and arbitrarily short endpoint extensions.
Translation invariance
Lebesgue measure respects the additive structure of the line.
Proposition. For every \(a\in\mathbb R\), \(m(E+a)=m(E)\) for Lebesgue measurable \(E\).
Proof. Translating interval covers preserves total length, so \(m^*(E+a)=m^*(E)\). Translation also preserves the Carath?odory criterion, hence measurable sets remain measurable and their measures agree.
Null sets and completion
Lebesgue measure contains more sets than the Borel sigma-algebra.
Proposition. Lebesgue measurable sets are exactly the completion of the Borel sigma-algebra: \[ \mathcal L =\{B\triangle N:B\in\mathcal B(\mathbb R),\ N\subset Z,\ m(Z)=0\}. \]
Proof. Completeness gives inclusion from right to left. Conversely, outer regularity supplies, for measurable \(E\), open sets \(U_n\supset E\) with \(m(U_n\setminus E)<2^{-n}\). The Borel set \(B=\bigcap_nU_n\) contains \(E\), and \[ m(B\setminus E)\le m(U_n\setminus E)<2^{-n} \] for every \(n\), so \(B\setminus E\) is null. Thus \(E\) differs from a Borel set by a subset of a null Borel set.
Remark. The Cantor set has measure zero and cardinality \(2^{\aleph_0}\), so all of its \(2^{2^{\aleph_0}}\) subsets are Lebesgue measurable. There are only \(2^{\aleph_0}\) Borel subsets of \(\mathbb R\); consequently most subsets of the Cantor set are not Borel. Completion is therefore a substantial enlargement, not a cosmetic convention.