Why Complete a Measure Space?
A measure space is complete when every subset of every measurable null set is measurable. This condition matters because arguments about “almost everywhere” behavior naturally single out arbitrary subsets of null sets; without completeness, such a subset need not belong to the sigma-algebra. The completion adds all these subsets while preserving the measure of the sets that were already measurable.
Let \((X,\mathcal{M},\mu)\) be any measure space: \(\mathcal{M}\) is a sigma-algebra on \(X\), and \(\mu:\mathcal{M}\to[0,\infty]\) is a measure. No finiteness, sigma-finiteness, or completeness assumption is imposed. A measurable null set is a set \(Z\in\mathcal{M}\) with \(\mu(Z)=0\). The construction will make every subset of every such \(Z\) measurable.
The sandwich description says that a completed-measurable set lies between two original measurable sets whose difference is null. Equivalently, \(A\in\overline{\mathcal{M}}\) exactly when \(A=E\cup N\) for some \(E,Z\in\mathcal{M}\) with \(\mu(Z)=0\) and \(N\subseteq Z\). Indeed, from a sandwich take \(N=A\setminus E\subseteq F\setminus E\); conversely, if \(A=E\cup N\) with \(N\subseteq Z\), use \(E\subseteq A\subseteq E\cup Z\), whose outer-minus-inner difference is contained in \(Z\).
The Construction Is a Measure Space
Proof. First, \(\varnothing\in\overline{\mathcal{M}}\), using \(E=F=\varnothing\). If \(A\in\overline{\mathcal{M}}\) has a sandwich \(E\subseteq A\subseteq F\), then \(F^c\subseteq A^c\subseteq E^c\), and \(E^c\setminus F^c=F\setminus E\) is null. Hence \(A^c\in\overline{\mathcal{M}}\). Now let \(A_n\in\overline{\mathcal{M}}\) have sandwiches \(E_n\subseteq A_n\subseteq F_n\), where \(E_n,F_n\in\mathcal{M}\) and \(\mu(F_n\setminus E_n)=0\). Then \[ \bigcup_n E_n\subseteq\bigcup_n A_n\subseteq\bigcup_n F_n. \] If \(x\in(\bigcup_nF_n)\setminus(\bigcup_nE_n)\), then for some \(k\), \(x\in F_k\), while \(x\notin E_k\). Thus \[ \left(\bigcup_nF_n\right)\setminus\left(\bigcup_nE_n\right) \subseteq\bigcup_n(F_n\setminus E_n), \] which is null by countable subadditivity. Therefore \(\bigcup_n A_n\in\overline{\mathcal{M}}\), proving that \(\overline{\mathcal{M}}\) is a sigma-algebra. Every \(A\in\mathcal{M}\) has the sandwich \(A\subseteq A\subseteq A\), so \(\mathcal{M}\subseteq\overline{\mathcal{M}}\).
We next show that the proposed value \(\overline{\mu}(A)\) does not depend on the choice of measurable inner set. Suppose \(E\subseteq A\subseteq F\) and \(E'\subseteq A\subseteq F'\), with all four sets in \(\mathcal{M}\) and \(\mu(F\setminus E)=\mu(F'\setminus E')=0\). If \(x\in E\setminus E'\), then \(x\in A\subseteq F'\), so \(x\in F'\setminus E'\). If \(x\in E'\setminus E\), then \(x\in F\setminus E\). Hence \[ E\triangle E'\subseteq(F\setminus E)\cup(F'\setminus E'), \] a measurable null set. The measurable sets \(E\) and \(E'\) therefore have equal measure: each of \(E\setminus E'\) and \(E'\setminus E\) is null, so removing either difference does not change the measure. Thus \(\mu(E)=\mu(E')\), and \(\overline{\mu}\) is well-defined. In particular, for \(A\in\mathcal{M}\), choosing \(E=F=A\) gives \(\overline{\mu}(A)=\mu(A)\).
It remains to prove countable additivity. Let \(A_n\in\overline{\mathcal{M}}\) be pairwise disjoint, and choose sandwiches \(E_n\subseteq A_n\subseteq F_n\) as above. For \(i\ne j\), disjointness gives \(E_i\cap E_j\subseteq F_i\setminus E_i\), so \(E_i\cap E_j\) is null. Define \(B_1=E_1\) and, for \(n\ge2\), define \(B_n=E_n\setminus\bigcup_{i<n}E_i\). These sets are measurable and pairwise disjoint. The difference \(E_n\setminus B_n\) is contained in the finite union \(\bigcup_{i<n}(E_n\cap E_i)\), so it is null and \(\mu(B_n)=\mu(E_n)\). Also, \(\bigcup_nB_n=\bigcup_nE_n\). The union \(\bigcup_n A_n\) has inner set \(\bigcup_n E_n\) and outer set \(\bigcup_n F_n\); their difference is null by the same countable-union argument used above. Using countable additivity of \(\mu\) on the disjoint \(B_n\), we obtain \[ \overline{\mu}\left(\bigcup_n A_n\right) =\mu\left(\bigcup_nE_n\right) =\sum_n\mu(B_n) =\sum_n\mu(E_n) =\sum_n\overline{\mu}(A_n). \] This holds with infinite values allowed, so \(\overline{\mu}\) is a measure.
Finally, suppose \(A\subseteq Z\), where \(Z\in\overline{\mathcal{M}}\) and \(\overline{\mu}(Z)=0\). By the definition of the completion measure, there is \(E\in\mathcal{M}\) with \(E\subseteq Z\) and \(\mu(E)=0\). Since \(E\subseteq Z\), the measurable set \(E\) is itself an inner set for \(Z\); the zero value follows from \(\overline{\mu}(Z)=0\). More directly, take a sandwich \(E_0\subseteq Z\subseteq F_0\): \(\mu(E_0)=0\), and \(\mu(F_0\setminus E_0)=0\), so \(\mu(F_0)=0\). Now \(A\subseteq F_0\), and \(\varnothing\subseteq A\subseteq F_0\) is a sandwich with null difference. Thus \(A\in\overline{\mathcal{M}}\). This is completeness. \(\square\)
Worked Completions
Worked Example: Completing a Finite Measure Space
Let \(X=\{1,2,3\}\), \(\mathcal{M}=\{\varnothing,\{1\},\{2,3\},X\}\), and define \(\mu(\varnothing)=\mu(\{2,3\})=0\) and \(\mu(\{1\})=\mu(X)=1\). This is a measure: the sigma-algebra consists of the unions of the partition sets \(\{1\}\) and \(\{2,3\}\), and their assigned weights are \(1\) and \(0\). The measurable null sets are \(\varnothing\) and \(\{2,3\}\). Every subset of \(\{2,3\}\) is therefore included in the completion.
Consequently, \[ \overline{\mathcal{M}}=\{\varnothing,\{1\},\{2\},\{3\},\{2,3\},\{1,2\},\{1,3\},X\}=\mathcal{P}(X). \] For each new set, its measure is zero if it is a subset of \(\{2,3\}\), and one if it contains \(1\). For example, \(\overline{\mu}(\{1,3\})=1\), since \(\{1\}\subseteq\{1,3\}\subseteq X\) and \(\mu(X\setminus\{1\})=\mu(\{2,3\})=0\). Also \(\overline{\mu}(\{3\})=0\), using \(\varnothing\subseteq\{3\}\subseteq\{2,3\}\). Check: the extension agrees on all original sets: \(\overline{\mu}(\{1\})=1\), \(\overline{\mu}(\{2,3\})=0\), and \(\overline{\mu}(X)=1\). The completed space is complete because its sigma-algebra is all of \(\mathcal{P}(X)\).
Worked Example: Completing Borel Lebesgue Measure on the Real Line
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and let \(\mu\) be Lebesgue measure restricted to the Borel sets. This measure space is sigma-finite, but the completion construction does not require sigma-finiteness. Let \(C\) be the middle-thirds Cantor set. At stage \(n\), it is covered by \(2^n\) intervals of length \(3^{-n}\), whose total length is \((2/3)^n\). Since this tends to zero, \(C\) has Lebesgue measure zero; it is also Borel.
For any subset \(S\subseteq C\), whether or not \(S\) is Borel, the sandwich \(\varnothing\subseteq S\subseteq C\) shows that \(S\in\overline{\mathcal{M}}\), and \(\overline{\mu}(S)=0\). In fact some such \(S\) are not Borel: the Borel subsets of \(\mathbb{R}\) have cardinality at most that of \(\mathbb{R}\), since they have countable descriptions, whereas \(\mathcal{P}(C)\) has strictly larger cardinality by Cantor’s theorem and \(C\) has the cardinality of \(\mathbb{R}\). Thus the completion can add genuinely new sets. Check: \(C\) is the measurable outer set, \(\varnothing\) is the measurable inner set, and \(\mu(C\setminus\varnothing)=0\), so the definition assigns measure zero to every such \(S\).
Worked Example: Completing Counting Measure on the Natural Numbers
Let \(X=\mathbb{N}\), \(\mathcal{M}=\mathcal{P}(\mathbb{N})\), and let \(\mu\) be counting measure, so \(\mu(A)\) is the number of elements of \(A\) when \(A\) is finite and \(+\infty\) when \(A\) is infinite. The only measurable null set is \(\varnothing\): every nonempty set contains a point and therefore has measure at least one. Its only subset is \(\varnothing\), which is already measurable. Hence the completion adds no sets: \(\overline{\mathcal{M}}=\mathcal{P}(\mathbb{N})\), and \(\overline{\mu}=\mu\).
Check: if \(A\subseteq\mathbb{N}\) is finite with \(k\) elements, its measure remains \(k\); if it is infinite, its measure remains \(+\infty\). The space was already complete because every subset of its only null set, \(\varnothing\), belongs to \(\mathcal{M}\).
The Completion Is the Smallest Complete Extension
Proof. Let \(A\in\overline{\mathcal{M}}\), and choose \(E,F\in\mathcal{M}\) with \(E\subseteq A\subseteq F\) and \(\mu(F\setminus E)=0\). Since \(F\setminus E\in\mathcal{M}\subseteq\mathcal{N}\) and \(\nu(F\setminus E)=\mu(F\setminus E)=0\), completeness of \((X,\mathcal{N},\nu)\) implies that every subset of \(F\setminus E\) belongs to \(\mathcal{N}\). In particular, \(A\setminus E\in\mathcal{N}\). As \(E\in\mathcal{N}\), the identity \(A=E\cup(A\setminus E)\) shows \(A\in\mathcal{N}\). Also \(E\subseteq A\) and \(A\setminus E\subseteq F\setminus E\), so \(\nu(A)=\nu(E)+\nu(A\setminus E)=\mu(E)=\overline{\mu}(A)\). Thus every complete extension agreeing with \(\mu\) must contain the completion, and must give it the same measure. \(\square\)
A Common Mistake
A measurable null set is not the same thing as all its subsets being measurable. In the finite example, \(\{2,3\}\) is already in \(\mathcal{M}\), but \(\{2\}\) and \(\{3\}\) are not. The incorrect argument “the null set \(\{2,3\}\) is measurable, so its subsets are measurable” confuses membership of a set with closure under taking arbitrary subsets. A sigma-algebra is closed under complements and countable unions, not arbitrary subsets. The completion specifically adds those subsets, assigning them measure zero; it does not alter the measures of the original measurable sets.
Check Your Understanding
Use the sandwich definition and the completeness property to check each conclusion.
- For \(A\in\overline{\mathcal{M}}\), explain why two different measurable inner sets used to define \(\overline{\mu}(A)\) must have equal measure.
- If \(Z\in\mathcal{M}\), \(\mu(Z)=0\), and \(S\subseteq Z\), give a sandwich proving \(S\in\overline{\mathcal{M}}\) and compute \(\overline{\mu}(S)\).
- In the finite example, determine \(\overline{\mu}(\{1,2\})\) and state a measurable sandwich that verifies it.
- Why does completing counting measure on \(\mathbb{N}\) add no sets?
- Explain why the completion of Borel Lebesgue measure on \(\mathbb{R}\) includes every subset of the Cantor set.