Why Intersections Matter
A family of sets may be intended to serve as a collection of measurable sets, but it need not already be a sigma-algebra. One way to complete it is to include it in sigma-algebras and take their intersection. This works because the defining closure properties of a sigma-algebra are preserved under intersection. The result is not merely some sigma-algebra containing the original family: it is contained in every such sigma-algebra, so it is the smallest one.
This is a question about sets, not yet about a measure or a measure space. Fix an underlying set \(X\), and write \(\mathcal{P}(X)\) for the collection of all subsets of \(X\). Every sigma-algebra considered below is a subcollection of \(\mathcal{P}(X)\) on this same \(X\). Recall that a sigma-algebra \(\mathcal{A}\) contains \(X\), is closed under complements relative to \(X\), and is closed under countable unions. In particular, it contains \(\varnothing=X^c\).
Intersections Preserve the Sigma-Algebra Axioms
Proof. Set \(\mathcal{A}=\bigcap_{i\in I}\mathcal{A}_i\). Since every \(\mathcal{A}_i\) contains \(X\), \(X\) belongs to their intersection, so \(X\in\mathcal{A}\). Now take \(E\in\mathcal{A}\). By the definition of intersection, \(E\in\mathcal{A}_i\) for every \(i\in I\). Each \(\mathcal{A}_i\) is closed under complements, so \(E^c\in\mathcal{A}_i\) for every \(i\). Thus \(E^c\in\mathcal{A}\). Finally, let \((E_n)_{n\ge1}\) be any sequence in \(\mathcal{A}\). For every \(i\), all of the sets \(E_n\) belong to \(\mathcal{A}_i\), so closure of \(\mathcal{A}_i\) under countable unions gives \(\bigcup_{n=1}^{\infty}E_n\in\mathcal{A}_i\). This holds for every \(i\), hence \(\bigcup_{n=1}^{\infty}E_n\in\mathcal{A}\). The three sigma-algebra axioms hold for \(\mathcal{A}\). \(\square\)
Worked Example: Intersecting Two Sigma-Algebras
Take \(X=\{a,b,c\}\), and define \(\mathcal{A}_1=\{\varnothing,\{a\},\{b,c\},X\}\) and \(\mathcal{A}_2=\{\varnothing,\{b\},\{a,c\},X\}\). Find their intersection and verify the result.
For \(\mathcal{A}_1\), the complements of \(\varnothing,\{a\},\{b,c\},X\) are respectively \(X,\{b,c\},\{a\},\varnothing\), so complements remain in \(\mathcal{A}_1\). For any sequence of members of \(\mathcal{A}_1\), its union is \(\varnothing\) if all the members are empty, \(\{a\}\) if every member is either empty or \(\{a\}\) and at least one is \(\{a\}\), \(\{b,c\}\) if every member is either empty or \(\{b,c\}\) and at least one is \(\{b,c\}\), and \(X\) if a member is \(X\) or if both \(\{a\}\) and \(\{b,c\}\) occur. Thus \(\mathcal{A}_1\) is a sigma-algebra. The same check applies to \(\mathcal{A}_2\), with \(\{b\}\) and \(\{a,c\}\) in place of \(\{a\}\) and \(\{b,c\}\).
The sets common to the two displayed collections are exactly \(\varnothing\) and \(X\). Therefore \(\mathcal{A}_1\cap\mathcal{A}_2=\{\varnothing,X\}\). This is a sigma-algebra: the complement of \(\varnothing\) is \(X\), the complement of \(X\) is \(\varnothing\), and a union of any sequence of these two sets is again either \(\varnothing\) or \(X\). The calculation illustrates that an intersection can be strictly smaller than either of the sigma-algebras being intersected.
Constructing the Smallest Sigma-Algebra
Given \(\mathcal{E}\subseteq\mathcal{P}(X)\), consider the collection of every sigma-algebra on \(X\) that contains \(\mathcal{E}\). This collection is nonempty: \(\mathcal{P}(X)\) is a sigma-algebra and contains every subset of \(X\), including every member of \(\mathcal{E}\). Intersecting all these sigma-algebras therefore gives a well-defined sigma-algebra. The next result proves that it has exactly the desired minimality property.
Proof. Let \(\mathfrak{S}\) be the collection inside the intersection, so \(\mathfrak{S}=\{\mathcal{A}:\mathcal{A}\text{ is a sigma-algebra on }X\text{ and }\mathcal{E}\subseteq\mathcal{A}\}\). It is nonempty because \(\mathcal{P}(X)\in\mathfrak{S}\). By the intersection theorem, \(\mathcal{G}=\bigcap_{\mathcal{A}\in\mathfrak{S}}\mathcal{A}\) is a sigma-algebra on \(X\). If \(E\in\mathcal{E}\), then \(E\in\mathcal{A}\) for every \(\mathcal{A}\in\mathfrak{S}\), so \(E\in\mathcal{G}\). Thus \(\mathcal{G}\) contains \(\mathcal{E}\). Also, its definition as an intersection gives \(\mathcal{G}\subseteq\mathcal{A}\) for every \(\mathcal{A}\in\mathfrak{S}\). Hence it is a smallest sigma-algebra containing \(\mathcal{E}\). If \(\mathcal{H}\) were another smallest one, then the minimality of \(\mathcal{G}\) would give \(\mathcal{G}\subseteq\mathcal{H}\), and the minimality of \(\mathcal{H}\) would give \(\mathcal{H}\subseteq\mathcal{G}\). Therefore \(\mathcal{H}=\mathcal{G}\), proving uniqueness. We define \(\sigma(\mathcal{E})=\mathcal{G}\). \(\square\)
The construction also gives a practical test: to prove \(\sigma(\mathcal{E})\subseteq\mathcal{A}\), it is enough to verify that \(\mathcal{A}\) is a sigma-algebra containing \(\mathcal{E}\). Conversely, every member of \(\mathcal{E}\) must belong to \(\sigma(\mathcal{E})\), and any sigma-algebra containing \(\mathcal{E}\) must contain all of \(\sigma(\mathcal{E})\). The construction does not require finding a list of all the generated sets before it can be defined.
Worked Example: Generating a Sigma-Algebra from One Set
Let \(X=\{1,2,3,4\}\) and \(\mathcal{E}=\{\{1,2\}\}\). Determine \(\sigma(\mathcal{E})\).
Write \(A=\{1,2\}\), so \(A^c=\{3,4\}\). Consider \(\mathcal{C}=\{\varnothing,A,A^c,X\}\). Complements interchange \(\varnothing\) with \(X\) and \(A\) with \(A^c\), so \(\mathcal{C}\) is closed under complements. A union of members of \(\mathcal{C}\) is \(\varnothing\) if all are empty, is \(A\) if the only nonempty set occurring is \(A\), is \(A^c\) if the only nonempty set occurring is \(A^c\), and is \(X\) if \(X\) occurs or both \(A\) and \(A^c\) occur. Thus \(\mathcal{C}\) is a sigma-algebra containing \(\mathcal{E}\).
Every sigma-algebra on \(X\) containing \(\mathcal{E}\) must contain \(A\). It must also contain \(A^c\) by complement closure, and \(\varnothing\) and \(X\) by the sigma-algebra axioms. It therefore contains all four members of \(\mathcal{C}\). Since \(\mathcal{C}\) is itself such a sigma-algebra, the intersection construction gives \(\sigma(\mathcal{E})=\mathcal{C}=\{\varnothing,\{1,2\},\{3,4\},X\}\). As a check, the four listed sets are distinct and include both required complements.
Worked Example: Two Generators Produce Every Subset
Let \(X=\{1,2,3\}\), \(A=\{1,2\}\), \(B=\{2,3\}\), and \(\mathcal{E}=\{A,B\}\). Find the sigma-algebra generated by \(\mathcal{E}\).
Any sigma-algebra containing \(A\) and \(B\) must contain their complements \(A^c=\{3\}\) and \(B^c=\{1\}\). It must also contain \(A\cap B\): indeed, \(A\cap B=(A^c\cup B^c)^c\), and the right-hand side is formed using complements and a finite union, both permitted by the sigma-algebra axioms. Here \(A\cap B=\{2\}\). Thus any such sigma-algebra contains each singleton \(\{1\},\{2\},\{3\}\). Closure under finite unions, which follows from closure under countable unions, then forces it to contain every subset of \(X\).
To see the last claim explicitly, the subsets obtained by choosing none, one, two, or all three of the singletons are \(\varnothing\), \(\{1\},\{2\},\{3\}\), \(\{1,2\},\{1,3\},\{2,3\}\), and \(X\). These are all \(2^3=8\) subsets of \(X\). Since \(\mathcal{P}(X)\) is itself a sigma-algebra containing \(A\) and \(B\), it follows that \(\sigma(\mathcal{E})=\mathcal{P}(X)\). Check the generators: \(A=\{1,2\}\) and \(B=\{2,3\}\) are among the eight listed sets, as required.
Empty Families and a Common Mistake
The family \(\mathcal{E}\) of generators is allowed to be empty. In that case the smallest sigma-algebra on \(X\) is \(\{\varnothing,X\}\): this collection is a sigma-algebra, and every sigma-algebra on \(X\) contains both \(\varnothing\) and \(X\). If \(X=\varnothing\), these two symbols denote the same set, so the collection has just one member, \(\{\varnothing\}\), and the conclusion remains valid. By contrast, the intersection theorem above was stated for a nonempty indexed collection of sigma-algebras. The intersection of an empty collection is sometimes defined to be \(\mathcal{P}(X)\), but that convention is separate from the proof using a nonempty index set.
Common mistake: replacing intersection by union. The union of sigma-algebras need not be a sigma-algebra. In the first worked example, \(\mathcal{A}_1\cup\mathcal{A}_2\) contains \(\{a\}\), since \(\{a\}\in\mathcal{A}_1\), and it contains \(\{b\}\), since \(\{b\}\in\mathcal{A}_2\). But it does not contain \(\{a,b\}\), which belongs to neither displayed sigma-algebra. Therefore this union is not closed under finite unions and is not a sigma-algebra. The correct construction of the smallest sigma-algebra containing given sets is to intersect all sigma-algebras that contain them, not to take their union.
These results are useful whenever a sigma-algebra is specified indirectly by a collection of sets it must contain. The intersection construction supplies existence without guessing the answer; the minimality property then allows proofs of inclusion by checking only that a candidate collection is a sigma-algebra containing the prescribed sets.
Check Your Understanding
Use the definitions and minimality property to justify each answer.
- Why must all sigma-algebras in an intersection theorem be on the same underlying set \(X\)?
- Suppose \(\mathcal{A}_i\) is a sigma-algebra on \(X\) for every \(i\) in a nonempty index set. Which step in the proof establishes closure of \(\bigcap_i\mathcal{A}_i\) under countable unions?
- For \(X=\{u,v\}\) and \(\mathcal{E}=\{\{u\}\}\), determine \(\sigma(\mathcal{E})\) and verify its complements.
- If \(\mathcal{E}\subseteq\mathcal{A}\) and \(\mathcal{A}\) is a sigma-algebra on \(X\), what inclusion relating \(\sigma(\mathcal{E})\) and \(\mathcal{A}\) follows from minimality?
- Give a reason that the union of two sigma-algebras on the same \(X\) need not be a sigma-algebra.