Two Different Kinds of Closure
Some arguments about sets begin with a small family and need to control what happens when its members are combined. Two useful kinds of family are pi-systems and lambda-systems. A pi-system is governed by intersections; a lambda-system is governed by complements and countable disjoint unions. These conditions are not interchangeable. In particular, intersection closure alone does not imply that a family is a lambda-system.
Fix a nonempty set \(X\). The families below are collections of subsets of \(X\), so they are contained in the power set \(\mathcal{P}(X)\). Complements are always relative to \(X\): the complement of \(A\subseteq X\) is \(A^c=X\setminus A\). We use the convention that a pi-system is nonempty and closed under intersections of two of its members. A lambda-system must contain \(X\), be closed under complements relative to \(X\), and be closed under countable unions of pairwise disjoint members.
The restrictions are worth reading literally. A pi-system need not contain \(X\) or \(\varnothing\); its requirement is only about intersections of members it already contains. A lambda-system must contain \(X\), and its union condition applies to pairwise disjoint members, not to arbitrary countable unions. The two definitions therefore call for separate checks.
Interval Families on the Real Line
Worked Example: Closed Bounded Intervals Form a Pi-System When Empty Is Included
Let \(X=\mathbb{R}\) and set \(\mathcal{P}=\{\varnothing\}\cup\{[a,b]:a,b\in\mathbb{R},\ a\le b\}\). We test two arbitrary members. If either is \(\varnothing\), their intersection is \(\varnothing\), which belongs to \(\mathcal{P}\). Otherwise write them as \([a,b]\) and \([c,d]\). A point belongs to both exactly when it is at least \(\max(a,c)\) and at most \(\min(b,d)\). Thus, if \(\max(a,c)\le\min(b,d)\), their intersection is the interval \([\max(a,c),\min(b,d)]\); if \(\max(a,c)>\min(b,d)\), it is \(\varnothing\). In either case the intersection is in \(\mathcal{P}\), so \(\mathcal{P}\) is a pi-system.
It is not a lambda-system on \(X=\mathbb{R}\): \(\mathbb{R}\notin\mathcal{P}\), so the first lambda-system condition fails. Adding \(\mathbb{R}\) would not be enough by itself. For example, the complement in \(\mathbb{R}\) of \([0,1]\) is \((-\infty,0)\cup(1,\infty)\), which is not a member of the family of closed bounded intervals together with \(\varnothing\) and \(\mathbb{R}\). Check: the two intervals \([0,2]\) and \([1,3]\) intersect in \([1,2]\), while \([0,1]\) and \([2,3]\) intersect in \(\varnothing\). Both results are in \(\mathcal{P}\), as the general calculation predicts.
Worked Example: Initial Closed Rays
Again let \(X=\mathbb{R}\), and consider \(\mathcal{H}=\{\varnothing,\mathbb{R}\}\cup\{(-\infty,a]:a\in\mathbb{R}\}\). The intersection of two nonempty proper rays is \([(-\infty,a])\cap(-\infty,b]=(-\infty,\min(a,b)]\). An intersection involving \(\varnothing\) is \(\varnothing\), and intersecting with \(\mathbb{R}\) leaves the other set unchanged. All these results belong to \(\mathcal{H}\), so \(\mathcal{H}\) is a pi-system.
To test the lambda-system condition on complements, take \((-\infty,0]\in\mathcal{H}\). Its complement relative to \(\mathbb{R}\) is \((0,\infty)\), which is neither \(\varnothing\), \(\mathbb{R}\), nor an initial closed ray. Hence \(\mathcal{H}\) is not a lambda-system. This single failed condition is enough; there is no need to test its disjoint-union condition once complement closure has failed.
Proof. By the definition of a sigma-algebra, \(X\in\mathcal{A}\), and \(\mathcal{A}\) is closed under complements and countable unions. It is nonempty because it contains \(X\). For \(A,B\in\mathcal{A}\), the set \(A\cap B\) is in \(\mathcal{A}\), since \(A\cap B=(A^c\cup B^c)^c\) and the right-hand side is obtained using complements and a finite, hence countable, union. This proves that \(\mathcal{A}\) is a pi-system. It already contains \(X\) and is closed under complements. If \(A_1,A_2,\ldots\in\mathcal{A}\) are pairwise disjoint, then their countable union is in \(\mathcal{A}\) by countable-union closure. Thus all three lambda-system conditions hold. \(\square\)
This proposition gives a useful direction of implication, not an equivalence: a family satisfying just the pi-system condition need not be a sigma-algebra or a lambda-system. The interval examples show why the extra requirements must be checked rather than inferred from intersection closure.
Classifying Families on Finite Spaces
Worked Example: A Pi-System on Three Points That Is Not a Lambda-System
Let \(X=\{1,2,3\}\), and take \(\mathcal{P}=\{\varnothing,\{1\},\{1,2\},X\}\). To check the pi-system condition, list the intersections among the three nonempty proper members: \(\{1\}\cap\{1\}=\{1\}\), \(\{1\}\cap\{1,2\}=\{1\}\), and \(\{1,2\}\cap\{1,2\}=\{1,2\}\). Intersecting any member with \(X\) leaves it unchanged; intersecting any member with \(\varnothing\) gives \(\varnothing\). Every result is in \(\mathcal{P}\), so this family is a pi-system.
It contains \(X\), but it is not a lambda-system: the complement of \(\{1\}\) relative to \(X\) is \(\{2,3\}\), and \(\{2,3\}\notin\mathcal{P}\). Thus the failure is precisely in the complement condition. Check: \(\{1\}\cap\{1,2\}=\{1\}\) passes the pi-system test, whereas \(X\setminus\{1\}=\{2,3\}\) fails the lambda-system test. These calculations illustrate the different operations the definitions require.
Worked Example: Unions of Blocks of a Partition
Let \(X=\{1,2,3\}\) and partition it into blocks \(B_1=\{1,3\}\) and \(B_2=\{2\}\). Consider the family of all unions of these blocks: \(\mathcal{D}=\{\varnothing,\{1,3\},\{2\},X\}\). It contains \(X\). Complements interchange \(\varnothing\) with \(X\), and interchange \(\{1,3\}\) with \(\{2\}\), so every complement remains in \(\mathcal{D}\).
For the disjoint-union test, each member of \(\mathcal{D}\) is a union of blocks from the two-block partition. In a pairwise disjoint sequence of such members, a block cannot occur in two different members: if it did, those two members would share every point in that block, contradicting disjointness. The union of the sequence is therefore still a union of some of \(B_1,B_2\), and belongs to \(\mathcal{D}\). This argument includes sequences containing \(\varnothing\), which contributes no block. Thus \(\mathcal{D}\) is a lambda-system. It is also a pi-system: the intersection of two unions of blocks is the union of exactly the blocks present in both, so it belongs to \(\mathcal{D}\). Directly, the intersections of its members are again among \(\varnothing,\{1,3\},\{2\},X\).
Check: the disjoint members \(\{1,3\}\) and \(\{2\}\) have union \(X\in\mathcal{D}\), and the complement of \(\{2\}\) is \(\{1,3\}\in\mathcal{D}\). These checks agree with the general block argument.
A Reliable Membership Test
For any proposed family, write the conditions separately and check them in the ambient set \(X\). A useful order is to first test whether the family is nonempty and intersection-closed if you are testing for a pi-system. For a lambda-system, test \(X\), then complements, then countable pairwise disjoint unions. A single counterexample to any required condition disproves membership. For a finite underlying set, the possible members are readily listed, but the definition still quantifies over every permitted choice of sets or sequence.
Write down \(X\), because complements depend on it. For example, the complement of \([0,1]\) in \([0,2]\) differs from its complement in \(\mathbb{R}\).
Choose arbitrary \(A,B\) from the family and verify that \(A\cap B\) remains in it. Do not assume that \(\varnothing\) or \(X\) is present unless the definition or the family supplies it.
Check \(X\), complements relative to \(X\), and unions of pairwise disjoint sequences. Closure under arbitrary unions is not one of the defining conditions.
A common mistake is to reason that a family is a lambda-system merely because intersections of its members stay in the family. For the finite family \(\{\varnothing,\{1\},\{1,2\},\{1,2,3\}\}\) on \(X=\{1,2,3\}\), all pairwise intersections do stay in the family. The incorrect conclusion “therefore it is a lambda-system” overlooks complements: \(X\setminus\{1\}=\{2,3\}\), which is absent. The correction is to test the lambda-system conditions independently; this family is a pi-system but not a lambda-system.
Check Your Understanding
For each question, specify the underlying set and apply the relevant closure conditions directly.
- On \(X=\mathbb{R}\), is \(\{\varnothing,\mathbb{R}\}\cup\{[a,\infty):a\in\mathbb{R}\}\) a pi-system? Is it a lambda-system?
- For \(\mathcal{P}=\{\varnothing,[0,1],[1,2]\}\) on \(X=\mathbb{R}\), identify a pair whose intersection shows whether \(\mathcal{P}\) is a pi-system.
- On \(X=\{a,b,c\}\), test whether \(\{\varnothing,\{a\},\{b,c\},X\}\) is a lambda-system and a pi-system.
- Explain why the disjoint-union condition for a lambda-system does not state closure under all countable unions.
- State which fact about sigma-algebras guarantees that every sigma-algebra is a lambda-system, and which fact guarantees that it is also a pi-system.