Closure Properties of Lambda-Systems

From the Axioms to Useful Closure Rules

Lambda-systems are set families with enough closure to support many arguments about generated sigma-algebras, but they need not be sigma-algebras themselves. Their defining union condition applies to disjoint sets, not to arbitrary sets. A useful first step is to extract other closure rules from the axioms and to keep track of exactly where disjointness is required.

Fix a set \(X\). A family \(\mathcal{D}\subseteq\mathcal{P}(X)\), where \(\mathcal{P}(X)\) is the collection of all subsets of \(X\), is called a lambda-system on \(X\) if it satisfies the following conditions:

These axioms immediately include the empty set: since \(X\in\mathcal{D}\), taking its complement gives \(\varnothing\in\mathcal{D}\). The countable-union axiom allows empty terms, so it also covers finite disjoint unions by padding a finite list with empty sets. It does not assert closure under arbitrary countable unions.

Nested Differences

Proof. Since \(A\subseteq B\), the sets \(A\) and \(X\setminus B\) are disjoint: a point in \(A\) belongs to \(B\), so it cannot belong to \(X\setminus B\). Both sets belong to \(\mathcal{D}\), by the assumption \(A\in\mathcal{D}\) and the complement axiom applied to \(B\). The disjoint-union axiom therefore gives \(A\cup(X\setminus B)\in\mathcal{D}\). Its complement in \(X\) is \[ X\setminus\bigl(A\cup(X\setminus B)\bigr)=B\setminus A. \] The complement axiom now gives \(B\setminus A\in\mathcal{D}\). This also covers \(A=B\), when the difference is \(\varnothing\), and \(A=\varnothing\), when it is \(B\). \(\square\)

The containment condition matters: this proposition says nothing about \(B\setminus A\) when \(A\) and \(B\) are unrelated members. Nor does it establish closure under intersections. It is specifically the nested difference that the lambda-system axioms provide.

Increasing Unions and an Equivalent Formulation

A sequence \(A_1\subseteq A_2\subseteq\cdots\) is called increasing. Lambda-systems are closed under increasing countable unions, even though they are not necessarily closed under all countable unions. To obtain this rule, convert the increasing sequence into disjoint pieces.

Proof. Define \(C_1=A_1\) and, for \(n\ge2\), define \(C_n=A_n\setminus A_{n-1}\). The nested-difference proposition shows that \(C_n\in\mathcal{D}\) for each \(n\ge2\); also \(C_1\in\mathcal{D}\) by assumption. The sets \(C_n\) are pairwise disjoint: if \(i<j\), then \(C_i\subseteq A_i\subseteq A_{j-1}\), while \(C_j=A_j\setminus A_{j-1}\), so \(C_i\cap C_j=\varnothing\). For each \(N\), the union of the first \(N\) pieces is \(A_N\), since \(C_1=A_1\) and each later piece adds exactly \(A_n\setminus A_{n-1}\). Consequently \(\bigcup_{n=1}^{\infty}C_n=\bigcup_{n=1}^{\infty}A_n\). The disjoint-union axiom applied to the \(C_n\) proves the claim. \(\square\)

The standard axioms can also be replaced by a formulation using nested differences and increasing unions. This equivalence helps separate the roles of the closure conditions.

Proof. Suppose first that \(\mathcal{D}\) is a lambda-system. The nested-difference proposition proves (ii), and the increasing-union proposition proves (iii); (i) is an axiom.

Conversely, suppose (i)–(iii) hold. Setting \(A=\varnothing\) is not yet justified, so first use (ii) with \(A=B=X\) to obtain \(X\setminus X=\varnothing\in\mathcal{D}\). For any \(A\in\mathcal{D}\), use (ii) with \(A\subseteq X\) to get \(X\setminus A\in\mathcal{D}\). Thus the complement axiom holds. Now let \(A_1,A_2,\ldots\in\mathcal{D}\) be pairwise disjoint. We show by induction that \(S_N=\bigcup_{n=1}^{N}A_n\in\mathcal{D}\). The base case is \(S_1=A_1\). If \(S_N\in\mathcal{D}\), then \(A_{N+1}\subseteq X\setminus S_N\), because \(A_{N+1}\) is disjoint from each of \(A_1,\ldots,A_N\). The complement \(X\setminus S_N\) is in \(\mathcal{D}\), as just proved, and (ii) gives \((X\setminus S_N)\setminus A_{N+1}\in\mathcal{D}\). Its complement in \(X\), also in \(\mathcal{D}\), is \(S_N\cup A_{N+1}=S_{N+1}\). Hence every \(S_N\in\mathcal{D}\). The sequence \(S_1\subseteq S_2\subseteq\cdots\) is increasing, so (iii) gives \(\bigcup_N S_N=\bigcup_n A_n\in\mathcal{D}\). This is the disjoint-union axiom, completing the proof. \(\square\)

What the Disjointness Requirement Does Not Give

Even-cardinality subsets of a four-element universe form a lambda-system, but not a sigma-algebra. For example, \(A=\{1,2\}\) and \(B=\{2,3\}\) both belong to that family, while \(A\cup B=\{1,2,3\}\) has odd cardinality and does not. The sets overlap, so the disjoint-union axiom cannot be applied. This is not a defect in the proof of closure: arbitrary unions are simply not among the lambda-system axioms.

A Common Mistake

A common error is to infer that a lambda-system is closed under unions just because it contains both sets being united. In the even-cardinality example, the calculation \(|\{1,2\}\cup\{2,3\}|=2+2=4\) is wrong: it counts the shared element \(2\) twice. The correct calculation is \(2+2-1=3\), so the union is not in the lambda-system. The union axiom applies only when the sets are pairwise disjoint; in that case, cardinalities add without subtracting overlaps. For nested sets, the justified operation is instead the difference \(B\setminus A\), proved above.

These closure rules are useful because they let one decompose increasing sequences into disjoint increments and manipulate nested sets without assuming full sigma-algebra closure. When a later argument needs arbitrary unions or intersections, it must establish those properties separately rather than treating them as consequences of the lambda-system axioms.