From Finite Closure to Countable Closure
Let \(X\) be a set, and let \(\mathcal{F}\) be a collection of subsets of \(X\). Closure under complements and finite unions is not enough for \(\mathcal{F}\) to be a sigma-algebra: a sequence of sets can have a union that is not obtainable using only finitely many unions. A sigma-algebra requires closure under countable unions, which is the key additional condition.
This tutorial concerns collections of sets, not a measure space: no measure or measurable functions are assumed. We write \(E^c=X\setminus E\) for the complement of \(E\) relative to \(X\). A countable family may be indexed by the positive integers; finite unions are included by allowing some of the sets in the sequence to be empty.
The set \(X\) in the first condition matters: closure under complements is defined relative to \(X\). The countable-union condition applies to every sequence of members, including a sequence that repeats sets or contains empty sets. It does not require the sets to be disjoint or increasing.
Basic Examples and Non-Examples
Worked Example: The Power Set
For any set \(X\), its power set \(\mathcal{P}(X)\) is the collection of all subsets of \(X\). We check the three conditions directly. First, \(X\subseteq X\), so \(X\in\mathcal{P}(X)\). If \(E\in\mathcal{P}(X)\), then \(E\subseteq X\), and consequently \(E^c=X\setminus E\subseteq X\); hence \(E^c\in\mathcal{P}(X)\). Finally, if \(E_n\in\mathcal{P}(X)\) for every \(n\), then each \(E_n\subseteq X\). Any \(x\in\bigcup_{n=1}^{\infty}E_n\) belongs to at least one \(E_n\), and so belongs to \(X\). Thus \(\bigcup_{n=1}^{\infty}E_n\subseteq X\), which puts the union in \(\mathcal{P}(X)\). Therefore \(\mathcal{P}(X)\) is a sigma-algebra.
Check: if \(X=\{a,b\}\), then \(\mathcal{P}(X)=\{\varnothing,\{a\},\{b\},X\}\). The complement of \(\{a\}\) is \(\{b\}\), and the union of any sequence drawn from these four sets is one of the same four sets. If \(X=\varnothing\), its power set is \(\{\varnothing\}\), which also satisfies all three conditions.
Worked Example: The Trivial Sigma-Algebra
For any set \(X\), consider \(\mathcal{T}=\{\varnothing,X\}\). It contains \(X\), and taking complements exchanges its two members: \(\varnothing^c=X\) and \(X^c=\varnothing\). Now take any sequence \(E_n\in\mathcal{T}\). If every \(E_n=\varnothing\), its union is \(\varnothing\). If at least one \(E_n=X\), its union is \(X\). In either case, \(\bigcup_n E_n\in\mathcal{T}\), so \(\mathcal{T}\) is a sigma-algebra.
Check: when \(X=\varnothing\), the two displayed symbols \(\varnothing\) and \(X\) denote the same set, so \(\mathcal{T}=\{\varnothing\}\). It still contains \(X\), is closed under complements, and is closed under countable unions.
Worked Example: Unions of Blocks in a Partition
Let \(X=\{1,2,3,4,5,6\}\), partitioned into the three disjoint, nonempty blocks \(A=\{1,2\}\), \(B=\{3,4\}\), and \(C=\{5,6\}\). Consider the collection of all unions of these blocks, including the empty union. Its members are \(\varnothing,A,B,C,A\cup B,A\cup C,B\cup C,X\).
The collection contains \(X=A\cup B\cup C\). The complement of any union of blocks is the union of the blocks not selected: for example, \((A\cup C)^c=B\), so complements remain in the collection. For countable-union closure, take any sequence \(E_n\) from the collection. Each \(E_n\) selects some of \(A,B,C\). The union \(\bigcup_n E_n\) contains exactly those blocks selected at least once. Since there are only three blocks, the result is one of the eight listed sets.
For a concrete sequence, let \(E_1=A\), \(E_2=B\cup C\), and \(E_n=\varnothing\) for \(n\ge3\). Then \(\bigcup_{n=1}^{\infty}E_n=A\cup B\cup C=X\), which belongs to the collection. The argument applies to every sequence, not just this one, so this collection is a sigma-algebra.
Why Countable Closure Is a Separate Requirement
Worked Example: Finite and Cofinite Sets
Let \(X=\mathbb{N}=\{1,2,3,\ldots\}\), and let \(\mathcal{C}\) consist of all finite subsets of \(X\) and all subsets whose complements in \(X\) are finite. Such sets are called cofinite when their complements are finite.
This collection contains \(X\), which is cofinite, and \(\varnothing\), which is finite. Taking a complement exchanges finite and cofinite sets. It is also closed under finite unions: a union of two finite sets is finite; a union containing a cofinite set is cofinite, because its complement is contained in the finite complement of that cofinite set. Repeating this argument gives closure under any finite union.
But for each \(n\ge1\), the singleton \(E_n=\{2n\}\) is finite and so belongs to \(\mathcal{C}\). Their countable union is the set of positive even integers, \(E=\{2,4,6,\ldots\}\). This set is infinite, so it is not finite. Its complement, the positive odd integers \(\{1,3,5,\ldots\}\), is also infinite, so \(E\) is not cofinite. Thus \(E\notin\mathcal{C}\), even though every \(E_n\in\mathcal{C}\). Therefore \(\mathcal{C}\) is not a sigma-algebra.
Check: both \(E\) and \(E^c\) contain infinitely many distinct integers: \(2,4,6,\ldots\) belong to \(E\), while \(1,3,5,\ldots\) belong to \(E^c\). This verifies both parts of the claim that \(E\) is neither finite nor cofinite.
The example shows the logical gap: checking unions of two sets, and then iterating that check finitely many times, does not test unions of arbitrary sequences. To establish the countable condition, the argument must start with an arbitrary sequence and show that its union belongs to the collection.
A Useful Equivalent Closure Test
Proof. Suppose first that \(\mathcal{M}\) is a sigma-algebra, and take a sequence \(E_n\in\mathcal{M}\). Each \(E_n^c\) belongs to \(\mathcal{M}\) by complement closure. Countable-union closure gives \(\bigcup_{n=1}^{\infty}E_n^c\in\mathcal{M}\). Taking its complement again gives a member of \(\mathcal{M}\), and De Morgan's identity identifies that complement as \(\bigcap_{n=1}^{\infty}E_n\). Thus \(\mathcal{M}\) is closed under countable intersections.
Conversely, suppose \(\mathcal{M}\) contains \(X\), is closed under complements, and is closed under countable intersections. Given \(E_n\in\mathcal{M}\), each \(E_n^c\in\mathcal{M}\), so \(\bigcap_{n=1}^{\infty}E_n^c\in\mathcal{M}\). Its complement belongs to \(\mathcal{M}\), and De Morgan's identity gives \(\left(\bigcap_{n=1}^{\infty}E_n^c\right)^c=\bigcup_{n=1}^{\infty}E_n\). Hence \(\mathcal{M}\) is closed under countable unions and satisfies all the sigma-algebra conditions. \(\square\)
As a consequence, every sigma-algebra contains \(\varnothing\), since \(\varnothing=X^c\), and is closed under countable intersections. It is also closed under differences: if \(E,F\in\mathcal{M}\), then \(F^c\in\mathcal{M}\), and \(E\setminus F=E\cap F^c\in\mathcal{M}\). Here the intersection is a countable intersection with the sequence \(E,F,E,F,\ldots\), so it follows from the proposition.
A Common Closure Mistake
Pitfall: after checking that a collection contains \(X\), is closed under complements, and is closed under finite unions, it is tempting to conclude that it is a sigma-algebra. That conclusion is invalid unless countable-union closure has also been checked. For \(\mathcal{C}\) above, each \(\{2n\}\) is an allowed set, and every finite union \(\{2\}\cup\{4\}\cup\cdots\cup\{2k\}\) is finite and allowed. The incorrect calculation is to infer from these finite unions that \(\bigcup_{n=1}^{\infty}\{2n\}\) belongs to \(\mathcal{C}\). The correction is to identify the infinite union exactly: it is the infinite, non-cofinite set of positive even integers, so it is not in \(\mathcal{C}\).
When testing a proposed sigma-algebra, keep the conditions separate: check membership of \(X\), complements relative to \(X\), and the union of an arbitrary countable sequence. A successful check of finite unions is useful, but it cannot replace the last test.
Check Your Understanding
For each question, distinguish what follows from finite closure from what requires countable closure.
- Give the three defining conditions for a sigma-algebra on \(X\). Which condition directly tests countable unions?
- For \(X=\{a,b,c\}\), list all members of the power set and explain why it is a sigma-algebra.
- Let \(X=\mathbb{N}\) and \(E_n=\{2n\}\). Why does \(\bigcup_{n=1}^{\infty}E_n\) show that the finite-cofinite collection is not a sigma-algebra?
- Use De Morgan's identity to explain why closure under complements and countable unions implies closure under countable intersections.
- For the partition \(\{\{1,2\},\{3,4\},\{5,6\}\}\), identify the union of the blocks \(\{1,2\}\) and \(\{5,6\}\), and then identify its complement.