Sigma-Algebras

From Sets to Measurable Collections

The previous tutorial used compactness and continuity to control sequences of points and function values. Measure theory asks a different kind of question: which subsets of a space can be assigned a consistent notion of size or probability? The answer begins by specifying a carefully organized collection of subsets. A sigma-algebra is the structure that makes this possible.

Let \(X\) be a set, called the underlying space. Its subsets are the objects we want to collect. Some familiar operations on subsets are complements, unions, and intersections. A sigma-algebra is designed to remain stable under complements and under countably many unions. The word “countably” is essential: sequences of sets arise naturally when taking limits, and the collection must include the sets formed by those operations.

Definition: A sigma-algebra on a set \(X\) is a collection \(\mathcal{F}\) of subsets of \(X\) such that:

\(X\in\mathcal{F}\);
if \(A\in\mathcal{F}\), then its complement \(X\setminus A\) belongs to \(\mathcal{F}\);
if \(A_1,A_2,\ldots\in\mathcal{F}\), then \(\bigcup_{n=1}^{\infty}A_n\in\mathcal{F}\).

The members of \(\mathcal{F}\) are called measurable sets (with respect to \(\mathcal{F}\)).

The symbol \(\mathcal{F}\) denotes a family whose elements are themselves sets: \(\mathcal{F}\subseteq\mathcal{P}(X)\), where \(\mathcal{P}(X)\) is the power set of \(X\), the collection of all subsets of \(X\). The first condition ensures that the whole space is included. Taking its complement then also ensures that \(\varnothing\in\mathcal{F}\).

A countable union allows finitely many sets as well: to express \(A_1\cup\cdots\cup A_k\), use the sequence \(A_1,\ldots,A_k,\varnothing,\varnothing,\ldots\). We will establish that \(\varnothing\) belongs to the collection below. This convention lets the countable closure rule cover both finite and infinite unions without separate conditions.

Basic Examples and a Near Miss

Worked Example: The Smallest and Largest Sigma-Algebras

For any set \(X\), consider \(\{\varnothing,X\}\). It contains \(X\). The complement of \(\varnothing\) is \(X\), and the complement of \(X\) is \(\varnothing\). A countable sequence of sets from this collection has union \(X\) if at least one term is \(X\), and has union \(\varnothing\) otherwise. Thus \(\{\varnothing,X\}\) is a sigma-algebra.

At the other extreme, \(\mathcal{P}(X)\) is a sigma-algebra: complements and unions of subsets of \(X\) are still subsets of \(X\). These two examples work for every \(X\). The first distinguishes only the whole space from the empty set; the second includes every subset. Other sigma-algebras lie between them, recording different amounts of information about \(X\).

Worked Example: A Sigma-Algebra from a Partition

Let \(X=\{a,b,c\}\), partitioned into the two disjoint blocks \(B_1=\{a,b\}\) and \(B_2=\{c\}\). Consider the collection of all unions of these blocks:

\mathcal{F}=\{\varnothing,\{a,b\},\{c\},X\}.

This list is exhaustive: a union of blocks chooses neither block, just \(B_1\), just \(B_2\), or both. The complement in \(X\) of each listed set is also listed: the complements of \(\varnothing,X,\{a,b\},\{c\}\) are respectively \(X,\varnothing,\{c\},\{a,b\}\). Any union of a sequence of members of \(\mathcal{F}\) is a union of some of the two blocks, so it is again in \(\mathcal{F}\). Therefore \(\mathcal{F}\) is a sigma-algebra. It can distinguish whether a point lies in \(\{a,b\}\) or in \(\{c\}\), but it cannot distinguish \(a\) from \(b\): the singleton \(\{a\}\) is not in \(\mathcal{F}\).

Finite closure alone is not enough. A family may be closed under complements and finite unions while failing the countable-union condition. This distinction is one reason sigma-algebras are stronger than collections defined using only finite operations.

Worked Example: The Finite-or-Cofinite Family Is Not a Sigma-Algebra

Take \(X=\mathbb{N}=\{1,2,3,\ldots\}\), and let \(\mathcal{C}\) consist of all finite subsets of \(\mathbb{N}\) together with all subsets whose complements in \(\mathbb{N}\) are finite. Such sets are called cofinite. The collection contains \(\mathbb{N}\), and taking a complement swaps finite sets with cofinite sets. It is also closed under finite unions: the union of two finite sets is finite, and any union involving a cofinite set is cofinite.

For each \(n\), the singleton \(\{2n\}\) belongs to \(\mathcal{C}\), since it is finite. But their countable union is the set of positive even integers:

\bigcup_{n=1}^{\infty}\{2n\}=\{2,4,6,\ldots\}.

This union is not finite, and its complement, the set of positive odd integers, is also infinite. Therefore the union is not cofinite either, so it does not belong to \(\mathcal{C}\). The family fails the countable-union condition and is not a sigma-algebra.

Consequences of the Closure Rules

Although the definition names complements and countable unions, these operations imply other useful closure properties. In particular, countable intersections remain measurable. The proof makes explicit why the complement condition must accompany the union condition.

Theorem (Closure Under Countable Intersections): If \(\mathcal{F}\) is a sigma-algebra on \(X\) and \(A_n\in\mathcal{F}\) for every \(n\in\mathbb{N}\), then \(\bigcap_{n=1}^{\infty}A_n\in\mathcal{F}\).

Proof. Since each \(A_n\in\mathcal{F}\), the complement \(X\setminus A_n\) belongs to \(\mathcal{F}\). The countable-union condition then gives \(\bigcup_{n=1}^{\infty}(X\setminus A_n)\in\mathcal{F}\). Taking its complement, which is also in \(\mathcal{F}\), and applying De Morgan’s law yields

X\setminus\bigcup_{n=1}^{\infty}(X\setminus A_n) =\bigcap_{n=1}^{\infty} A_n.

Hence \(\bigcap_{n=1}^{\infty} A_n\in\mathcal{F}\), as required. \(\square\)

This argument is a general method: convert an intersection into the complement of a union of complements, use the operations supplied by the definition, and then identify the result by De Morgan’s law. The same idea also handles more elaborate set expressions.

Theorem (Closure Under Relative Differences): If \(A,B\in\mathcal{F}\), then \(A\setminus B\in\mathcal{F}\).

Proof. Since \(B\in\mathcal{F}\), its complement \(X\setminus B\) belongs to \(\mathcal{F}\). By the countable-intersection theorem, the intersection of \(A\) and \(X\setminus B\) belongs to \(\mathcal{F}\). But

A\cap(X\setminus B)=A\setminus B.

Therefore \(A\setminus B\in\mathcal{F}\). \(\square\)

Worked Example: Combining Countable Operations

Suppose \(A_n\in\mathcal{F}\) for every \(n\), and \(B\in\mathcal{F}\). The set of points that belong to at least one \(A_n\) but do not belong to \(B\) is

\left(\bigcup_{n=1}^{\infty}A_n\right)\setminus B.

The union belongs to \(\mathcal{F}\) by its defining countable-union property. The relative-difference theorem therefore shows that the displayed set belongs to \(\mathcal{F}\) as well. Equivalently, it is \(\bigcup_{n=1}^{\infty}(A_n\setminus B)\). To check this identity, a point belongs to either side exactly when it lies in some \(A_n\) and does not lie in \(B\). Each \(A_n\setminus B\) is also in \(\mathcal{F}\), so the countable union on the right belongs to \(\mathcal{F}\) directly.

Building the Smallest Sigma-Algebra Containing Given Sets

Often a problem specifies certain subsets that must be measurable, without listing every measurable set. We can ask for the smallest sigma-algebra that contains those prescribed sets. “Smallest” here means contained in every other sigma-algebra with that property. The construction uses an intersection of collections.

Theorem (Existence of the Generated Sigma-Algebra): Let \(\mathcal{A}\) be any collection of subsets of \(X\). There is a smallest sigma-algebra on \(X\) containing \(\mathcal{A}\). It is called the sigma-algebra generated by \(\mathcal{A}\), and is denoted \(\sigma(\mathcal{A})\).

Proof. Consider the collection \(\mathfrak{S}\) of all sigma-algebras on \(X\) that contain \(\mathcal{A}\). This collection is nonempty because \(\mathcal{P}(X)\) is a sigma-algebra and contains every subset of \(X\), hence contains \(\mathcal{A}\). Define

\mathcal{G}=\bigcap_{\mathcal{F}\in\mathfrak{S}}\mathcal{F}.

Thus a set belongs to \(\mathcal{G}\) exactly when it belongs to every sigma-algebra in \(\mathfrak{S}\). Since each such sigma-algebra contains \(X\), we have \(X\in\mathcal{G}\). If \(E\in\mathcal{G}\), then \(E\) belongs to every \(\mathcal{F}\in\mathfrak{S}\); each of those collections contains \(X\setminus E\), so \(X\setminus E\in\mathcal{G}\). Finally, if \(E_n\in\mathcal{G}\) for every \(n\), then all the \(E_n\) belong to every \(\mathcal{F}\in\mathfrak{S}\). Each \(\mathcal{F}\) contains their union, so \(\bigcup_{n=1}^{\infty}E_n\in\mathcal{G}\). This proves that \(\mathcal{G}\) is a sigma-algebra.

Every member of \(\mathfrak{S}\) contains \(\mathcal{A}\), so \(\mathcal{G}\) contains \(\mathcal{A}\). Also, by its construction as an intersection, \(\mathcal{G}\subseteq\mathcal{F}\) for every \(\mathcal{F}\in\mathfrak{S}\). It is therefore the smallest sigma-algebra containing \(\mathcal{A}\). This also proves uniqueness: two smallest such sigma-algebras must each be contained in the other, and hence must be equal. \(\square\)

The intersection in this proof is an intersection of families of sets, not an intersection of individual measurable subsets. This distinction matters: \(\sigma(\mathcal{A})\) is itself a collection of subsets of \(X\). The theorem provides a construction even when explicitly listing all sets generated by \(\mathcal{A}\) would be difficult.

Why Countable Closure Matters

Countable unions and intersections let us form sets described by “at least one” or “every” condition across a sequence. In analysis, limits are defined through infinitely many indexed conditions. For example, convergence of a sequence of real-valued functions is often expressed using countable unions and intersections of sets involving their values. A collection closed only under finite operations may fail to contain the resulting limit set.

Sigma-algebras also set the domain on which measures are defined. A measure assigns sizes to sets while respecting countable disjoint unions; probability is a measure whose total size is one. Not every subset of every space is necessarily included in the collection chosen for a measure. The sigma-algebra specifies which sets are available for measurement, while the measure specifies their sizes. These are separate parts of the structure.

On \(\mathbb{R}\), an important example is the Borel sigma-algebra: the sigma-algebra generated by all open subsets of \(\mathbb{R}\). The generation theorem guarantees that this is a well-defined smallest sigma-algebra containing the open sets. Its members include open sets and all sets obtainable from them using complements and countable unions, such as countable intersections. This description is more useful than trying to list every Borel set individually.

A common pitfall is to treat “closed under unions” as though it automatically meant closed under countable unions. It does not. The finite-or-cofinite example shows that a family may satisfy complement and finite-union rules and still fail as a sigma-algebra. Another pitfall is to think that every sigma-algebra must contain every subset of \(X\); only the power set sigma-algebra has that property. The choice of collection is part of the mathematical model.

Check Your Understanding

Use the closure rules and constructions developed here to answer the following questions.

Why does the definition imply that \(\varnothing\) belongs to every sigma-algebra on \(X\)?
How does De Morgan’s law prove closure under countable intersections?
Why does the finite-or-cofinite family on \(\mathbb{N}\) fail to be a sigma-algebra even though it is closed under complements and finite unions?
In the generated sigma-algebra theorem, why is the family being intersected nonempty?
What does “smallest sigma-algebra containing \(\mathcal{A}\)” mean in terms of inclusion?

What You'll Learn