Examples of Sigma-Algebras

From the Definition to Examples

The definition of a sigma-algebra gives three closure requirements, but examples help show how those requirements shape a collection of sets. Some sigma-algebras contain almost no sets beyond those forced by the definition; others contain every subset of the space. Between these extremes are families that encode a partition, a notion of smallness, or the information retained by a function.

Throughout, complements are relative to the specified underlying set $X$. We will use the results from the previous tutorials that every sigma-algebra contains $\varnothing$, is closed under finite unions and relative differences, and is closed under countable intersections. When checking a proposed example, it is important to verify countable closure, not merely the corresponding finite closure.

The Smallest and Largest Examples

Example (The Trivial Sigma-Algebra): For any set $X$, the family $\{\varnothing,X\}$ is a sigma-algebra on $X$. It is called the trivial sigma-algebra.

To check this, the family contains $X$. The complement of $\varnothing$ is $X$, and the complement of $X$ is $\varnothing$. A sequence chosen from this family has union $\varnothing$ if every term is empty, and union $X$ if at least one term is $X$. Thus its countable union also belongs to the family.

Example (The Power-Set Sigma-Algebra): The power set $\mathcal{P}(X)$, consisting of all subsets of $X$, is a sigma-algebra on $X$.

Indeed, $X$ is a subset of itself. The complement in $X$ of any subset of $X$ is another subset of $X$, and the union of any sequence of subsets of $X$ is still a subset of $X$. These examples work for finite and infinite underlying sets alike.

Worked Example: Comparing the Two Extremes

Let $X=\{p,q,r\}$. The trivial sigma-algebra is $\{\varnothing,X\}$. It does not contain $\{p\}$, so it is not the family of all subsets of $X$. The power-set sigma-algebra contains all eight subsets:

\mathcal{P}(X)= \{\varnothing,\{p\},\{q\},\{r\}, \{p,q\},\{p,r\},\{q,r\},X\}.

For example, the complement of $\{p\}$ in $X$ is $\{q,r\}$, and both sets are in $\mathcal{P}(X)$. Any union of members of $\mathcal{P}(X)$ is a subset of $X$, so it too is in $\mathcal{P}(X)$. Thus these are two different sigma-algebras on the same space.

Sets That Are Countable or Cocomountable

A useful example classifies sets according to whether they, or their complements, are countable. Here, “countable” includes finite sets. A set whose complement is countable is called cocountable.

Theorem (Countable-Cocountable Sigma-Algebra): For any set $X$, the family $$ \mathcal{C}=\{A\subseteq X: A\text{ is countable or }X\setminus A\text{ is countable}\} $$ is a sigma-algebra on $X$.

Proof. The set $X$ is in $\mathcal{C}$, since $X\setminus X=\varnothing$ is countable. If $A\in\mathcal{C}$, then either $A$ is countable or $X\setminus A$ is countable. The complement of $A$ is $X\setminus A$, so in either case that complement is countable or has countable complement. Hence $X\setminus A\in\mathcal{C}$.

Now take a sequence $A_n\in\mathcal{C}$. If every $A_n$ is countable, then $\bigcup_{n=1}^{\infty}A_n$ is countable, so it belongs to $\mathcal{C}$. Otherwise, at least one set, say $A_j$, is cocountable, meaning $X\setminus A_j$ is countable. Since $A_j\subseteq\bigcup_n A_n$, we have

X\setminus\bigcup_{n=1}^{\infty}A_n \subseteq X\setminus A_j.

A subset of a countable set is countable. Therefore the union is cocountable and belongs to $\mathcal{C}$. This verifies all three defining conditions. $\square$

Worked Example: A Countable Space and an Uncountable Space

If $X=\mathbb{N}$, every subset of $X$ is countable. Consequently every subset is in $\mathcal{C}$, and $\mathcal{C}=\mathcal{P}(\mathbb{N})$.

By contrast, let $X=\mathbb{R}$ and $A=\mathbb{Z}$. The set $A$ is countable, so $A\in\mathcal{C}$. Its complement is cocountable and is also in $\mathcal{C}$. But the interval $(0,1)$ is neither countable nor cocountable: it is uncountable, and its complement in $\mathbb{R}$ is uncountable. Thus $(0,1)\notin\mathcal{C}$. On $\mathbb{R}$, this sigma-algebra is a proper subfamily of $\mathcal{P}(\mathbb{R})$.

Building a Sigma-Algebra from a Partition

A partition of $X$ is a collection of nonempty, pairwise disjoint subsets of $X$ whose union is $X$. Its members are called blocks. A family of sets can be formed by allowing unions of blocks: such a set contains each block either in full or not at all.

Theorem (Sigma-Algebra of Unions of Partition Blocks): Let $\mathcal{Q}$ be a partition of $X$, and let $\mathcal{F}_{\mathcal{Q}}$ be the family of all unions of subcollections of $\mathcal{Q}$, including the empty union. Then $\mathcal{F}_{\mathcal{Q}}$ is a sigma-algebra on $X$.

Proof. The union of all blocks is $X$, so $X\in\mathcal{F}_{\mathcal{Q}}$. The empty union is $\varnothing$, so it belongs to the family as well.

Let $A\in\mathcal{F}_{\mathcal{Q}}$. Then $A$ is the union of some subcollection of blocks. Because the blocks partition $X$, every block not used in this union is contained in $X\setminus A$, and every block used in the union is disjoint from $X\setminus A$. Therefore $X\setminus A$ is exactly the union of the blocks not used to form $A$. Hence $X\setminus A\in\mathcal{F}_{\mathcal{Q}}$.

Finally, take a sequence $A_n\in\mathcal{F}_{\mathcal{Q}}$. Each $A_n$ is a union of blocks. Their union contains precisely those blocks that occur in at least one of these unions. It is therefore itself a union of blocks, so $\bigcup_{n=1}^{\infty}A_n\in\mathcal{F}_{\mathcal{Q}}$. All three conditions hold, proving the result. $\square$

Worked Example: A Sigma-Algebra from Three Blocks

Let $X=\{a,b,c,d,e\}$, partitioned into the blocks $Q_1=\{a,b\}$, $Q_2=\{c\}$, and $Q_3=\{d,e\}$. Each set in the resulting sigma-algebra is a union of some of these three blocks. For instance, choosing $Q_1$ and $Q_3$ gives $\{a,b,d,e\}$, and choosing no blocks gives $\varnothing$. The full family is

\{\varnothing,\{a,b\},\{c\},\{d,e\}, \{a,b,c\},\{a,b,d,e\},\{c,d,e\},X\}.

There are eight sets because each of the three blocks can be either included or excluded. For a direct check, the complement of $\{a,b,c\}$ is $\{d,e\}$, also in the family. A union of any sequence of these sets selects some collection of the same three blocks, and hence remains in the family. The partition theorem guarantees the same closure properties for partitions with any number of blocks.

Pulling a Sigma-Algebra Back Through a Function

Another construction starts with a function and a sigma-algebra on its target. The inverse image of a set records which points in the domain are mapped into that set. This construction is important because it transfers a family of measurable sets from one space to another without requiring the function to be one-to-one or onto.

Theorem (Inverse Images Form a Sigma-Algebra): Let $f:X\to Y$, and let $\mathcal{G}$ be a sigma-algebra on $Y$. Then $$ \mathcal{F}=\{f^{-1}(B):B\in\mathcal{G}\} $$ is a sigma-algebra on $X$.

Proof. Since $Y\in\mathcal{G}$, we have $f^{-1}(Y)=X$, so $X\in\mathcal{F}$. For any $B\in\mathcal{G}$, the complement identity

X\setminus f^{-1}(B)=f^{-1}(Y\setminus B)

shows that the complement of $f^{-1}(B)$ belongs to $\mathcal{F}$, because $Y\setminus B\in\mathcal{G}$.

Now let $f^{-1}(B_n)\in\mathcal{F}$ for every $n$, where $B_n\in\mathcal{G}$. For any $x\in X$, membership in the inverse image of a union means that $f(x)$ belongs to at least one of the sets in that union. Thus

\bigcup_{n=1}^{\infty}f^{-1}(B_n) = f^{-1}\left(\bigcup_{n=1}^{\infty}B_n\right).

The union on the right inside the inverse image belongs to $\mathcal{G}$, by its countable-union closure. Its inverse image therefore belongs to $\mathcal{F}$. We have verified the defining conditions on $X$, so $\mathcal{F}$ is a sigma-algebra. $\square$

Worked Example: Inverse Images under a Two-Valued Function

Let $X=\{1,2,3,4\}$, $Y=\{u,v\}$, and define $f(1)=u$, $f(2)=u$, $f(3)=v$, and $f(4)=v$. Take the power-set sigma-algebra $\mathcal{G}=\mathcal{P}(Y)$. The inverse images of its four members are

f^{-1}(\varnothing)=\varnothing,\quad f^{-1}(\{u\})=\{1,2\},\quad f^{-1}(\{v\})=\{3,4\},\quad f^{-1}(Y)=X.

The resulting family $\{\varnothing,\{1,2\},\{3,4\},X\}$ is a sigma-algebra on $X$. For example, the complement of $\{1,2\}$ is $\{3,4\}$, and the union of the two nonempty proper members is $X$. The function groups together points with the same value, so the inverse-image sigma-algebra consists exactly of unions of those two groups.

The Borel Sigma-Algebra and a Common Pitfall

On $\mathbb{R}$, the Borel sigma-algebra is the smallest sigma-algebra containing every open subset of $\mathbb{R}$. Its existence follows from the Existence of the Generated Sigma-Algebra theorem established earlier in the course. This is a central example: it contains open sets and also the sets obtained from them by complements and countable unions or intersections.

The collection of open subsets of $\mathbb{R}$ is not itself a sigma-algebra. For example, $(0,1)$ is open, but its complement $\mathbb{R}\setminus(0,1)=(-\infty,0]\cup[1,\infty)$ is not open. The Borel sigma-algebra includes that complement, as well as countable unions of open sets. Thus “the sigma-algebra containing the open sets” is not the same as “the family of open sets.”

The examples above illustrate several ways to recognize or construct a sigma-algebra. The trivial and power-set examples are useful benchmarks; the countable-cocountable example organizes sets by size; partitions give families whose members are unions of entire blocks; and inverse images preserve the sigma-algebra structure. In every case, countable-union closure is essential. Checking only complements and finite unions is not enough to establish that a family is a sigma-algebra.

Check Your Understanding

Use the examples and proofs in this tutorial to answer the following questions.

Why is the trivial sigma-algebra closed under countable unions?
In the countable-cocountable proof, why does a union containing one cocountable set remain cocountable?
How does the complement of a union of partition blocks relate to the blocks not used in that union?
For a function $f:X\to Y$, what inverse-image identity verifies closure under countable unions?
Why does the family of open subsets of $\mathbb{R}$ fail to be a sigma-algebra?

What You'll Learn