Definition of a Sigma-Algebra

Reading the Definition Precisely

The previous tutorial introduced sigma-algebras and established several consequences of their closure rules. Here we make the definition precise as a set-theoretic object and develop an equivalent way to state its countable-union condition. This alternative formulation is useful because many results in measure theory concern unions of disjoint sets.

A sigma-algebra is not a set of points in the underlying space. It is a collection of subsets of that space. If $X$ is the underlying set, then a sigma-algebra $\mathcal{F}$ on $X$ satisfies $\mathcal{F}\subseteq\mathcal{P}(X)$, where $\mathcal{P}(X)$ is the power set of $X$. The pair $(X,\mathcal{F})$ is called a measurable space. The elements of $\mathcal{F}$ are its measurable sets.

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

$X\in\mathcal{F}$;
if $A\in\mathcal{F}$, then $X\setminus A\in\mathcal{F}$;
if $A_n\in\mathcal{F}$ for every $n\in\mathbb{N}$, then $\bigcup_{n=1}^{\infty}A_n\in\mathcal{F}$.

The complement in the second condition is always taken relative to $X$.

The ambient set matters: the complement of $A$ is not meaningful here until $X$ is specified. For example, if $A=\{1\}$, its complement in $X=\{1,2,3\}$ is $\{2,3\}$, while its complement in $Y=\{1,2,3,4\}$ is $\{2,3,4\}$. Thus the same collection of subsets can satisfy the definition on one underlying space but fail to be a sigma-algebra on another.

The countable-union condition concerns a sequence of members of $\mathcal{F}$, with repetitions allowed. It includes finite unions once we know that $\varnothing\in\mathcal{F}$: after listing finitely many sets, fill the rest of the sequence with empty sets. The next result records these basic consequences explicitly.

Theorem (Empty Set and Finite Unions): If $\mathcal{F}$ is a sigma-algebra on $X$, then $\varnothing\in\mathcal{F}$, and the union of any finite collection of members of $\mathcal{F}$ belongs to $\mathcal{F}$.

Proof. Since $X\in\mathcal{F}$, the complement condition gives $X\setminus X=\varnothing\in\mathcal{F}$. Now let $A_1,\ldots,A_k\in\mathcal{F}$, where $k\geq 1$. Define a sequence by taking its first $k$ terms to be $A_1,\ldots,A_k$ and all later terms to be $\varnothing$. Every term belongs to $\mathcal{F}$, so countable-union closure gives

\bigcup_{n=1}^{\infty}A_n = A_1\cup\cdots\cup A_k \in\mathcal{F},

where the equality uses the fact that all terms after $k$ are empty. The union of no sets is $\varnothing$, which also belongs to $\mathcal{F}$. This proves the claim. $\square$

Checking the Role of the Ambient Space

Worked Example: A Missing Relative Complement

Let $X=\{0,1,2\}$ and consider

\mathcal{G}=\{\varnothing,X,\{0\}\}.

The family contains $X$, and it contains $\varnothing$. But the complement of $\{0\}$ relative to $X$ is $X\setminus\{0\}=\{1,2\}$, which is not in $\mathcal{G}$. Therefore $\mathcal{G}$ is not a sigma-algebra on $X$. It would not help to take the complement relative to some larger set: the definition requires the complement relative to the stated underlying space.

Worked Example: The Same Family on Two Spaces

Let $Y=\{1,2,3,4\}$ and $\mathcal{H}=\{\varnothing,Y,\{1,2\},\{3,4\}\}$. The complements in $Y$ of these four sets are, respectively, $Y,\varnothing,\{3,4\},\{1,2\}$, all in $\mathcal{H}$. Any union of a sequence chosen from $\mathcal{H}$ is either empty, one of the two listed pairs, or all of $Y$. Hence $\mathcal{H}$ is a sigma-algebra on $Y$.

Now regard the same collection as a family of subsets of $Z=\{1,2,3,4,5\}$. It does not contain $Z$, so it fails the first defining condition on $Z$. Even if we added $Z$, complements would have to be recomputed relative to $Z$: for instance, $Z\setminus\{1,2\}=\{3,4,5\}$. This example shows why the phrase “on $X$” is part of the definition, not optional context.

Replacing Arbitrary Unions by Disjoint Unions

A sequence of sets need not be disjoint. However, any countable sequence can be replaced by a pairwise disjoint sequence without changing its union. This conversion explains why countable disjoint unions can serve as an alternative test for the countable-union axiom.

Theorem (Disjointification of a Sequence): Let $A_n\in\mathcal{F}$ for each $n$, where $\mathcal{F}$ is a sigma-algebra on $X$. Define $$ B_1=A_1,\qquad B_n=A_n\setminus\bigcup_{k=1}^{n-1}A_k\quad(n\geq 2). $$ Then each $B_n\in\mathcal{F}$, the sets $B_n$ are pairwise disjoint, and $\bigcup_{n=1}^{\infty}B_n=\bigcup_{n=1}^{\infty}A_n$.

Proof. For $n=1$, $B_1=A_1\in\mathcal{F}$. For $n\geq2$, the finite union $\bigcup_{k=1}^{n-1}A_k$ belongs to $\mathcal{F}$ by the finite-union result. The Closure Under Relative Differences theorem from the previous tutorial then implies $B_n\in\mathcal{F}$.

To check disjointness, take integers $i<j$. By definition, $B_i\subseteq A_i$, while $B_j$ excludes every point in $\bigcup_{k=1}^{j-1}A_k$, which includes $A_i$. Hence $B_i\cap B_j=\varnothing$.

It remains to verify equality of the unions. Since $B_n\subseteq A_n$ for every $n$, we have $\bigcup_n B_n\subseteq\bigcup_n A_n$. Conversely, take $x\in\bigcup_n A_n$. There is at least one index $n$ such that $x\in A_n$. Let $m$ be the least such index. Then $x\notin A_k$ for every $k<m$, so $x\in A_m\setminus\bigcup_{k=1}^{m-1}A_k=B_m$. Thus $x\in\bigcup_n B_n$. The two unions are equal, as required. $\square$

Theorem (Disjoint-Union Criterion for a Sigma-Algebra): Suppose $\mathcal{F}\subseteq\mathcal{P}(X)$, $X\in\mathcal{F}$, $X\setminus A\in\mathcal{F}$ whenever $A\in\mathcal{F}$, and $A\cup B\in\mathcal{F}$ whenever $A,B\in\mathcal{F}$. Then $\mathcal{F}$ is a sigma-algebra if and only if the union of every countable pairwise disjoint sequence of members of $\mathcal{F}$ belongs to $\mathcal{F}$.

Proof. If $\mathcal{F}$ is a sigma-algebra, its countable-union condition applies to every sequence, and therefore to every pairwise disjoint sequence.

For the converse, suppose unions of countable pairwise disjoint sequences in $\mathcal{F}$ belong to $\mathcal{F}$. Take any sequence $A_n\in\mathcal{F}$, not necessarily disjoint, and define $B_1=A_1$ and $B_n=A_n\setminus\bigcup_{k=1}^{n-1}A_k$ for $n\geq2$. For $n\geq2$, $B_n=X\setminus\bigl((X\setminus A_n)\cup\bigcup_{k=1}^{n-1}A_k\bigr)$, so finite-union and complement closure give $B_n\in\mathcal{F}$. These sets are pairwise disjoint, and the least-index argument shows that $\bigcup_n B_n=\bigcup_n A_n$. The assumed disjoint-union condition gives $\bigcup_n A_n\in\mathcal{F}$. Together with $X\in\mathcal{F}$ and closure under complements, this is exactly the definition of a sigma-algebra. $\square$

Worked Example: Disjointifying Overlapping Sets

Let $X=\{a,b,c,d,e\}$, and let $\mathcal{F}$ consist of all unions of the blocks $\{a,b\}$, $\{c\}$, and $\{d,e\}$. Consider $A_1=\{a,b,c\}$ and $A_2=\{c,d,e\}$. Both are unions of blocks and hence belong to $\mathcal{F}$. They overlap at $c$, so they are not disjoint. Their disjointification is

B_1=A_1=\{a,b,c\},\qquad B_2=A_2\setminus A_1=\{d,e\}.

Both $B_1$ and $B_2$ are still unions of the designated blocks, they are disjoint, and $B_1\cup B_2=\{a,b,c,d,e\}=A_1\cup A_2$. Thus the overlap can be removed while preserving the union. The same construction works for a countable sequence, using the union of all earlier sets at each step.

Why the Equivalent Formulation Matters

The two versions of the definition serve different purposes. The arbitrary-union version is direct and often easiest when verifying a family. The disjoint-union version is especially natural when adding sizes: if a measure is defined on $\mathcal{F}$, its defining additivity property concerns countable pairwise disjoint unions. Disjointification provides the link between arbitrary sequences of measurable sets and disjoint sequences.

A common error is to check only finite unions. The definition requires closure under countably many unions, and finite closure does not establish it. Another is to treat “disjoint” as an extra requirement in the original definition. It is not: disjointness is needed only in the equivalent criterion, and the disjointification theorem explains why that criterion suffices. Finally, the complement condition is always relative to the underlying space $X$.

Check Your Understanding

Use the definition and the disjoint-union criterion to answer the following questions.

What are the two different kinds of objects denoted by $X$ and $\mathcal{F}$ in a measurable space $(X,\mathcal{F})$?
Why does every sigma-algebra contain the empty set?
For $A_1,A_2,\ldots\in\mathcal{F}$, how is the disjointified set $B_n$ defined?
Why does disjointification preserve the union of the original sequence?
What assumptions, besides closure under disjoint countable unions, are needed for the disjoint-union criterion?

What You'll Learn