Sigma-Algebras Generated by Partitions

Partitions as Building Blocks

A partition groups the points of a set into disjoint, nonempty pieces. When the partition has finitely or countably many pieces, the sigma-algebra it generates has a particularly concrete description: its measurable sets are precisely the unions of those pieces. Thus, membership in the generated sigma-algebra depends only on whether an entire block is included, not on how a set selects points inside a block.

We work with sets and sigma-algebras; no measure is needed here. A sigma-algebra on $X$ is a collection $\mathcal{M}$ containing $X$, closed under complements relative to $X$ and countable unions. For a family $\mathcal{E}$ of subsets of $X$, $\sigma(\mathcal{E})$ is the smallest sigma-algebra containing every member of $\mathcal{E}$. Equivalently, it is the intersection of all sigma-algebras on $X$ that contain $\mathcal{E}$.

Definition: A partition of a nonempty set $X$ is a family $\mathcal{P}=\{P_i:i\in I\}$ of nonempty subsets of $X$ such that distinct blocks are disjoint and $\bigcup_{i\in I}P_i=X$. The index set $I$ may be finite or countably infinite. The sigma-algebra generated by the partition is $\sigma(\mathcal{P})$. For $X=\varnothing$, we use the empty family as its partition.

For a finite or countable index set $I$, choosing a collection of blocks is the same as choosing a subset $J\subseteq I$. The resulting union is $\bigcup_{i\in J}P_i$, with the empty choice giving $\varnothing$. The theorem below says these are all the sets in $\sigma(\mathcal{P})$.

The Generated Sigma-Algebra

Theorem: Let $X$ be a set and $\mathcal{P}=\{P_i:i\in I\}$ a finite or countable partition of $X$. Then $$ \sigma(\mathcal{P})=\left\{\bigcup_{i\in J}P_i:J\subseteq I\right\}. $$ For the empty set with the empty partition, the right side consists of $\varnothing=X$.

Proof. First suppose $X\ne\varnothing$. Let $\mathcal{U}$ denote the family on the right side. It contains $X$, by taking $J=I$, and it contains $\varnothing$, by taking $J=\varnothing$. If $E=\bigcup_{i\in J}P_i$, then the blocks partition $X$, so $$ E^c=\bigcup_{i\in I\setminus J}P_i. $$ Thus $\mathcal{U}$ is closed under complements. If $E_n=\bigcup_{i\in J_n}P_i$ for each positive integer $n$, then $$ \bigcup_{n=1}^{\infty}E_n =\bigcup_{n=1}^{\infty}\bigcup_{i\in J_n}P_i =\bigcup_{i\in \bigcup_{n=1}^{\infty}J_n}P_i, $$ which belongs to $\mathcal{U}$. Therefore $\mathcal{U}$ is a sigma-algebra. It contains each block $P_i$, so the minimality of $\sigma(\mathcal{P})$ gives $\sigma(\mathcal{P})\subseteq\mathcal{U}$.

For the reverse inclusion, let $J\subseteq I$. Because $I$ is finite or countable, $J$ is finite or countable. Hence $\bigcup_{i\in J}P_i$ is a finite or countable union of members of $\mathcal{P}$, and so it belongs to $\sigma(\mathcal{P})$. This proves $\mathcal{U}\subseteq\sigma(\mathcal{P})$. If $X=\varnothing$, the empty family generates the sigma-algebra $\{\varnothing\}$, which is also the right side under the stated convention. The equality follows in all cases. $\square$

The countability assumption enters in the reverse inclusion: a sigma-algebra is closed under countable unions, not arbitrary unions. For a finite or countable partition, every choice of blocks is a countable choice, so the closure rule suffices.

Worked Example: A Partition of a Finite Set

Let $X=\{1,2,3,4,5,6,7,8\}$ and take the partition $$ A=\{1,2\},\qquad B=\{3,5,7\},\qquad C=\{4,6,8\}. $$ Find $\sigma(\{A,B,C\})$, and decide whether $E=\{1,2,4,6,8\}$ belongs to it.

There are three blocks, so each generated set is a union corresponding to one of the $2^3=8$ choices of blocks. Listing them gives $$ \sigma(\{A,B,C\})= \{\varnothing,A,B,C,A\cup B,A\cup C,B\cup C,X\}. $$ In elements, $A\cup C=\{1,2,4,6,8\}$, so $E=A\cup C$ belongs to the sigma-algebra. Its complement is $B=\{3,5,7\}$, which is also a union of blocks. Check: $E\cap B=\varnothing$, and $E\cup B=X$, so the proposed complement is correct. A set such as $\{1,4,6,8\}$ is not in the generated sigma-algebra: it includes only part of $A$, rather than all or none of that block.

Worked Example: Residue Classes in the Integers

Partition $\mathbb{Z}$ into the three congruence classes modulo $3$: $$ C_0=\{3k:k\in\mathbb{Z}\},\qquad C_1=\{3k+1:k\in\mathbb{Z}\},\qquad C_2=\{3k+2:k\in\mathbb{Z}\}. $$ Describe the generated sigma-algebra and test $F=C_0\cup C_2$.

Every integer has exactly one remainder $0$, $1$, or $2$ modulo $3$, so these are nonempty, pairwise disjoint blocks whose union is $\mathbb{Z}$. The theorem gives the eight unions $$ \varnothing,\ C_0,\ C_1,\ C_2,\ C_0\cup C_1,\ C_0\cup C_2,\ C_1\cup C_2,\ \mathbb{Z}. $$ Since $F$ is exactly the union of $C_0$ and $C_2$, $F\in\sigma(\{C_0,C_1,C_2\})$. Its complement is $C_1$. For instance, $5=3(1)+2$ belongs to $C_2$, and hence to $F$, while $4=3(1)+1$ belongs to $C_1=F^c$. Check: the remainder classes are disjoint and cover every integer, so $F\cap F^c=\varnothing$ and $F\cup F^c=\mathbb{Z}$.

Blocks Are the Atoms

A set $A$ in a sigma-algebra is called an atom if $A\ne\varnothing$ and every set in the sigma-algebra that is contained in $A$ is either $\varnothing$ or $A$. Atoms are the smallest nonempty measurable pieces: within one atom, the sigma-algebra cannot distinguish any smaller nonempty subset.

Proposition: For a finite or countable partition $\mathcal{P}=\{P_i:i\in I\}$ of $X$, the atoms of $\sigma(\mathcal{P})$ are exactly the blocks $P_i$.

Proof. Fix $i\in I$. The block $P_i$ is nonempty and belongs to $\sigma(\mathcal{P})$. If $D\in\sigma(\mathcal{P})$ and $D\subseteq P_i$, the theorem writes $D$ as a union of blocks. Any block in that union other than $P_i$ is disjoint from $P_i$, so it cannot contribute a point to $D$. Thus either no block is selected and $D=\varnothing$, or $P_i$ is selected and $D=P_i$. Hence $P_i$ is an atom.

Conversely, let $D$ be any atom of $\sigma(\mathcal{P})$. The set $D$ is nonempty, and the theorem expresses it as a union of blocks. Choose $x\in D$. There is a unique block $P_i$ containing $x$, because the blocks are disjoint and cover $X$. Since $D$ is a union of blocks, $P_i\subseteq D$; also $P_i\in\sigma(\mathcal{P})$ and $P_i\ne\varnothing$. The definition of atom applied to $P_i\subseteq D$ gives $P_i=D$. Thus every atom is a block. If $X=\varnothing$, there are no atoms and no blocks, so the assertion also holds. $\square$

Worked Example: A Partition of the Positive Integers

Partition $\mathbb{N}=\{1,2,3,\ldots\}$ into $P_0=\{1\}$ and $P_n=\{2n,2n+1\}$ for $n\ge1$. Describe the generated sigma-algebra and decide whether $G=\{1,4,5,8,9\}$ belongs to it.

The blocks are disjoint: $P_0$ contains only $1$, and for $n\ge1$ the pair $P_n$ consists of the consecutive integers $2n$ and $2n+1$. Together they cover $\mathbb{N}$. The theorem says that a set belongs to the generated sigma-algebra exactly when it is a union of whole blocks. Here $$ G=P_0\cup P_2\cup P_4 =\{1\}\cup\{4,5\}\cup\{8,9\}, $$ so $G\in\sigma(\{P_n:n\ge0\})$. Its complement is the union of all the remaining blocks: $$ G^c=P_1\cup P_3\cup\bigcup_{n=5}^{\infty}P_n. $$ The displayed union is countable, as required. Check: $2\in P_1$, so $2\notin G$; $4,5\in P_2$, so both are in $G$. A set containing $4$ but not $5$ cannot be in this sigma-algebra, because those two integers lie in the same block.

Why Countability Matters

It is tempting to apply the union description to partitions of any size. The collection of all unions of blocks is always a sigma-algebra: complements select the remaining blocks, and countable unions of unions are still unions of blocks. But when there are uncountably many blocks, that collection can be strictly larger than the sigma-algebra generated by the blocks. Sigma-algebra closure only guarantees countable unions, so it may not produce every union.

Common mistake. For the partition of $\mathbb{R}$ into singleton blocks $\{\{x\}:x\in\mathbb{R}\}$, one might say that every subset of $\mathbb{R}$ is generated, because every subset is a union of singletons. This uses an arbitrary union, not necessarily a countable one. In fact, the generated sigma-algebra consists exactly of the countable and cocountable subsets of $\mathbb{R}$.

To verify the correction, let $\mathcal{C}$ be the family of countable or cocountable subsets of $\mathbb{R}$, where cocountable means having countable complement. It contains every singleton. It is a sigma-algebra: complements interchange countable and cocountable sets; a countable union of countable sets is countable; and if any set in a countable union is cocountable, the complement of the union is a subset of that set's countable complement, hence countable. Therefore $\sigma(\{\{x\}:x\in\mathbb{R}\})\subseteq\mathcal{C}$. Conversely, every countable set is a countable union of singleton blocks, and every cocountable set is the complement of a countable set. Both kinds of sets belong to the generated sigma-algebra, proving equality. For example, $\mathbb{Q}$ is generated because it is countable, but an uncountable set with uncountable complement is not in this sigma-algebra.

The finite-or-countable theorem avoids this issue: when there are only finitely or countably many blocks, every subcollection of blocks can itself be united using the allowed countable-union operation. The partition description is useful whenever a sigma-algebra records only which block contains a point. In that setting, the blocks give both the complete list of smallest measurable pieces and a direct test for whether a proposed set is measurable.

Check Your Understanding

Use the definition of generated sigma-algebra and the block-union description to answer the questions.

For the partition $\{\{a,b\},\{c\},\{d,e\}\}$ of $\{a,b,c,d,e\}$, list all sets in the generated sigma-algebra.
For a partition of $\mathbb{Z}$ into the even and odd integers, is the set of positive even integers in the generated sigma-algebra? Explain using whole blocks.
Prove that the complement of any union of partition blocks is a union of the remaining blocks.
For the partition $P_0=\{1\}$, $P_n=\{2n,2n+1\}$ of $\mathbb{N}$, determine whether $\{2,3,6,7\}$ is in the generated sigma-algebra.
Why does the theorem not imply that every union of blocks belongs to the generated sigma-algebra when the partition is uncountable?

Sigma-Algebras Generated by Partitions

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