1.2Sigma Algebras

Updated 6 Sep 2026Chapter (PDF)

We define families of sets that will serve as the domain of a measure with our ideal properties. In this way, a measure won't work for every subset of a space, but for a select few.

[1.2.1]Definition(Algebra,

\sigma

-Algebra)#

Let $X$ be a nonempty set. An algebra $\mathcal{A}$ of sets on $X$ is a nonempty collection of subsets on $X$ that is closed under complementation and finite intersections. A $\sigma$ -algebra is an algebra that is closed under countable intersections.

Note that $\bigcap E_j = (\bigcup E_j^c)^c$ , and so algebras (resp. $\sigma$ -algebras) are closed under finite (resp. countable) unions as well. Note that any algebra contains $\emptyset, X$ , since for any $E \in \mathcal{A}$ , we have that $\emptyset = E \cap E^c \in \mathcal{A}$ and $X = E \cup E^c \in \mathcal{A}$ .

Note that if an algebra is closed under countable disjoint unions, then it is also a full $\sigma$ -algebra. Indeed, for any countable sequence of sets $\{E_k\}_{k = 1}^\infty$ , note that we may produce a new sequence

F_k = E_k \setminus \left[ \bigcup_{j = 1}^{k - 1} E_k \right] = E_k \cap \left( \bigcup_{j = 1}^{k - 1} E_k \right)^c.

Then $\{F_k\}_{k = 1}^\infty$ is a sequence of disjoint sets such that $\left(\bigcup_{k = 1}^\infty E_k = \bigcup_{k = 1}^\infty F_k\right) \in \mathcal{A}$ , making the algebra a $\sigma$ -algebra. The technique of taking a sequence of sets and producing a sequence of disjoint sets with the same union is an important one!

[1.2.2]Example(Examples of

\sigma

-Algebras)#

Let $X$ be a set. Trivially, $\mathcal{P}(\bR)$ and $\{\emptyset, X\}$ are $\sigma$ -algebras on $X$ . We also have the $\sigma$ -algebra formed by

\mathcal{A} = \{E: E \text{ is countable or } E^c \text{ is countable} \},

called the $\sigma$ -algebra of countable or co-countable sets.

Note that the intersection of any family of $\sigma$ -algebras on $X$ is a $\sigma$ -algebra itself. With this in mind, for any $\mathcal{E} \subseteq \mathcal{P}(X)$ , we can define the smallest $\sigma$ -algebra containing $\mathcal{E}$ by intersecting all $\sigma$ -algebras that contain $\mathcal{E}$ (recall that $\mathcal{P}(X)$ is its own $\sigma$ -algebra). We say this is the $\sigma$ -algebra generated by $\mathcal{E}$ , denoted $\M(\mathcal{E})$ .

[1.2.3]Theorem#

If $\mathcal{E} \subseteq \M(\mathcal{F})$ , then $\M(\mathcal{E}) \subseteq \M(\mathcal{F})$ .

Let $(X, \Tau)$ be any topological space. Then note that $\Tau \subseteq \mathcal{P}(X)$ , and so there is a $\sigma$ -algebra that can be generated by the topology. This is called a Borel $\sigma$ -algebra, denoted $\B_X$ . Since a $\sigma$ -algebra is closed under complementation, countable unions, and countable intersections, note we have a variety of sets in a Borel $\sigma$ -algebra. For example, a Borel $\sigma$ -algebra contains the countable intersection of open sets (called a $G_\delta$ set), the countable union of closed sets (called a $F_\sigma$ set), the countable union of $G_\delta$ sets ( $G_{\sigma \delta}$ ), the countable intersection of $F_\sigma$ sets ( $F_{\delta \sigma}$ ), and more. Note that we don't consider the countable union of open sets or countable intersection of closed sets, since these are already open and closed.

[1.2.4]Theorem#

The Borel $\sigma$ -algebra generated by $\bR$ 's standard topology, $\B_\bR$ , can be generated by the following sets.

The open intervals $\{(a, b): a < b \in \bR\}$ , and the closed intervals $\{[a, b]: a < b \in \bR\}$ .
The open rays $\{(-\infty, a): a \in \bR\}$ or $\{(a, \infty): a \in \bR\}$ , and the closed rays $\{(-\infty, a]: a \in \bR\}$ and $\{[a, \infty): a \in \bR\}$ .
The half-open intervals $\{(a, b]: a < b \in \bR\}$ or $\{[a, b): a < b \in \bR\}$ .

Proof.

$(1)$ : Note that all arbitrary unions of open intervals reduce to countable unions, since $\bR$ is second-countable. Then the standard topology on $\bR$ is a subset of the $\sigma$ -algebra generated by the set of all open intervals, and thus so is $\B_\bR$ . Conversely, the set of open intervals is a subset of $\B_\bR$ , and so open intervals generate $\B_\bR$ . Note all open intervals $(a, b)$ can be written as $\bigcup_{k = 1}^\infty [a - 1/n, b + 1/n]$ , and so the set of all closed intervals also generates $\B_\bR$ .

$(2)/(3)$ : Open and closed sets are generated by these sets under finite intersections.

[1.2.5]Definition(Product

\sigma

-Algebra)#

Let $\{X_\lambda\}_{\lambda \in \Lambda}$ be a collection of sets, and let $X = \prod_{\lambda \in \Lambda X_\lambda}$ . Let $\M_\lambda$ be a $\sigma$ -algebra on each $X_\lambda$ . We can define a $\sigma$ -algebra on $X$ , called the product $\sigma$ -algebra, which is generated by

\{\pi_\lambda^{-1}(E_\lambda): E_\lambda \in \M_\lambda \}_{\lambda \in \Lambda}.

We denote the product $\sigma$ -algebra by $\bigotimes_{\lambda \in \Lambda} \M_\lambda$ , writing it instead as $\bigotimes_{k = 1}^n \M_k$ or $\bigotimes_{k = 1}^\infty \M_k$ if $\Lambda$ is countable.

[1.2.6]Theorem(Characterization of Product

\sigma

-Algebra for Countable Indexing Sets)#

Let $\{X_\lambda\}_{\lambda \in \Lambda}$ be a family of sets with product $X = \prod_{\lambda \in \Lambda} X_\lambda$ . For each $\lambda \in \Lambda$ , let $\M_\lambda$ be a $\sigma$ -algebra endowed on $X_\lambda$ . If $\Lambda$ is countable, then the product $\sigma$ -algebra $\bigotimes_{\lambda \in \Lambda} \M_\lambda$ can be generated by

A = \Bigg\{ \prod_{\lambda \in \Lambda} E_\lambda: E_\lambda \in \M_\lambda \Bigg\}.

Proof.

Note that $\bigotimes_{\lambda \in \Lambda} \M_\lambda$ is generated by

B = \{\pi_\lambda^{-1}(E_\lambda): E_\lambda \in \M_\lambda\}_{\lambda \in \Lambda}.

We show $\M(A) = \M(B)$ .

First, we show $A \subseteq \M(B) \implies \M(A) \subseteq \M(B)$ . Let $\prod_{\lambda \in \Lambda} E_\lambda$ be any element of $A$ . Then note that

\prod_{\lambda \in \Lambda} E_\lambda = \bigcap_{\lambda \in \Lambda} \left(E_\lambda \times \prod_{\beta \in \Lambda \setminus \{\lambda\}} X\right) = \bigcap_{\lambda \in \Lambda} \pi_\lambda^{-1}(E_\lambda).

Note that a $\sigma$ -algebra is closed under countable unions, and so this is why this element of $A$ lies in $B$ . Thus $A \subseteq \M(B) \implies \M(A) \subseteq \M(B)$ .

Finally, we show $B \subseteq A$ , which implies $\M(B) \subseteq \M(A)$ . Let $\pi_{\lambda_0}^{-1}(E_{\lambda_0})$ be an arbitrary element of $B$ (for some fixed $\lambda_0 \in \Lambda$ ). Note that

\pi_{\lambda_0}^{-1}(E_{\lambda_0}) = \prod_{\lambda \in \Lambda} F_\lambda

where $F_{\lambda_0} = E_{\lambda_0}$ and $F_\lambda = X_\lambda$ for all $\lambda \ne \lambda_0$ . Since $X_\lambda \in \M_\lambda$ trivially, this product belongs to $A$ . Thus $B \subseteq A$ , meaning $\M(B) \subseteq \M(A)$ .

[1.2.7]Remark(Intuition Behind Product

\sigma

-Algebras)#

Let's consider the generating sets for a $\sigma$ -algebra with a countable indexing set $\Lambda$ . First, by definition, we have that

\bigotimes_{\lambda \in \Lambda} \M_\lambda = \M(\{\pi_\lambda^{-1}(E_\lambda): E_\lambda \in \M_\lambda \}_{\lambda \in \Lambda}).

Each $\pi_k^{-1}(E_k) = E_k \times \prod_{\beta \ne \lambda} X_\beta$ . By definition, this restricts one “dimension” of the $\sigma$ -algebra to the set $E_k$ , and the other “dimensions” can be anything. This is similar to how the general element of a subbasis for the product topology is a specific open set in one “dimension”, while the rest are free to be anything. We call these sets cylinder sets, since they are free to vary in all dimensions but one.

Now note the theorem from before. Since $\Lambda$ is countable, note that

\bigotimes_{\lambda \in \Lambda} \M_\lambda = \M\left( \Bigg\{\prod_{\lambda \in \Lambda} E_\lambda: E_\lambda \in \M_\lambda \Bigg\} \right).

Then we have that a general element of this generating set is restricted in all coordinates. We call these sets rectangle sets, since they are intersections of restricted sets in each axis.

The crucial idea is that a rectangle is the intersection of $|\Lambda|$ 1D cylinder sets. Since $\sigma$ -algebras are closed only under countable intersections, a rectangle set can only be represented in terms of cylinder sets for a countable indexing set, as an uncountable indexing set introduces too many restrictions.

[1.2.8]Theorem#

Let $\{X_\lambda\}_{\lambda \in \Lambda}$ be a family of sets with product $X = \prod_{\lambda \in \Lambda} X_\lambda$ . For each $\lambda \in \Lambda$ , let $\M_\lambda$ be a $\sigma$ -algebra endowed on $X_\lambda$ , where each $\M_\lambda$ is generated by $\mathcal{E}_\lambda$ . Then $\bigotimes_{\lambda \in \Lambda} \M_\lambda$ is generated by

A = \{\pi_\lambda^{-1}(E_\lambda): \lambda \in \Lambda, E_\lambda \in \mathcal{E}_\lambda\}.

If $\Lambda$ is countable and each $X_\lambda \in \mathcal{E}_\lambda$ , then the product $\sigma$ -algebra is also generated by

C = \Bigg\{ \prod_{\lambda \in \Lambda} E_\lambda: E_\lambda \in \mathcal{E}_\lambda \Bigg\}.

Proof.

Let $B = \{\pi_\lambda^{-1}(E_\lambda): \lambda \in \Lambda, E_\lambda \in \M_\lambda\}$ . Immediately, we have that $A \subseteq B \subseteq \M(B)$ , and so $\M(A) \subseteq \M(B)$ . Now we prove the reverse inclusion. Let $\pi_\lambda^{-1}(E_\lambda)$ be any element of $B$ , where $E_\lambda \in \M_\lambda$ for some fixed $\lambda \in \Lambda$ . Note that the inverse image commutes with all set-theoretic operations, including the ones for which a $\sigma$ -algebra is closed under (complementation, countable union, countable intersection). Thus

\M(\{ \pi_\lambda^{-1}(E_\lambda): E_\lambda \in \mathcal{E}_\lambda \}) = \{ \pi_\lambda^{-1}(E_\lambda): E_\lambda \in \M(\mathcal{E}_\lambda) \} = \{ \pi_\lambda^{-1}(E_\lambda): E_\lambda \in \M_\lambda \}.

Since $\{ \pi_\lambda^{-1}(E_\lambda): E_\lambda \in \mathcal{E}_\lambda \} \subseteq A$ , the left-hand side is contained in $\M(A)$ . Thus $\pi_\lambda^{-1}(E_\lambda) \in \M(A)$ , so $B \subseteq \M(A) \implies \M(B) \subseteq \M(A)$ , meaning $A$ generates the product $\sigma$ -algebra.

Finally, we show that $C$ also generates the product $\sigma$ -algebra. Any element of $C$ can be written as a countable intersection of cylinders in $A$ :

\prod_{\lambda \in \Lambda} E_\lambda = \bigcap_{\lambda \in \Lambda} \pi_\lambda^{-1}(E_\lambda) \in \M(A),

which gives $C \subseteq \M(A) \implies \M(C) \subseteq \M(A)$ . Conversely, any element of $A$ belongs to $C$ since $\pi_\lambda^{-1}(E_\lambda) = E_\lambda \times \prod_{\beta \ne \lambda} X_\beta$ (using that $X_\beta \in \mathcal{E}_\beta$ ), which gives $A \subseteq C \implies \M(A) \subseteq \M(C)$ . Thus $\M(C) = \M(A)$ , completing the proof.

[1.2.9]Theorem#

Let $(X_1, \Tau_1), \dots, (X_n, \Tau_n)$ be arbitrary topological spaces, and let $X = \prod_{k = 1}^n X_k$ be the product in its product topology $\Tau$ . Then

\bigotimes_{k = 1}^n \B_{X_k} \subseteq \bigotimes \B_X.

If each $X_k$ is second-countable, then

\bigotimes_{k = 1}^n \B_{X_k} = \bigotimes \B_X.

Proof.

Note that the topology $\Tau_k$ on $X_k$ generates $\B_{X_k}$ . Thus

\{\pi_k^{-1}(U_k): U_k \in \Tau_k\}_{k = 1}^n

generates $\bigotimes_{k = 1}^n \B_{X_k}$ . Since $\pi_k^{-1}(U_k)$ is open in $X$ for any $k \in \{1, \dots, n\}$ and $U_k \in \Tau_k$ , note that the generator for $\bigotimes_{k = 1}^n \B_{X_k}$ is a subset of the topology on $X$ , meaning it is a subset of $\B_X$ . Thus $\bigotimes_{k = 1}^n \B_{X_k} \subseteq \B_X$ .

We now prove the converse if each $X_k$ is second-countable. We show $\B_X \subseteq \bigotimes_{k = 1}^n \B_{X_k}$ . Note that the topology on $X$ generates $\B_X$ , so we show the topology on $X$ is a subset of $\bigotimes_{k = 1}^n \B_{X_k}$ . Since each $X_k$ is second-countable, there is a countable basis $\mathcal{U}_k$ . Since the collection of sets is finite, note that

\mathcal{B} = \Bigg\{ \prod_{k = 1}^n U_k : U_k \in \mathcal{U}_k \Bigg\}

is a basis for the product $X$ . Let $W \in \Tau$ be any open set in $X$ . Then note $W$ is the union of arbitrary basis elements from $\mathcal{B}$ , but since it is countable, this is a countable union. That is,

W = \bigcup_{j = 1}^\infty \left( \prod_{k = 1}^\infty U_{j, k} \right) \quad U_{j, k} \in \mathcal{U_k}.

Since $W$ is a countable union of elements that generate $\bigotimes_{k = 1}^n \B_{X_k}$ , we note that the topology on $X$ is a subset of the product $\sigma$ -algebra, and so $\B_X \subseteq \bigotimes_{k = 1}^n \B_{X_k}$ . With the previous result, we reach equality.

[1.2.10]Corollary#

$\B_{\bR^n} = \bigotimes_{k = 1}^n \B_{\bR}$ .

[1.2.11]Definition(Elementary Family)#

For any set $X$ , a collection $\mathcal{E}$ of subsets of $X$ is called an elementary family if it satisfies the following conditions.

$\emptyset \in \mathcal{E}$ .
$\mathcal{E}$ is closed under finite intersections.
For any $E \in \mathcal{E}$ , $E$ can be written as a finite disjoint union of sets from $\mathcal{E}$ .

[1.2.12]Theorem#

Let $X$ be any set together with an elementary algebra $\mathcal{E}$ . Define $\mathcal{A}$ by

\mathcal{A} = \Bigg\{ \bigcup_{k = 1}^n E_k: E_k \in \mathcal{E} \text{ is pairwise disjoint} \Bigg\}_{n \in \bZ_+},

the collection of all finite disjoint unions in $\mathcal{E}$ . Then $\mathcal{A}$ forms an algebra.

Proof.

Immediate by induction.