Sigma-Algebras
0.1 \(\sigma\)-algebras
Definition 1 (\(\sigma\)-algebra (\(\sigma\)-field, Borel field)) A \(\sigma\)-algebra [1] (also called a \(\sigma\)-field [2] or a Borel field [3], Definition 1.2.1) on a set \(S\) is a collection \(\mathcal{S}\) of subsets of \(S\) that satisfies:
- \(\mathcal{S}\) contains \(S\) itself.
- For each set \(A\) in \(\mathcal{S}\), \(\mathcal{S}\) contains its complement \(S \setminus A\).
- For each sequence \(A_1, A_2, \ldots\) of sets in \(\mathcal{S}\), \(\mathcal{S}\) contains their union \(\bigcup_{i=1}^{\infty} A_i\).
Remark 1 (“Borel” in “Borel field”). The “Borel” in “Borel field” is a different use of “Borel” from the Borel \(\sigma\)-algebra on \(\mathbb{R}\), which is one particular \(\sigma\)-algebra.
Example 1 (\(\sigma\)-algebras for die rolls) For the die rolls \(D = {\mathopen{}\left\{1, 2, 3, 4, 5, 6\right\}\mathclose{}}\) (sets of die rolls), the collection of all subsets of \(D\) is a \(\sigma\)-algebra on \(D\), since complements and unions of subsets of \(D\) are again subsets of \(D\). So is the smaller collection
\[{\mathopen{}\left\{\emptyset, {\mathopen{}\left\{2, 4, 6\right\}\mathclose{}}, {\mathopen{}\left\{1, 3, 5\right\}\mathclose{}}, D\right\}\mathclose{}}\]
which contains \(D\), the complement of each of its sets, and every union of its sets: for example, \({\mathopen{}\left\{2, 4, 6\right\}\mathclose{}} \cup {\mathopen{}\left\{1, 3, 5\right\}\mathclose{}} = D\).
Example 2 (A collection that is not a \(\sigma\)-algebra) The collection \({\mathopen{}\left\{\emptyset, {\mathopen{}\left\{1\right\}\mathclose{}}, D\right\}\mathclose{}}\) is not a \(\sigma\)-algebra on \(D = {\mathopen{}\left\{1, 2, 3, 4, 5, 6\right\}\mathclose{}}\): it contains \({\mathopen{}\left\{1\right\}\mathclose{}}\) but not its complement \(D \setminus {\mathopen{}\left\{1\right\}\mathclose{}} = {\mathopen{}\left\{2, 3, 4, 5, 6\right\}\mathclose{}}\).
Example 3 (A \(\sigma\)-algebra is closed under complements and countable unions) A \(\sigma\)-algebra \(\mathcal{S}\) on a set \(S\) is closed under complements in \(S\) and under unions of sequences of its sets, because Definition 1 requires both. The collection in Example 2 is not closed under complements.
Theorem 1 (Closure properties of a \(\sigma\)-algebra) If \(\mathcal{S}\) is a \(\sigma\)-algebra on a set \(S\), then \(\mathcal{S}\) contains:
- the empty set \(\emptyset\);
- the union \(A_1 \cup \cdots \cup A_n\) of any finitely many sets \(A_1, \ldots, A_n\) in \(\mathcal{S}\);
- the intersection \(\bigcap_{i=1}^{\infty} A_i\) of any sequence \(A_1, A_2, \ldots\) of sets in \(\mathcal{S}\).
So \(\mathcal{S}\) is closed under finite unions and under intersections of sequences of its sets.
Proof. Empty set. \(\mathcal{S}\) contains \(S\), so it contains the complement of \(S\), which is \(S \setminus S = \emptyset\).
Finite unions. Extend \(A_1, \ldots, A_n\) to a sequence by setting
\[ \begin{aligned} A_{n+1} &= A_{n+2} \\ &= \cdots \\ &= \emptyset; \end{aligned} \]
each of these sets is in \(\mathcal{S}\), by the first part. Then:
\[ \begin{aligned} A_1 \cup \cdots \cup A_n &= A_1 \cup \cdots \cup A_n \cup \emptyset \cup \emptyset \cup \cdots && \text{(a union with } \emptyset \text{ adds no elements)} \\ &= \bigcup_{i=1}^{\infty} A_i && \text{(} A_i = \emptyset \text{ for } i > n \text{)} \end{aligned} \]
and \(\mathcal{S}\) contains \(\bigcup_{i=1}^{\infty} A_i\) by the union rule of Definition 1.
Countable intersections. Each \(A_i\) is a subset of \(S\), so an element of \(S\) is outside \(S \setminus A_i\) exactly when it is in \(A_i\); that is, \(S \setminus (S \setminus A_i) = A_i\). So, by De Morgan’s laws:
\[ \begin{aligned} S \setminus \bigcup_{i=1}^{\infty} (S \setminus A_i) &= \bigcap_{i=1}^{\infty} \mathopen{}\left(S \setminus (S \setminus A_i)\right)\mathclose{} && \text{(De Morgan's laws)} \\ &= \bigcap_{i=1}^{\infty} A_i && \text{(} S \setminus (S \setminus A_i) = A_i \text{)} \end{aligned} \]
Each \(S \setminus A_i\) is in \(\mathcal{S}\) by the complement rule, so their union is in \(\mathcal{S}\) by the union rule, and the complement of that union is in \(\mathcal{S}\) by the complement rule again.
Theorem 2 (All subsets form a \(\sigma\)-algebra) For any set \(S\), the collection of all subsets of \(S\) is a \(\sigma\)-algebra on \(S\).
Proof. Each condition of Definition 1 holds:
- \(S\) is a subset of itself.
- For each subset \(A\) of \(S\), the complement \(S \setminus A\) contains only elements of \(S\), so it is a subset of \(S\).
- For each sequence \(A_1, A_2, \ldots\) of subsets of \(S\), every element of \(\bigcup_{i=1}^{\infty} A_i\) is in some \(A_i\) and so in \(S\); the union is a subset of \(S\).
Theorem 3 (An intersection of \(\sigma\)-algebras is a \(\sigma\)-algebra) Let \(\mathcal{S}_j\) be a \(\sigma\)-algebra on a set \(S\) for each \(j\) in a nonempty set of indices \(J\). Then their intersection \(\bigcap_{j \in J} \mathcal{S}_j\), the collection of subsets of \(S\) that are in every \(\mathcal{S}_j\), is a \(\sigma\)-algebra on \(S\).
Proof. Each condition of Definition 1 holds:
- \(S\) is in every \(\mathcal{S}_j\), so it is in the intersection.
- If \(A\) is in the intersection, then \(A\) is in every \(\mathcal{S}_j\), so its complement \(S \setminus A\) is in every \(\mathcal{S}_j\), and so it is in the intersection.
- If \(A_1, A_2, \ldots\) are in the intersection, then they are in every \(\mathcal{S}_j\), so their union \(\bigcup_{i=1}^{\infty} A_i\) is in every \(\mathcal{S}_j\), and so it is in the intersection.
Definition 2 (\(\sigma\)-algebra generated by a collection) Let \(\mathcal{C}\) be a collection of subsets of a set \(S\). The \(\sigma\)-algebra generated by \(\mathcal{C}\), written \({\sigma\mathopen{}\left(\mathcal{C}\right)\mathclose{}}\), is the intersection of all the \(\sigma\)-algebras on \(S\) that contain every set in \(\mathcal{C}\).
Remark 2 (The smallest \(\sigma\)-algebra containing \(\mathcal{C}\)). \({\sigma\mathopen{}\left(\mathcal{C}\right)\mathclose{}}\) is a \(\sigma\)-algebra by Theorem 3 (the collection of all subsets of \(S\) is one of the \(\sigma\)-algebras intersected, by Theorem 2, so there is at least one). It contains every set in \(\mathcal{C}\), and it is contained in every other \(\sigma\)-algebra on \(S\) that does, so it is the smallest \(\sigma\)-algebra on \(S\) that contains \(\mathcal{C}\).
Example 4 (The \(\sigma\)-algebra generated by the even rolls) For the die rolls \(D = {\mathopen{}\left\{1, 2, 3, 4, 5, 6\right\}\mathclose{}}\), let \(\mathcal{C}= {\mathopen{}\left\{{\mathopen{}\left\{2, 4, 6\right\}\mathclose{}}\right\}\mathclose{}}\). Every \(\sigma\)-algebra on \(D\) that contains \({\mathopen{}\left\{2, 4, 6\right\}\mathclose{}}\) must also contain \(D\), the complement \({\mathopen{}\left\{1, 3, 5\right\}\mathclose{}}\), and \(\emptyset\) (Definition 1 and Theorem 1). The collection \({\mathopen{}\left\{\emptyset, {\mathopen{}\left\{2, 4, 6\right\}\mathclose{}}, {\mathopen{}\left\{1, 3, 5\right\}\mathclose{}}, D\right\}\mathclose{}}\) is already a \(\sigma\)-algebra (Example 1), so it is \({\sigma\mathopen{}\left(\mathcal{C}\right)\mathclose{}}\).
Definition 3 (Borel \(\sigma\)-algebra) The Borel \(\sigma\)-algebra on \(\mathbb{R}\), written \(\mathcal{B}\), is the \(\sigma\)-algebra generated by (Definition 2) the collection of all intervals.
Example 5 (The Borel \(\sigma\)-algebra contains \(\mathbb{R}\) and the half-lines) The interval \((-\infty, 0)\) is in \(\mathcal{B}\) by Definition 3. So is its complement \([0, \infty)\), because \(\mathcal{B}\) is a \(\sigma\)-algebra (Definition 1). The set \(\mathbb{R}\) is in \(\mathcal{B}\) too: a \(\sigma\)-algebra on \(\mathbb{R}\) contains \(\mathbb{R}\) itself.
Definition 4 (Borel set) A Borel set [4] is a set in the Borel \(\sigma\)-algebra \(\mathcal{B}\).
Example 6 (Some Borel sets)
- Every interval, such as \([0, 1]\), \((2, 5)\), or \([0, \infty)\), is a Borel set.
- A single point \({\mathopen{}\left\{a\right\}\mathclose{}} = [a, a]\) is an interval, so it is a Borel set.
- The integers are a Borel set: they are the union of the sequence of single points \({\mathopen{}\left\{0\right\}\mathclose{}}, {\mathopen{}\left\{1\right\}\mathclose{}}, {\mathopen{}\left\{-1\right\}\mathclose{}}, {\mathopen{}\left\{2\right\}\mathclose{}}, {\mathopen{}\left\{-2\right\}\mathclose{}}, \ldots\) (countable and uncountable sets lists them this way), and \(\mathcal{B}\) contains that union by Definition 1.
- The complement \(\mathbb{R}\setminus [0, 1]\) is a Borel set, because \(\mathcal{B}\) contains the complement of each of its sets.