Let \(X\) be a set. A \(\sigma\)-algebra, \(\mathcal{A}\), is a subset of the power set \(\mathcal{P}(X)\) that satisfies the following:

1. \(X \in \mathcal{A}\)
2. If \(A \in \mathcal{A}\) then \(A^C \in \mathcal{A}.\)
3. If \(A_1, A_2, \dots\) is a countable sequence of sets in \(\mathcal{A},\) then \(\bigcup_{i=1}^\infty A_i \in \mathcal{A}\).

Let \(X\) be a set and \(\mathcal{S} \subset \mathcal{P}(X).\) The \(\sigma\)-algebra generated by \(\mathcal{S}\), written \(\sigma(\mathcal{S})\), is the smallest \(\sigma\)-algebra on \(X\) that contains \(\mathcal{S}.\)

The following claims shows that for any set \(\mathcal{S},\) \(\sigma(\mathcal{S})\) exists and is unique.

Claim: If we let \(\mathfrak{A}\) be the set of all \(\sigma\)-algebras on \(X,\) then \[\sigma(\mathcal{S}) = \bigcap_{\mathcal{A} \in \mathfrak{A}, \mathcal{S} \subset \mathcal{A}} \mathcal{A}\]

Proof: To make notation simpler, define \(\mathcal{B}\) to be \[\mathcal{B} = \bigcap_{\mathcal{A} \in \mathfrak{A}, \mathcal{S} \subset \mathcal{A}} \mathcal{A}\]

Let's first check that \(\mathcal{A} \in \mathfrak{A}, \mathcal{S} \subset \mathcal{A}\) is non-empty, so there is something in the intersection. The power set \(\mathcal{P}(X)\) satisfies \(\mathcal{P}(X) \in \mathfrak{A}, \mathcal{S} \subset \mathcal{P}(X)\) so the intersection is non-empty for any set \(\mathcal{S}.\)

Suppose \(\mathcal{A}\) is a \(\sigma\)-algebra on \(X\) and \(\mathcal{S} \subset \mathcal{A}.\) Then \(\mathcal{A}\) is one of the sets that were in the interersection that defines \(\mathcal{B}\) so \(\mathcal{B} \subset \mathcal{A}.\) In other words, \(\mathcal{B}\) is smaller than (or equal to) \(\mathcal{A}.\)

Since every \(\sigma\)-algebra in the intersection of the definition of \(\mathcal{B}\) contains \(\mathcal{S},\) \(\mathcal{S} \subset \mathcal{B}.\) So, \(\mathcal{B}\) is a set containing \(\mathcal{S}\) that is smaller than or equal to every \(\sigma\)-algebra that contains \(\mathcal{S}.\) We only need to show \(\mathcal{B}\) is itself a \(\sigma\)-algebra.

1. Since \(X\) is in every \(\sigma\)-algebra, it is in every \(\sigma\)-algebra in the intersection. So, \(X \in \mathcal{B}.\)
2. If \(B \in \mathcal{B}\) then \(B\) is in every \(\sigma\)-algebra in the intersection. Since \(\sigma\)-algebras are closed over complements, \(B^C\) is in every \(\sigma\)-algebra in the intersection. Therefore, \(B^C \in \mathcal{B}.\)
3. If \(A_1, A_2, \dots \in \mathcal{B}\) then \(A_1, A_2, \dots\) are in every \(\sigma\)-algebra in the intersection. Since \(\sigma\)-algebras are closed over countable unions, \(A_1 \cup A_2 \cup \dots\) is in every \(\sigma\)-algebra in the intersection. Therefore, \(A_1 \cup A_2 \cup \dots \in \mathcal{B}.\)

Since \(\mathcal{B}\) satisfies 1 through 3 above, it is a \(\sigma\)-algebra.