Existence and Uniqueness of Sigma-Algebra Generated by Collection of Subsets

This definition makes sense because of the following result:

Theorem
Given a nonempty set $$X$$ and a collection $$\mathcal{S} \subseteq \mathcal{P}(X)$$ of subsets of $$X$$, there exists a unique $$\sigma$$-algebra $$\mathcal{A}$$ over $$X$$ which satisfies the following two properties:


 * 1) $$\mathcal{S} \subseteq \mathcal{A}$$.
 * 2) If $$\mathcal{A}'$$ is any $$\sigma$$-algebra over $$X$$ such that $$\mathcal{S} \subseteq \mathcal{A}'$$, then $$\mathcal{A} \subseteq \mathcal{A}'$$.

We express the above by saying that $$\mathcal{A}$$ is the smallest $$\sigma$$-algebra which contains $$\mathcal{S}$$.

Proof
There is at least one $$\sigma$$-algebra which contains $$\mathcal{S}$$, namely, the power set $$\mathcal{P}(X)$$. Then, consider the intersection of all $$\sigma$$-algebras $$\mathcal{A}'$$ over $$X$$ such that $$\mathcal{S} \subseteq \mathcal{A}'$$, and call it $$\mathcal{A}$$.

As this is a nonempty intersection of $$\sigma$$-algebras, it is a sigma-algebra, and it must contain $$\mathcal{S}$$, as $$\mathcal{S}$$ is contained in every $$\sigma$$-algebra of which we have taken the intersection. Furthermore, it is the smallest with such property (i.e., it satisfies property 2), as we have taken the intersection of all of them.

Finally, there is only one $$\sigma$$-algebra which satisfies the properties 1 and 2 above, for assume there are two of them, $$\mathcal{A}_1$$ and $$\mathcal{A}_2$$. Then, by property 2, $$\mathcal{A}_1 \subseteq \mathcal{A}_2$$, and also by property 2, $$\mathcal{A}_2 \subseteq \mathcal{A}_1$$. So, $$\mathcal{A}_1 = \mathcal{A}_2$$.