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

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 $$\mathcal{S}$$ is contained in every $$\sigma$$-algebra of which we have taken the intersection, $$\mathcal{A}$$ must contain $$\mathcal{S}$$.

As $$\mathcal{A}$$ is a nonempty intersection of $$\sigma$$-algebras, it is a $\sigma$-algebra.

Furthermore, it is the smallest with such property (i.e., it satisfies property 2), as we have taken the intersection of all of them.

Now assume there are two $$\sigma$$-algebra which satisfies the properties 1 and 2 above: $$\mathcal{A}_1$$ 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, $$\mathcal{A}_1 = \mathcal{A}_2$$.

Therefore there is only one $$\sigma$$-algebra which satisfies the properties 1 and 2 above.