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.