[Math] Smallest of sigma-algebra

measure-theory

Can anyone show me how to solve this question extracted from Michael Taylor's book:

If $\{\mathcal{F}_{\alpha} : \alpha \in A\}$ is the collection of all $\sigma$-algebras of subsets of $X$ that contain $\mathcal{C}$, show that $$\bigcap_{\alpha \in A} \mathcal{F}_{\alpha} = \mathcal{F}$$ is a $\sigma$-algebra of subsets of $X$, containing $\mathcal{C}$, and is in fact the smallest such $\sigma$-algebra. One says $\mathcal{F}$ is the $\sigma$-algebra generated by $\mathcal{C}$ and writes $\mathcal{F} = \sigma(\mathcal{C})$

Best Answer

Recall the definition of a $\sigma$-algebra: it is a collection of subsets of some particular $X$, such that:

  1. It is closed under countable unions;
  2. It is closed under taking complement (relative to $X$).

From this, countable intersections follow by DeMorgan's Laws.

Now suppose that $X_i\in\mathcal F$ for $i\in\mathbb N$. So for all $\alpha\in A$ we have $X_i\in\mathcal F_\alpha$. Since $\mathcal F_\alpha$ is a $\sigma$-algebra we have that $\bigcap X_i\in\mathcal F_\alpha$ for all $\alpha\in A$ and therefore $\bigcap X_i\in\bigcap\mathcal F_\alpha=\mathcal F$.

For complements, the principle is the same.

Furthermore, the same idea works to show that $\mathcal F$ contains $\mathcal C$. Lastly we need to show that this is indeed the smallest:

Suppose that $\mathcal S$ is a $\sigma$-algebra which contains $\mathcal C$, then for some $\alpha\in A$ we have $\mathcal S=\mathcal F_\alpha$, so it took part in the intersection which generated $\mathcal F$, therefore $\mathcal F\subseteq\mathcal S$.

Further Reading:

  1. The $\sigma$-algebra of subsets of $X$ generated by a set $\mathcal{A}$ is the smallest sigma algebra including $\mathcal{A}$
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