[Math] How to show the sigma algebra generated by a set of subsets C is equal to the sigma algebra generated by the algebra generated by C

algebrasmeasure-theory

Suppose I have a set of subsets $\mathcal{C}$ of a set $\Omega$. I can generate a $\sigma$-algebra using this set, namely

\begin{align} \displaystyle \sigma(\mathcal{C}) = \bigcap_{\mathcal{F}~\sigma – algebra,~ a(\mathcal{C}) ~ \subseteq~ \mathcal{F}} \mathcal{F}. \end{align}
Likewise, I can generate an algebra using $\mathcal{C}$ as follows:
\begin{align} \displaystyle a(\mathcal{C}) = \bigcap_{\mathcal{C} ~\subseteq ~\mathcal{A}} \mathcal{A}, \textrm{where}~ \mathcal{A}~\textrm{is an algebra}. \end{align}
I want to show that $\sigma(a(\mathcal{C})) = \sigma(\mathcal{C}).$

$\textbf{Thoughts:}$

  • To show that $\sigma(a(\mathcal{C})) \subseteq \sigma(\mathcal{C})$, we need that every $\sigma$-algebra $\mathcal{F}$ such that $\mathcal{C} \subseteq \mathcal{F}$ also satisfies $a(\mathcal{C}) \subseteq \mathcal{F}.$
  • Will the vice versa of the above work to show the opposite inclusion?

Thus, my question is, how do I show the above fact, and is there something similar to show the opposite inclusion?

Or even better, is there a far 'slicker' way to prove the whole thing?

I'm fairly new to this side of maths, so if this is 'obvious' in some way, go easy on me! 😛

Many thanks,

Sam

Best Answer

With the definitions as you have them, the proof is almost trivial. Just show first that if $\mathcal{A}$ is an algebra, then $a(\mathcal{A}) = \mathcal{A}$. In particular, for any $\mathcal{C}$, $a(a(\mathcal{C})) = a(\mathcal{C})$.

Now according to your definition, you have \begin{align} \displaystyle \sigma(\mathcal{C}) = \bigcap_{\mathcal{F}~\sigma - algebra,~ a(\mathcal{C}) ~ \subseteq~ \mathcal{F}} \mathcal{F} \end{align}

\begin{align} \displaystyle \sigma(a(\mathcal{C})) = \bigcap_{\mathcal{F}~\sigma - algebra,~ a(a(\mathcal{C})) ~ \subseteq~ \mathcal{F}} \mathcal{F}. \end{align} What is the difference between the right-hand sides?