I have a question regarding sigma algebra generated by a set.
It is from 5.4b) from here:
http://stat.math.uregina.ca/~kozdron/Teaching/Regina/451Fall13/Handouts/451lecture05.pdf
Basically I know that given a sigma algebra (say $F= ${$F_1, F_2, …$} ) of a set $\Omega$, I have that for any subset $\Omega_i \subset \Omega$, the collection of sets $F \cap \Omega_i$ also being a sigma algebra (this result is from 5.4a)).
My question is if I know that a given set say $C$ generates the sigma algebra $F$, i.e. $\sigma(C)=F$, how do I show that the sets $F \cap \Omega_i$ is generated by $C \cap \Omega_i$? i.e. $\sigma(C \cap \Omega_i) = F\cap \Omega_i$ where $F \cap \Omega_i$ is a collection of subsets.
I was able to show that $\sigma (C \cap \Omega_i)$ is a subset of $F \cap \Omega_i$. Because $F \cap \Omega_i$ is a sigma algebra that contains the collection of sets $C \cap \Omega_i$ , so therefore obviously we have $\sigma(C \cap \Omega_i)$ being a subset of $F \cap \Omega_i$ since $\sigma(C \cap \Omega_i)$ denotes the INTERSECTION of all sigma algebra that contains the set $C \cap \Omega_i$.
So I have $\sigma(C \cap \Omega_i) \subseteq F\cap\Omega_i$.
My problem is to show the other way i.e. show that $F\cap\Omega_i \subseteq \sigma(C \cap \Omega_i) $.
I was able to start with this:
I have that $F \cap\Omega_i \subseteq \sigma(C) \cap \Omega_i$ because $F$ is a subset of $\sigma(C)$ , and hence $F \cap\Omega_i \subseteq \sigma(C) \cap \Omega_i$, but that is not really what I want to show. I think I need little more step to prove $F\cap\Omega_i \subseteq \sigma(C \cap \Omega_i) $, but I got kind of stuck.
Could someone give me some hints. Also if the steps above so far are all correct?
Best Answer
Let it be that $\langle\Omega,\mathcal F\rangle$ is a measurable space and that $f:\Omega'\to\Omega$ denotes a function.
Moreover let it be that $\mathcal C\subseteq\mathcal F$ such that $\mathcal F=\sigma(\mathcal C)$.
Then it can be shown that:$$f^{-1}(\sigma(\mathcal C))=\sigma(f^{-1}(\mathcal C))\tag1$$
In the special case where $\Omega'\subseteq\Omega$ and $f$ denotes the inclusion application of $(1)$ results in:$$\{\Omega'\cap F\mid F\in\mathcal F\}=\sigma\left(\{\Omega'\cap C\mid C\in\mathcal C\}\right)$$ which is exactly what you are after.
Familiarity with $(1)$ is in my view a "must" in measure theory.
For a proof of it have a look at this answer.
edit:
Let $\Omega'\subseteq\Omega$ and let's denote $\mathcal{C}'=\left\{ A\cap\Omega'\mid A\in\mathcal{C}\right\} $ and $\mathcal{F}'=\left\{ A\cap\Omega'\mid A\in\mathcal{F}\right\} $ where $\mathcal{C}\subseteq\wp\left(\Omega\right)$ and $\mathcal{F}=\sigma\left(\mathcal{C}\right)$.
It is not difficult to prove that $\mathcal{F}'$ is a $\sigma$-algebra, and this with $\mathcal{C}'\subseteq\mathcal{F}'$. This allows the conclusion $\sigma\left(\mathcal{C}'\right)\subseteq\mathcal{F}'$ and this part you had done yourself allready.
Now let $\mathcal{A}:=\left\{ A\in\wp\left(\Omega\right)\mid A\cap\Omega'\in\sigma\left(\mathcal{C}'\right)\right\} $.
Again it is not difficult to prove that $\mathcal{A}$ is a $\sigma$-algebra, and this with $\mathcal{C}\subseteq\mathcal{A}$. This allows the conclusion $\mathcal{F}=\sigma\left(\mathcal{C}\right)\subseteq\mathcal{A}$. That means that $A\cap\Omega'\in\sigma\left(\mathcal{C}'\right)$ for every $A\in\mathcal{F}$ and states that $\mathcal{F}'\subseteq\sigma\left(\mathcal{C}'\right)$.
If $f:\Omega'\to\Omega$ denotes the inclusion then $\mathcal C'=f^{-1}(\mathcal C)$ and $\mathcal F'=f^{-1}(\mathcal F)=f^{-1}(\sigma(\mathcal C))$ so in the first part it is shown that $\sigma(f^{-1}(\mathcal C))\subseteq f^{-1}(\sigma(\mathcal C))$ and in the second part that $f^{-1}(\sigma(\mathcal C))\subseteq\sigma(f^{-1}(\mathcal C))$.
For completeness: $$f^{-1}(\mathcal C):=\{f^{-1}(A)\mid A\in\mathcal C\}=\{\{\omega\in\Omega'\mid f(\omega)\in A\}\mid A\in\mathcal C\}\tag2$$
So if $f:\Omega'\to\Omega$ is prescribed by $\omega\mapsto\omega$ (i.e. inclusion) then we can expand $(2)$ with:$$=\{\{\omega\in\Omega'\mid \omega\in A\}\mid C\in\mathcal C\}=\{A\cap\Omega'\mid A\in \mathcal C\}=\mathcal C'$$