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?


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.


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:$$=\{\{\omega\in\Omega'\mid \omega\in A\}\mid C\in\mathcal C\}=\{A\cap\Omega'\mid A\in \mathcal C\}=\mathcal C'$$

  • $\begingroup$ Hi drhab, could you elaborate little bit more about what you mean when you say "f denotes the inclusion....", what does "f denotes the inclusion" means? sorry for my stupidity. Also I am just wondering the function "f" here could means any "set operations", "mapping" , because so far I have only learned that f as a function can map something from the domain to the range. But does it mean the function "f" is actually more than mapping? for example, the function "f" could mean "set operations" like taking complement, taking union, taking intersection? $\endgroup$ – john_w Aug 26 '17 at 7:42
  • $\begingroup$ because I got kind of lost how to use "inclusion", and application of (1) to arrive at the equation you have there. Sorry for my stupidity. $\endgroup$ – john_w Aug 26 '17 at 7:42
  • $\begingroup$ Here there are two sets $\Omega_0,\Omega$ and $f:\Omega_0\to\Omega$ is a function in the usual sense of the word. If $\Omega_0\subseteq\Omega$ then $f:\Omega_0\to\Omega$ is the inclusion of $\Omega_0$ if it is prescribed by $\omega\mapsto\omega$. In that case a preimage under $f$ of some $F\subseteq\Omega$ will take the shape $f^{-1}(F)=\{\omega\in\Omega_0\mid f(\omega)\in F\}=\{\omega\in\Omega_0\mid \omega\in F\}=\Omega_0\cap F$. $\endgroup$ – drhab Aug 26 '17 at 8:17
  • $\begingroup$ Hi drhab, thanks for your response. I understand little better. But it is kind of difficult to associate the proof of sigma algebra generated by a set with the "inclusion function". What I mean is it is kind of hard to understand the intuition behind why all of a sudden one would think of the "inclusion" function $\omega \mapsto \omega$ when trying to prove 5b). I think I need little more time to understand. Also you have shown what $f^{-1}(F)$ is , but I still have to figure out the right-hand side (i.e. why $\sigma (f^{-1}(C))$ equal the right side of your formula). I am still digesting. $\endgroup$ – john_w Aug 28 '17 at 3:50
  • $\begingroup$ I have added a direct proof. $\endgroup$ – drhab Aug 28 '17 at 9:35

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