$\sigma$ algebra on intersection classes

On Pg. 5 of Probability and Measure Theory the authors argue as follows:

If $\mathscr{C}$ is a class of subsets of $\Omega$ and $A \subset \Omega$, we denote by $\mathscr{C} \cap A$ the class $\{ B \cap A : B \in \mathscr{C}\}$. If the minimal $\sigma$-field over $\mathscr{C}$ is $\sigma(\mathscr{C}) = \mathscr{F}$ then the minimal $\sigma$-field of subsets of $A$, $\sigma_A(\mathscr{C} \cap A) = \mathscr{F} \cap A$.

They build the proof by arguing that $\mathscr{F} \cap A$ is a $\sigma$-field of subsets of $A$.

I don't see how that is obvious. First, $\mathscr{F}$ is a $\sigma$-field of $\mathscr{C}$ which has no relation to $A$. Second, a class of sets formed by intersection with $A$ is not guaranteed to contain $A$. So how is it that $\mathscr{F} \cap A$ is a $\sigma$-field of subsets of $A$?

Or did I misunderstand something here?

When the authors say "the minimal $\sigma$-field over $\mathscr{C}$", they mean the smallest $\sigma$-field of $\Omega$ containing $\mathscr{C}$ [they explain their terminology in the middle of page 4].

$\mathscr{F}\cap A$ contains the emptyset since $\mathscr{F}$ does. It contains $A$ since $\mathscr{F}$ contains $\Omega$. It is closed under complements since $A\setminus (F\cap A) = (\Omega\setminus F) \cap A$ for each $F \in \mathscr{F}$. It is closed under countable unions since $\cup_n (F_n\cap A) = (\cup_n F_n) \cap A$ for each collection $\{F_n\} \in \mathscr{F}$. It is closed under countable intersections since it is closed under countable unions and taking complements.

• That’s probably where I am confused. How come $\mathscr{F}$ contain $\Omega$? Is $\mathscr{C}$ “the” class of subsets of $\Omega$ or “a” class of subsets of $\Omega$? – Apoorv Khurasia Aug 5 '18 at 9:07
• @ApoorvKhurasia $\mathscr{C}$ is a class of subsets. $\mathscr{F}$ is the smallest sigma-field containing $\mathscr{C}$. In particular, $\mathscr{F}$ is a sigma-field, so it contains $\Omega$. My advice would be to ignore $\mathscr{C}$ and note that my answers shows the general claim that if $\mathscr{F}$ is a sigma-field, then $\mathscr{F}\cap A$ is a sigma-field for any $A \subseteq \Omega$. – mathworker21 Aug 5 '18 at 9:25
• how do we know $\Omega \in \mathscr{F}$? We only know that $\mathscr{F}$ is a $\sigma$-field over $\mathscr{C}$ not $\Omega$ with $\mathscr{C}$ being an arbitratry class of subsets of $\Omega$. Or perhaps I am missing something very fundamental. – Apoorv Khurasia Aug 6 '18 at 8:18
• I don't know what "$\mathscr{F}$ is a $\sigma$-field over $\mathscr{C}$" means. What I was assuming is that $\mathscr{F}$ is the smallest $\sigma$-field (over $\Omega$) containing $\mathscr{C}$. – mathworker21 Aug 6 '18 at 8:24
• @ApoorvKhurasia Look at the middle of page 4, where the author defines "minimal sigma-field over". That explains his (weird) terminology. – mathworker21 Aug 6 '18 at 10:18

In general if $f:X\to Y$ is a function and $\mathcal A$ denotes a $\sigma$-algebra on $Y$ then it can be shown that $f^{-1}(\mathcal A):=\{f^{-1}(A)\mid A\in\mathcal A\}$ is a $\sigma$-algebra on $X$.

Preimages are very coöperative by proving that (give it a try).

Your case is a special case where $\mathscr F=i^{-1}(\mathscr C)$ where $i:A\to\Omega$ denotes the inclusion.

It is a good thing to be familiar the more general rule.

There is also a rule that is even more general. I will not handle that here, but in case you are interested here you can find a proof of that.

• I finally managed to prove this and I found this theorem to be handy. Thanks. – Apoorv Khurasia Aug 12 '18 at 6:29
• Nice to hear that. You are welcome. – drhab Aug 12 '18 at 7:25