$\sigma$ algebra on intersection classes

Question

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?

Answer 1

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.

Answer 2

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.

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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.