Sigma algebra generated by class

Question

Let $\Omega$ be a nonempty set, $\mathscr A$ be a collection of subsets $\Omega$ and $B\subset\Omega$. Show that $\sigma(\mathscr A\cap B )=\sigma(\mathscr A)\cap B$.

How can i show this? What is the start point?

Answer 1

First, $\mathscr{A}\cap B=\{A\cap B : A\in\mathscr{A}\}$ is a collection of sets. We know $O=\sigma(\mathscr{A}\cap B)$ is the minimal $\sigma$-algebra containing the collection $\mathscr{A}\cap B$. In other words, $O$ is the smallest $\sigma$-algebra containing all sets of the form $A\cap B$. Can you see where to go from here?

Answer 2

For notational convenience, I'm going to rename $\Omega$ to $X$.

First of all, this is a special case of a more general result, that I've seen questions about twice in the past few days. Let $i:B\to X$ be the inclusion, then the statement is that $\newcommand{\sA}{\mathscr{A}}$ $\newcommand{\inv}{^{-1}}$ $\sigma(i\inv(\sA)) = i\inv\sigma(\sA)$. For a proof of the more general result that for any $f:X\to Y$, $\sigma(f\inv(\sA)) = f\inv\sigma(\sA)$, see here, here, or here.

Now to prove this special case.

To show $\sigma(\sA\cap B)\subseteq \sigma(\sA)\cap B$, it suffices to show that $\sigma(\sA)\cap B$ is a $\sigma$-algebra on $B$ containing $\sA\cap B$. (Because $\sigma(\sA\cap B)$ is the smallest such $\sigma$-algebra on $B$.)

To show that $\sigma(\sA\cap B)\supseteq \sigma(\sA)\cap B$, the simplest way is to show that any $\sigma$-algebra on $B$ containing $\sA\cap B$ must contain $\sigma(\sA)\cap B$. (Because $\sigma(\sA\cap B)$ is such a $\sigma$-algebra on $B$.)

To prove the first part, note that $\cap B$ distributes over arbitrary intersections and unions, and $(X\setminus A)\cap B = B\setminus (A\cap B)$, so intersecting with $B$ also preserves complements, so $\sigma(A)\cap B$ is a $\sigma$-algebra on $B$. It's clear that it contains $\sA\cap B$.

For the second part, suppose $\newcommand{\cA}{\mathcal{A}}\cA$ is a $\sigma$-algebra on $B$ containing $\sA\cap B$, consider $\newcommand{\cC}{\mathcal{C}}\cC:=\{C\subset X: C\cap B \in \cA\}$. You can check as in the first part that this is a $\sigma$-algebra on $X$, since intersection with $B$ respects unions, intersections and complements. Then $\sA\subseteq\cC$ by construction, so $\sigma(\sA)\subseteq\cC$, hence $\sigma(\sA)\cap B \subseteq \cC\cap B \subseteq \cA$ as desired.

Edit: Why is $\cC$ a $\sigma$-algebra?

Let $C_n$ be a countable collection of elements of $\cC$. Then we want to show that $\newcommand{\of}[1]{\left({#1}\right)}\newcommand{\NN}{\mathbb{N}}$ $$\of{\bigcup_{n\in\NN} C_n}\cap B\in\cA,$$ $$\of{\bigcap_{n\in\NN} C_n}\cap B\in\cA,$$ and if $C\in \cC$, then $(X\setminus C)\cap B \in\cA$. Now notice that $$\of{\bigcup_{n\in\NN} C_n}\cap B = \bigcup_{n\in\NN} C_n\cap B \in\cA, $$ and $$\of{\bigcap_{n\in\NN} C_n}\cap B = \bigcap_{n\in\NN} C_n\cap B \in\cA, $$ and finally that $(X\setminus C)\cap B = B\setminus (C\cap B) \in \cA$. Hence $\cC$ is closed under countable unions, countable intersections and complements, so $\cC$ is a $\sigma$-algebra on $X$.