0

Let E be a family of subsets of Omega. For subset B of Omega let's define $$E \cap B = \{A \cap B:A\in E\} (*)$$ prove that $$\sigma(E\cap B) = \sigma(E) \cap B$$

what I've tried so far is:

  1. (from right to left) let A be in intersection of sigma(E) and B. then we have: $$A \in (\bigcap_{E \subset L_1}\sigma(L_1)) \bigcap B = \bigcap_{E \subset L_1}{(\sigma(L1)\cap B)} =^{(1)} \bigcap_{(E \cap B)\subset L_1}{\sigma(L_1)} = \sigma(E\cap B)$$ but (1) equality seems weird to me and I can't strictly explain why it is right (I guess the (*) is used here), but I can't understand how. Also I got some problems with prove from left to right, any help would be appreciated!
Asaf Karagila
  • 393,674
myfakeaccount
  • 29

1 Answers1

1

It is handsome to go for a more general concept by proving the lemma:

If $f:X\to Y$ is a function and $\mathcal E$ is a family of subsets of $Y$ then:$$\sigma(f^{-1}(\mathcal E))=f^{-1}(\sigma(\mathcal E))$$

If this lemma is applied on the special case where $X=B\subseteq Y=\Omega$ and $f$ is the inclusion function of $B$ in $\Omega$ (i.e. is prescribed by $b\mapsto b$) then you arrive directly at:$$\sigma(\mathcal E\cap B)=\sigma(f^{-1}(\mathcal E))=f^{-1}(\sigma(\mathcal E))=\sigma(\mathcal E)\cap B$$

In order to prove the lemma first prove that in general:

  • If $\mathcal B$ is a $\sigma$-algebra on $Y$ then $f^{-1}(\mathcal B)$ is a $\sigma$-algebra on $X$.
  • If $\mathcal A$ is a $\sigma$-algebra on $X$ then $\{B|f^{-1}(B)\in\mathcal A\}$ is a $\sigma$-algebra on $Y$.

On base of the first bullet we find that $f^{-1}(\sigma(\mathcal E)))$ is a $\sigma$-algebra on $X$. Evidently it contains the collection $f^{-1}(\mathcal E)$ which justifies the conclusion:$$\sigma(f^{-1}(\mathcal E))\subseteq f^{-1}(\sigma(\mathcal E))$$

On base of the second bullet we find that $\{B|f^{-1}(B)\in\sigma(f^{-1}(\mathcal E))\}$ is a $\sigma$-algebra on $Y$. Evidently it contains the collection $\mathcal E$ which justifies the conclusion that it also contains the collection $\sigma(\mathcal E)$. That means exactly that: $f^{-1}(B)\in\sigma(f^{-1}(\mathcal E))$ for every $B\in\sigma(\mathcal E)$ or equivalently that:$$f^{-1}(\sigma(\mathcal E))\subseteq\sigma(f^{-1}(\mathcal E))$$

Proving the two bullets is not really difficult and I leave that to you.

To avoid eventual confusion: if $f:X\to Y$ is a function, $E$ is a subset of $Y$ and $\mathcal E$ is a collection of subsets of $Y$ then:

  • $f^{-1}(E)=\{x\in X|f(x)\in E\}$
  • $f^{-1}(\mathcal E)=\{f^{-1}(E)|E\in\mathcal E\}$
drhab
  • 151,093
  • in your second bullet "If A is a sigma-algebra then {...}" Do you mean that A is a sigma algebra on X? – myfakeaccount Feb 11 '24 at 16:46
  • Yes, I do mean that. I will edit. – drhab Feb 11 '24 at 17:41
  • you also say "Evidently it contains collection f^(-1)(epsilon) ... " right after the bullets. But how do we know that f^(-1)(Epsilon) is a collection? we do not know anything about the f function, and why does it justify prove from left to right then? – myfakeaccount Feb 11 '24 at 17:53
  • If $\mathcal E$ is a collection of subsets of $Y$ then $f^{-1}(\mathcal E):={f^{-1}(E)|E\in\mathcal E}$ is a collection of subsets of $X$. The fact that $\mathcal E\subseteq\sigma(\mathcal E)$ then justifies that $f^{-1}(\mathcal E)\subseteq f^{-1}(\sigma(\mathcal E))$ – drhab Feb 11 '24 at 19:40
  • "..we do not know anything about the $f$ function.." We know that it is a function $f:X\to Y$ and that $\mathcal E$ is a collection of subsets of $Y$. That is all we need here. – drhab Feb 11 '24 at 19:53