3

Let $f:Ω \toΩ'$, and let $G$ be a class of subsets of $Ω'$. Show that $\sigma(f^{-1}(G))=f^{-1}(\sigma(G))$.

It is not hard to verfify that $f^{-1}(\sigma(G))$ is a $\sigma$-algebra, and also $G\subset\sigma(G)$, so $f^{-1}(G)\subset f^{-1}(\sigma(G))$so we have $\sigma(f^{-1}(G))\subset f^{-1}(\sigma(G))$, but how to prove the other direction?

qmww987
  • 1,099

1 Answers1

4

Let $\Omega$ be a set and $\mathscr{G}$ a family of subsets of $\Omega'$; $f:\Omega \to \Omega'$ a function. First notice that $$\mathscr{G}\subseteq \mathscr{H}:=\{A \in \sigma(\mathscr{G}):f^{-1}(A)\in \sigma(f^{-1}(\mathscr{G}))\}\subseteq \sigma(\mathscr{G})$$ It remains to show that $\mathscr{H}$ is a $\sigma$-algebra:

(1). $f^{-1}(\Omega')=\Omega \in \sigma(f^{-1}(\mathscr{G}))$;

(2). If $B \in \mathscr{H}$, then $f^{-1}(B)\in \sigma(f^{-1}(\mathscr{G}))$. So $f^{-1}(B^c)=(f^{-1}(B))^c \in \sigma(f^{-1}(\mathscr{G}))$ and $B^c \in \mathscr{H}$;

(3) If $(B_n)_{n \in \mathbb{N}}\subseteq \mathscr{H}$; $f^{-1}(\cup_nB_n)=\cup_n f^{-1}(B_n)\in \sigma(f^{-1}(\mathscr{G}))$ so $\cup_n B_n \in \mathscr{H}$.

It follows that $\mathscr{H}=\sigma(\mathscr{G})$. So $\forall A \in \sigma(\mathscr{G}),f^{-1}(A)\in \sigma(f^{-1}(\mathscr{G}))$ and therefore $f^{-1}(\sigma(\mathscr{G}))\subseteq \sigma(f^{-1}(\mathscr{G}))$.

Snoop
  • 18,347
  • In the last step, we in fact need to show that for any $A\in f^{-1}(\sigma(\mathscr{G}))$, $A\in \sigma(f^{-1}(\mathscr{G}))$. We have $f(A)\in \sigma(\mathscr{G})$ so that $f^{-1}(f(A))\in \sigma(f^{-1}(\mathscr{G}))$. BUT $f^{-1}(f(A))$ not necessarily equals $A$. In other words, there might not exists $B\in \Omega'$ such that $A=f^{-1}(B)$ for $A\in f^{-1}(\Omega')$. For example $f:{a,b}\to {1}$ and $f(a)=f(b)=1$. Then $a\in f^{-1}({1})$ since $f(a)\in {1}$ but there does not exists $y$ such that $f^{-1}(y)=a$. – evenzhou Oct 14 '22 at 04:36
  • 1
    @evenzhou No, I think you're confused. The inclusion simply follows from the definition $f^{-1}(\sigma(\mathscr{G}))={f^{-1}(A):A \in \sigma(\mathscr{G})}$ – Snoop Oct 14 '22 at 13:03
  • Thank you so much! I have been thinking this for a long time. It turns out I confused the notation between $f^{-1}(A)={x\in X: f(x)\in A}$ and $f^{-1}(\mathscr{A})={f^{-1}(A): A\in \mathscr{A}}$. The meaning of inverse is totally different. – evenzhou Oct 14 '22 at 13:28
  • @evenzhou you're welcome! Leave an upvote if you like the answer :) – Snoop Oct 14 '22 at 13:31