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Let $T:X\to Y$ be a map. Show that $T^{-1}(\sigma(G))=\sigma(T^{-1}(G))$ holds for any families $G$ of subsets of $Y$.

I have shown one side: $T^{-1}(\sigma(G))\supseteq\sigma(T^{-1}(G))$. As $T^{-1}(\sigma(G))$ is still a sigma algebra and $T^{-1}(G)\subseteq T^{-1}(\sigma(G))$, the result follows.

On the other side, we need to prove for any $A\in T^{-1}(\sigma(G))$, i.e. $T(A)\in \sigma(G)$, we must have $A\in \sigma(T^{-1}(G))$. I think showing an element belongs to a $\sigma$-algebra is nearly impossible but the common strategy that construct $\{A: A\in \sigma(T^{-1}(G))\}$ does not work here.

Any ideas and proofs might help!

Similar questions are Preimage of generated $\sigma$-algebra and Prove the commutation of smallest sigma-algebra and inverse. BUT I think the answers are not complete!

The reason is illustrated as comments in @Gliven 's answer.

There is actually no mistake in the proof of the last implication. I have been confused on two notations of inverse: $f^{-1}(A)$ and $f^{-1}(\mathscr{A})$. The former means $f^{-1}(A)=\{x\in X: f(x)\in A\}$ but the latter is just a "notation", it means: $f^{-1}(\mathscr{A})=\{f^{-1}(A):A\in \mathscr{A}\}$ rather than $\{A: f(A)\in \mathscr{A}\}$.

In fact, $f^{-1}(\mathscr{A})\subseteq \{B: f(B)\in \mathscr{A}\}$ since for any $B\in f^{-1}(\mathscr{A})$ there exists $A\in \mathscr{A}$ such that $B=f^{-1}(A)$. Then $f(B)=f(f^{-1}(A))=A\in \mathscr{A}$. The inverse is not true since if $B\in \{B: f(B)\in \mathscr{A}\}$, then $f(B)\in \mathscr{A}$ but since $f^{-1}(f(B))$not necessarily equals $B$, there does not necessarily exist a set, say $A$, in $\mathscr{A}$ such that $B=f^{-1}(A)$.

evenzhou
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1 Answers1

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Hint: the collection $$ \mathcal{A}=\left\{A:T^{-1}(A)\in \sigma(T^{-1}(G))\right\} $$ should be sigma-algebra, which is guaranteed by the "good" properties of inverse map.

Apparently, we have $G\subset \mathcal{A}$. Then you are done :) since $\sigma(G)\subset \mathcal{A}$

Debrii
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  • I understand the proof up to $\sigma(G)\subset \mathcal{A}$. BUT this only shows that for any $B\in \sigma(G)$, $T^{-1}(B)\in \sigma(T^{-1}(G))$. There not necessarily exists a set $B$ such that $T^{-1}(B)=A$ for $A\in T^{-1}(\sigma(G))$. In other words, we need to prove for any $A\in T^{-1}(\sigma(G))$, $A\in\sigma(G)$ and we know that $T(A)\in \sigma(G)$ thus $T^{-1}T(A)\in \sigma(T^{-1}(G))$. However, $T^{-1}T(A)$ not necessarily equals $A$. – evenzhou Oct 14 '22 at 04:25
  • For example, $f: a\to 1, b\to 1$. Then $a\in T^{-1}({1})$ but there does not exist an element $y$ such that $a=f^{-1}(y)$. – evenzhou Oct 14 '22 at 04:28
  • I'm saying $\forall B\in \sigma(G),T^{-1}(B)\in \sigma(T^{-1}(G))$ is exactly what you want to show, isn;t it? – Debrii Oct 14 '22 at 14:29
  • @Gilven Your answer is correct. I have made a mistake as I explained in the edited Post. – evenzhou Oct 14 '22 at 15:28