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This question has also been asked here, but the answer there didn't help me.

I am trying to prove that, given some measurable space $(X, \Sigma)$, if $G$ is a collection of subset of $X$ such that $\sigma(G) = \Sigma$, then $$f^{-1}(\sigma(G)) = \sigma(f^{-1}(G)).$$ So far, I have been able to show that $\sigma(f^{-1}(G)) \subseteq f^{-1}(\sigma(G))$, but I am having trouble with the opposite inclusion. I have shown that $f^{-1}(G)$ is a $\sigma$-algebra, from which it follows that $f^{-1}(G) \subseteq \sigma(f^{-1}(G))$, but that's about all I have been able to show. How do I proceed from there?

The answer in the other question suggests the following approach: Prove that

  • $f^{-1}(G) \in \sigma(f^{-1}(G))$,
  • if $f^{-1}(A_i) \in \sigma(f^{-1}(G))$ for $i \in \mathbb{N}$ (i.e. a countable collection), then $f^{-1}(\bigcup_i A_i) \in \sigma(f^{-1}(G))$, and
  • if $f^{-1}(A) \in \sigma(f^{-1}(G))$ then $f^{-1}(Y \setminus A) \in \sigma(f^{-1}(G))$.

However, I am not sure how to actually prove this, and even then, I'm not sure how it follows from this that $f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$.

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  • $\begingroup$ $f^{-1}(G)$ is not a $\sigma$-algebra ($f^{1}(G)$ is a (sub)set; therefore it cannot be a $\sigma$-algebra since $\sigma$-algebras are - by definition- families of subsets). $f^{-1}(G) \in \sigma(f^{-1}(G))$ follows direclty from the definition of $\sigma(\cdots)$. And it might be helpful to add what exactly you do not understand about the linked answer. Otherwise it's pretty hard to write a helpful answer. $\endgroup$
    – saz
    Oct 25, 2015 at 15:22
  • $\begingroup$ @saz I had a few errors in my question. I have tried to explain what I don't understand in the linked answer. $\endgroup$
    – mrp
    Oct 25, 2015 at 17:18
  • $\begingroup$ But the linked answer mentions also that you should use that the preimage operation commutes which set algebra operations, i.e. $$f^{-1} (Y \backslash A) = X \backslash f^{-1}(A)$$ and $$f^{-1} \left( \bigcup_i A_i \right) = \bigcup_i f^{-1}(A_i)$$ and $$f^{-1} \left( \bigcap_i A_i \right) = \bigcap_i f^{-1}(A_i).$$ $\endgroup$
    – saz
    Oct 26, 2015 at 20:40
  • $\begingroup$ @saz I think I see it now. I'm able to prove the second and third bullet, but I still don't see how I'm supposed to prove $f^{-1}(G) \in \sigma(f^{-1}(G))$. I swear I'm not being willfully obtuse, I just don't get it. $\endgroup$
    – mrp
    Oct 26, 2015 at 21:48
  • $\begingroup$ If $G$ is a collection of sets (again; note that you changed it; at the beginning you assumed that $G$ is a set), then you have to show $f^{-1}(G) \subseteq \sigma(f^{-1}(G))$, i.e. any set $A \in \sigma^{-1}(G)$ satisfies $A \in \sigma(f^{-1}(G))$. $\endgroup$
    – saz
    Oct 27, 2015 at 6:13

3 Answers 3

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After working on it some more, I have come up with a proof. It follows the ideas of the question I linked, and uses the properties of preimage that saz listed in the comments. I also found out that the proof does not actually need the premise that $\sigma(\mathcal{G}) = \Sigma$, and it is in fact not even necessary that $X$ is a measurable space.

The proof is as follows. We wish to prove that $f^{-1}(\sigma(\mathcal{G})) \subseteq \sigma(f^{-1}(\mathcal{G}))$. First we define $$D = \{G \subseteq Y \mid f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))\}.$$ Observe that what we want to show will follow if we prove that $D$ is a $\sigma$-algebra and that $\mathcal{G} \subseteq D$. This is because it will imply that $\sigma(\mathcal{G}) \subseteq D$, which, by the definition of $D$, will imply that if $G \in \sigma(\mathcal{G})$, then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$. This further implies that $$f^{-1}(\sigma(\mathcal{G})) = \{f^{-1}(G) \mid G \in \sigma(\mathcal{G})\} \subseteq \sigma(f^{-1}(\mathcal{G})).$$ Hence, we now show that $D$ is a $\sigma$-algebra and that $\mathcal{G} \subseteq D$. First observe that $f^{-1}(\mathcal{G}) \subseteq \sigma(f^{-1}(\mathcal{G}))$. This implies that if $G \in \mathcal{G}$ then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$. By the definition of $D$, this further implies that $\mathcal{G} \subseteq D$. To show that $D$ is a $\sigma$-algebra, we verify properties of a $\sigma$-algebra.

  • Since $\emptyset = f^{-1}(\emptyset)$ and $\emptyset \in \sigma(f^{-1}(\mathcal{G}))$, we have $\emptyset \in D$.

  • Assume $G \in D$. Then $f^{-1}(G) \in \sigma(f^{-1}(\mathcal{G}))$. Since $\sigma(f^{-1}(\mathcal{G}))$ is closed under complement, we get $$X \setminus f^{-1}(G) = f^{-1}(Y) \setminus f^{-1}(G) = f^{-1}(Y \setminus G) = f^{-1}(G^c) \in \sigma(f^{-1}(\mathcal{G})),$$ which implies that $G^c \in D$.

  • Assume $G_1,G_2,\dots \in D$. Then for all $G_i$ we have $f^{-1}(G_1) \in \sigma(f^{-1}(\mathcal{G}))$. Since $\sigma(f^{-1}(\mathcal{G}))$ is closed under countable union, we get $$\bigcup_i f^{-1}(G_i) = f^{-1}(\bigcup_i G_i) \in \sigma(f^{-1}(\mathcal{G})),$$ which implies that $\bigcup_i G_i \in D$.

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  • $\begingroup$ Please check and correct your notation for all the different $G$s you use. It is very confusing. $\endgroup$ Jul 14, 2019 at 13:47
  • $\begingroup$ you're the man. that was excellent. $\endgroup$ Jan 4 at 10:19
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This is based on the following properties of $f^{-1}$: $$ f^{-1}(\bigcup_{i=1}^{\infty}G_i)=\bigcup_{i=1}^{\infty}f^{-1}(G_i)\tag1 $$ $$ f^{-1}(\bigcap_{i=1}^{\infty}G_i)=\bigcap_{i=1}^{\infty}f^{-1}(G_i)\tag2 $$ $$ f^{-1}(G_i^c)=f^{-1}(G_i)^c\tag3 $$ Suppose $A\in \sigma(G)$ and $\:A=\bigcup_{i=1}^{\infty}G_i,\:G_i\in G$, i.e. $A$ is formed by the countable union of sets in $G$. Then $$ f^{-1}(A)=f^{-1}(\bigcup_{i=1}^{\infty}G_i)=\bigcup_{i=1}^{\infty}f^{-1}(G_i)\in\sigma(f^{-1}(G)) $$ For $A\in \sigma(G)$ and $\:A=\bigcap_{i=1}^{\infty}G_i,\:G_i\in G$, i.e. $A$ is formed by the countable intersection of sets in $G$. Then $$ f^{-1}(A)=f^{-1}(\bigcap_{i=1}^{\infty}G_i)=\bigcap_{i=1}^{\infty}f^{-1}(G_i)\in\sigma(f^{-1}(G)) $$ For $A\in \sigma(G)$ and $\:A=G_i^c,\:G_i\in G$, i.e. $A$ is formed by the complement of set in $G$. Then $$ f^{-1}(A)=f^{-1}(G_i^c)=f^{-1}(G_i)^c\in \sigma(f^{-1}(G)) $$ $G_i\in \sigma(G)$ is again the countable union, intersection and complement of sets and so $f^{-1}(G_i)\in \sigma(f^{-1}(G))$. So we have $$f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$$

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  • $\begingroup$ But how would you show that $f^{-1}(G) \in \sigma(f^{-1}(G))$? $\endgroup$
    – mrp
    Oct 26, 2015 at 8:32
  • $\begingroup$ What do you mean? If G is a single set, then it is clear already $\endgroup$
    – hermes
    Oct 26, 2015 at 10:17
  • $\begingroup$ But $G$ is a collection of subsets. I can see why $f^{-1}(G) \subseteq \sigma(f^{-1}(G))$, but not why the same holds with $\in$. $\endgroup$
    – mrp
    Oct 26, 2015 at 10:34
  • $\begingroup$ G is the countable union, intersection and complement of open sets. Read proof again. $\endgroup$
    – hermes
    Oct 26, 2015 at 20:35
  • $\begingroup$ @mrp If $G$ is a collection of subsets, then it has to read $f^{-1}(G) \subseteq \sigma(f^{-1}(G))$ and not $f^{-1}(G) \in \sigma(f^{-1}(G))$. (Note that in the first version of your question you stated that $G$ is a set; not a collection of sets.) $\endgroup$
    – saz
    Oct 26, 2015 at 20:42
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mrp and hermes answers are great! I just want to add a solution which does not hinge on the construction of our $\sigma$-algebra as combinations of set operations but rather the definiton of our the sigma algebra of a set $G$ being the intersection of all sigma algebras containing $G$, or the smallest sigma algebra containing $G$.

$f^{-1}(\sigma(G))\subseteq \sigma (f^{-1}(G))$: Lets prove the contrapositive that $\neg (f^{-1}(\sigma(G))\supset \sigma (f^{-1}(G)))$. Assume $f^{-1}(\sigma(G))\supset \sigma (f^{-1}(G))$ and let $f^{-1}(\sigma(G))\setminus \sigma (f^{-1}(G))=\{f^{-1}(x_1),f^{-1}(x_2),...,f^{-1}(x_n),...\}$ for some $x_i \in \sigma(G)$. It follows that $f^{-1}(\sigma(G))\setminus \{f^{-1}(x_1),f^{-1}(x_2),...,f^{-1}(x_n),...\}=f^{-1}(\sigma(G)\setminus\{x_1,x_2,...,x_n,...\})$ this will ofcourse be the same as $\sigma(f^{-1}(G))$. Now as $\sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$ is smaller than $\sigma(G)$ we must either (1) have that it does not contain $G$ or (2) its not a $\sigma$-algebra, the equality $f^{-1}(\sigma(G)\setminus\{x_1,x_2,...,x_n,...\})=\sigma (f^{-1}(S))$ rules out (2) so it must be that $\sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$ is no $\sigma$-algebra, this means that there are sets $A_n$ in $\sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$ such that $\bigcup_n A_n\notin \sigma(G)\setminus\{x_1,x_2,...,x_n,...\}$(or some $A\setminus B$) this means that $f^{-1}(\bigcup_n A_n)=\bigcup_n f^{-1}(A_n)\notin f^{-1}(\sigma(G)\setminus \{ x_1, x_2,...\})$ but $f^{-1}(A_n)$ must be in $\sigma(f^{-1}(G))$ as we otherwise would have removed it so $\bigcup_n f^{-1}(A_n)\in \sigma(f^{-1}(G))=f^{-1}(\sigma(G)\setminus \{ x_1, x_2,...\}$ a contradiction so our assumption must be wrong. This proves the inclusion

$f^{-1}(\sigma(G))\supseteq \sigma (f^{-1}(G))$: here it is enough to remember that the preimage of a $\sigma$-ring being a $\sigma$-ring so $f^{-1}(\sigma (G))$ is a $\sigma$-ring and as $G\subseteq\sigma (G)$ we also have $f^{-1}(G)\subseteq f^{-1}(\sigma(G))$ so $f^{-1}(\sigma(G))$ is a $\sigma$-ring containing $f^{-1}(G)$ from which it follows that $f^{-1}(\sigma(G))\supseteq \sigma(f^{-1}(G))$ as $\sigma(f^{-1}(G))$ is the smallest $\sigma$-ring containing $f^{-1}(G)$ $\square$

I am still fairly new to this area(the course started just over a week ago) so my argument may be messy, sorry about that. But i do think this is a nice alternative.

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