# Proof that the preimage of generated $\sigma$-algebra is the same as the generated $\sigma$-algebra of preimage.

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 $f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$.

• $f^{-1}(G)$ 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.
– saz
Oct 25, 2015 at 15:22
• @saz I had a few errors in my question. I have tried to explain what I don't understand in the linked answer.
– mrp
Oct 25, 2015 at 17:18
• 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 $$f^{-1} \left( \bigcap_i A_i \right) = \bigcap_i f^{-1}(A_i).$$
– saz
Oct 26, 2015 at 20:40
• @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.
– mrp
Oct 26, 2015 at 21:48
• 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 $A \in \sigma(f^{-1}(G))$.
– saz
Oct 27, 2015 at 6:13

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 $\bigcup_i G_i \in D$.

• Please check and correct your notation for all the different $G$s you use. It is very confusing. Jul 14, 2019 at 13:47
• you're the man. that was excellent. Jan 4 at 10:19

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 $$f^{-1}(\sigma(G)) \subseteq \sigma(f^{-1}(G))$$

• But how would you show that $f^{-1}(G) \in \sigma(f^{-1}(G))$?
– mrp
Oct 26, 2015 at 8:32
• What do you mean? If G is a single set, then it is clear already Oct 26, 2015 at 10:17
• 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 $\in$.
– mrp
Oct 26, 2015 at 10:34
• G is the countable union, intersection and complement of open sets. Read proof again. Oct 26, 2015 at 20:35
• @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.)
– saz
Oct 26, 2015 at 20:42

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.