Preimage of generated $\sigma$-algebra

For some collection of sets $A$, let $\sigma(A)$ denote the $\sigma$-algebra generated by $A$.

Let $C$ be some collection of subsets of a set $Y$, and let $f$ be a function from some set $X$ to $Y$. I want to prove:

$$f^{-1}(\sigma(C))=\sigma(f^{-1}(C))$$

I could prove that
$$\sigma(f^{-1}(C)) \subset f^{-1}(\sigma(C))$$
since complements and unions are ‘preserved’ by function inverse. But how do I go the other way?

EDIT: One way to go the other way would be to argue that any set in $\sigma(C)$ must be built by repeatedly applying the complement, union and intersection operations to elements of $C$ and all these operations are preserved when taking the inverse. The problem I am facing with the approach is formalizing the word “repeatedly”.

[not-homework]

Answer

Yes, we can use transfinite induction to prove this (formalizing the word “repeatedly”). That would be the bottom-up approach. There is also a top-down approach, using the characterization of $\sigma(C)$ as the smallest $\sigma$-algebra containing $C$.

A key fact here is the the preimage operation commutes with all the set algebra operations: if $f \colon X \to Y$ then

  • $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$
  • $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$
  • $f^{-1}(A – B) = f^{-1}(A) – f^{-1}(B)$. In particular $f^{-1}(Y – A) = X – f^{-1}(A)$.

and so forth. So $f^{-1}\colon P(Y) \to P(X)$ is a lattice homomorphism on the powerset lattices, to use that terminology. More importantly for us, the preimage operation even commutes with infinite unions and intersections.

To show that $f^{−1}(\sigma(C)) \subseteq \sigma(f^{−1}(C))$, you could follow this strategy:

  • Show that $f^{-1}(C) \in \sigma(f^{-1}(C))$
  • Show that if $f^{-1}(A_i) \in \sigma(f^{-1}(C))$ for $i \in \omega$ then $f^{-1}(\bigcup A_i) \in \sigma(f^{-1}(C))$
  • Show that if $f^{-1}(A) \in \sigma(f^{-1}(C))$ then $f^{-1}(Y – A) \in \sigma(f^{-1}(C))$

The point here is that, if we let $D$ be the collection of sets $A$ such that $f^{-1}(A) \in \sigma(f^{-1}(C))$, then $D$ is itself a $\sigma$ algebra containing $C$, which means $\sigma(C) \subseteq D$. But by the definition of $D$ this means $f^{-1}(\sigma(C)) \subseteq \sigma(f^{-1}(C))$.

None of the three bullets will require taking forward images under $f$. For example, for the third one, let $A$ be as stated. This means $f^{-1}(A)$ is in $\sigma(f^{-1}(C))$, which means that $X – f^{-1}(A)$ is also in $\sigma(f^{-1}(C))$. But $f^{-1}(Y – A)$ is exactly $X – f^{-1}(A)$, so we see that $f^{-1}(Y – A)$ is indeed in $\sigma(f^{-1}(C))$.

The underlying point here is that the entire proof is algebraic and that a more general theorem is true: you can replace $f^{-1}$ with any other homomorphism of the powerset lattices that preserves countable unions.

Attribution
Source : Link , Question Author : Jyotirmoy Bhattacharya , Answer Author : Carl Mummert

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