I want to show the equivalence of two sigma algebras, namely the sigma algebra generated by a function $f:\Sigma\to\mathbb{\bar R}$ and $\cap\mathcal{F}\space$ such that $\mathcal{F}$ is a sigma algebra of sets of $\Sigma$ and $f$ is $(\mathcal{F},\mathcal{\bar B})$-measurable. $\sigma(f):=\{f^{-1}(B)|B\in\mathcal{\bar B}\}$, so in showing this equivalence, showing $\cap\mathcal{F}\subset\sigma(f)$ seems straightforward. However, I ran into trouble trying to prove the inclusion in the other direction.
Assuming that $X\in\sigma(f)$, this means that $X\in\{f^{-1}(B)|B\in\mathcal{\bar B}\}$. What does this tell me in order to arrive at the desired result? I can see why it must be part of one of the $\mathcal{F}$'s that are part of the intersection, but what tells me that it is actually part of the intersection itself? Thank you for your time.
Showing set inclusion for sigma algebra generated by function
measurable-functionsmeasure-theory
Best Answer
Verify that $\sigma (f) $ is a sigma algebra and that $f$ is measurable w.r.t. this sigma algebra. Hence $\cap\mathcal F$ is contained in this.
Intersection of a family of sets is always contained in any member of the family. Here $\sigma(f)$ is itself a member of the family.
Converse part: If $\mathcal F$ is any sigma algebra w.r.t. which $f$ is measurable then $f^{-1}(B) \in \mathcal F$ for every Borel set $B$ by definition of measurability. Hence $\sigma (f) \subset \mathcal F$. This is true for each $\mathcal F$ so $\sigma (f)$ is contained in the intersection of all that $\mathcal F$'s.