# Sigma algebra generated by random variable on a set with generators

Cross Validated Asked by Gabriel on July 28, 2020

I’m having trouble proving an intuitive result I found in these lecture notes I’m using for self-study (1.2.14 there).

Suppose $$X$$ is a $$(mathbb{S}, mathcal{S})$$-valued random variable (from $$(Omega, mathcal{F})$$), and furthermore $$mathcal{S} = sigma(mathcal{A})$$. If $$mathcal{F}^X$$ is the $$sigma$$-algebra generated by $$X$$ in $$Omega$$, we want to show that $$mathcal{F}^X = sigma({X^{-1}(A) : A in mathcal{A}})$$.

It’s easy to prove that $$mathcal{F}^X supset sigma({X^{-1}(A) : A in mathcal{A}})$$, by noticing that (i) $$mathcal{F}^X$$ is a $$sigma$$-algebra, and that (ii) it contains $${X^{-1}(A) : A in mathcal{A}}$$. But I believe I’m missing the right proof strategy for the other direction. Just appealing to the definitions and the tools developed so far (e.g. the $$pi-lambda$$ theorem) didn’t take me very far.

I think I get the spirit of the claim. Basically, it says that if you have a set of generators $$mathcal{A}$$ of $$mathcal{S}$$, to obtain $$mathcal{F}^X$$ you can either take the inverse images of all sets generated by $$mathcal{A}$$, or you can take the inverse images of just the sets in $$mathcal{A}$$ and then use those to generate a $$sigma$$-algebra. So, the order of the operations of "taking inverse images" and "generating a $$sigma$$-algebra" doesn’t matter. Is this understanding correct?

Any hint on a direction that might work for the proof would be extremely appreciated!