# Smallest Sigma Field Generated by Maps

(Resnick, A Probability Path p.83)$$\,\,\,$$If for each $$t$$ in some index set $$T$$$$$$X_t:\left(\Omega, \mathcal{B}\right)\mapsto\left(\Omega',\mathcal{B}'\right),$$$$ then we denote by $$$$\sigma(X_t,\, t\in T)=\bigvee_{t\,\in\, T}\sigma(X_t)$$$$ the smallest $$\sigma$$-algebra containing all $$\sigma(X_t)$$.

I want to prove that $$\sigma(X_t,\, t\in T)$$ is indeed a $$\sigma$$-algebra, because a countable union of $$\sigma$$-algebras need not be a $$\sigma$$-algebra even if they form a monotone sequence. However, I cannot find a way to prove the set is closed under countable union (or intersection). I would appreciate your help.

• $V_{t\in T}$ is not the union but is the smallest sigma algebra that contains all $\sigma(X_t).$ Commented Jul 18, 2019 at 22:51

By the definition we have $$\sigma\big(\{X_t\}_{t\in T} \big)=\sigma\bigg(\bigcup_{t\in T}\sigma(X_t) \bigg)$$ i.e. the smallest $$\sigma$$-algebra that contains all the $$\sigma$$-algebras $$\sigma(X_t)\;,t\in T$$. Observe that this is exactly the $$\sigma$$-algebra generated by the collection $$\{ X_t^{-1}(B')\;:B'\in\mathcal{B'}, t\in T\}$$.
As for your question, it is enough to show that the $$\sigma$$-algebra $$\sigma(\mathcal{A})$$ generated by a collection of sets is indeed a $$\sigma$$-algebra. That follows immediately from the fact that an arbitrary intersection of $$\sigma$$-algebras is a $$\sigma$$-algebra.