(Resnick, A Probability Path p.83)$\,\,\,$If for each $t$ in some index set $T$\begin{equation}X_t:\left(\Omega, \mathcal{B}\right)\mapsto\left(\Omega',\mathcal{B}'\right),\end{equation} then we denote by \begin{equation} \sigma(X_t,\, t\in T)=\bigvee_{t\,\in\, T}\sigma(X_t)\end{equation} 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.
Best Answer
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.