# Showing that the collection of events $\{\omega: X_i(\omega) \in B_i, i = 1,\dots,k\}$ generates $\sigma(X_1,\dots, X_k)$

Let $$(X_n)_{n\in\mathbb{N}}$$ be a sequence of independent real valued random variables on the probability space $$(\Omega, F, P)$$. I'm trying to show that the collection $$\mathcal{C}$$ of events $$\{\omega: X_i(\omega) \in B_i, i = 1,\dots,k\}$$, for $$B_i \in \mathcal{B}(\mathbb{R})$$, generates the sigma-algebra $$H_k := \sigma(X_1,\dots,X_k)$$. I've already shown that $$\mathcal{C}$$ is a $$\pi$$-system hence the next step of my proof would be to show that any $$X_i, i = 1,\dots,k$$, is $$\sigma(\mathcal{C})/\mathcal{B}(\mathbb{R})$$ measurable. But given that any element $$A \in \mathcal{C}$$ has the form $$A = \{\omega: X_i(\omega) \in B_i, i = 1,\dots,k\} = \bigcap_{i=1}^k\{\omega: X_i(\omega) \in B_i\} = \bigcap_{i=1}^kX_i^{-1}[B_i]$$, I don't really know how to proceed. We certainly know that for any fixed $$j$$, $$A = \bigcap_{i=1}^kX_i^{-1}[B_i] \subset X_j^{-1}[B_j]$$, but I can't really convince myself of the fact that then $$X_j^{-1}[B_j] \in \sigma(\mathcal{C})$$. How should I proceed with this proof?

If $$B_i=\mathbb R$$ for all $$i \neq j$$ and $$B_j=B$$ then $$\{\omega: X_i(\omega) \in B_i,i=1,2...,k\}=X_j^{-1}(B)$$ so $$X_j^{-1}(B)\in \mathcal C \subseteq \sigma (\mathcal C)$$.
• While this answer did answer my exact question, could you also give your thought for the converse inclusion. Namely, does it follow from the fact that a similar treatment for other indices shows that $\mathcal{C}$ contains all possible pairs of intersections of all preimages of $X_1,\dots,X_k$, so that $\mathcal{C} \subset H_k \implies \sigma(\mathcal{C}) \subset \sigma(H_k) = H_k$? Feb 20, 2022 at 9:54
• Any set in $\mathcal C$ is a finite intersection of sets of the form $X_j^{-1}(B)$ so it is in $H_K$. Feb 20, 2022 at 9:57