Problem: If $\{ X_i: \Omega \rightarrow R_i \}_1^n$ are measurable functions from $(\Omega, M_\Omega )$ into measurable spaces $(R_i,M_i)$, show that the joint map $(X_i)_i : \Omega \rightarrow \prod_1^n R_i $ is also measurable. $\prod R_i$ given product space.
My proof:
Let $s$ denote the joint map. $\pi_i: \prod R_i \rightarrow R_i$ denote the projection. $s^{-1} ( \pi_i^{-1}(B_i) ) = X_i^{-1}(B_i)$ for measurable sets $B_i \in M_{i}$. Let $S:= \{ E \subseteq \prod R_i \, : \, s^{-1}(E) \in M_{\Omega} \}$ is a $\sigma$-algebra containg $\{ \pi^{-1}(B_i) \, : \, B_i \in M_i \}_i$, this by definition contains the $\sigma$ algebra of the product space $\prod R_i$.
In my proof, I did not use finiteness. Is my proof wrong, or is this statement true for arbitrary indicies (?)
EDIT: To be clear on why I asked this question. I have seen one defines the joint distribution of a finite collection of random variable, $(X_i)$, by $$ P_{(X_i)}(E)= P((X_i)^{-1}(E)) $$ where $E \subseteq \prod R_i$. Then there is nothing wrong about defining it for arbitrary product? I am thinking this is because one cannot generalize the notion of "independence".
Well, your proof is not grammatically correct enough to call it "correct" or not. But the idea is correct, and works regardless of the index set of the product (so finiteness is indeed irrelevant). To check that $s$ is measurable, you just have to check that $s^{-1}(\pi_i^{-1}(B))$ is measurable for any measurable subset $B\subset R_i$. This is true because $s^{-1}(\pi_i^{-1}(B))=X_i^{-1}(B)$, and therefore is measurable since $X_i$ is measurable.