I have previously asked this question, and this question is just a generalization
For a set of random variable $X_{s} :(\Omega, \mathcal{F}) \to (I, P(I)), s\in S$, where I is a countable set. Is the sigma-algebra generated by this set of random variables $\sigma(\{ \{ X_{1} = i_{1}, X_{2} = i_{2}, \dots, X_n = i_{n}\}\mid i_{1}, i_{2},\dots i_{n} \in I\})$(Here we assume S is a finite set)? Would this generalize to S that is countable, or even to any set of S?
Even in the finite case, I am too sure whether it is true. With some algebraic manipulations, I get to $$\sigma\left(\bigcap_{1\leq s \leq n}\bigcup_{i \in I}\{ X_{s} = i \}\right)$$ is in the set. But I don't know how to proceed further.
Also, see here for the definition of sigma-algebra generated by a set of random variables. It is the smallest subset of $\mathcal{F}$ that makes all these random variables measurable.
Best Answer
The result is true if $S$ is finite, but not otherwise. If $S = \{1, \dots, n\}$, then
$$ \sigma(\{X_s \in A\}: s \in S, A \subseteq I)=\sigma(\{X_1=i_1, \dots, X_n=i_n\}:i_1, \dots, i_n \in I).$$
You can prove this by showing that both sides include each other; LHS $\subseteq$ RHS because $\sigma$-algebras are closed under countable union, and RHS $\subseteq$ LHS because $\sigma$-algebras are closed under finite intersection.
However, for $S = \{1, 2, \dots\},$ take $I = \{0, 1\},$ $\Omega = I^S$, and $X_s(\omega) = \omega_s$. Then
$$\sigma(\{X_1=i_1, X_2=i_2, \dots\}:i_1, i_2, \ldots \in I)$$
contains only sets which are countable or have countable complement, so is not equal to $\sigma(X_s:s\in S)$. (You can prove this using the $\pi$-$\lambda$ theorem.)