I am reading a stochastic analysis script and in one proof the following 'fact' is used.
Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space, $I \neq \emptyset$ an index set and $(S, \mathcal{S})$ a measurable space. For a process $X = (X_i)_{i \in I}$ consisting of $S$ valued $\mathcal{F}-\mathcal{S}-$measurable random variables the following is claimed (without proof):
Let $\mathcal{I} = \{\bigcap_{j \in J} X_j^{-1}(A_j)| J \subset I \text{ finite }, A_j \in \mathcal{S}\}$, then $\sigma(\mathcal{I}) = \sigma(X_i, i \in I) (= \sigma( \bigcup_{i \in I} X_i^{-1}(\mathcal{S})))$. The interesting direction is to show that every set $\bigcup_{i \in I} X_i^{-1}(A_i)$, with $A_i \in \mathcal{S}$ is an element of $\sigma(\mathcal{I})$.
For uncountable $I$, I wanted to proof it myself, but is this even correct?
For example: Consider $(\Omega, \mathcal{F}, \mathbb{P}) = ([0,1], \mathcal{B}, \lambda)$ and $(S, \mathcal{S}) = (\mathbb{R}, \mathcal{B})$ and $X_i = 1_{\{i\}}$, for $i \in I$, where $I$ is some nonmeasureable subset of $[0,1]$.
Now, we have $\sigma(\mathcal{I}) \subset \mathcal{B}$, but $\bigcup_{i \in I} X_i^{-1}(\{1\}) = \bigcup_{i \in I} \{i\} = I \in \sigma(X_i, i \in I)$. So, according to the above claim, we would have $I \in \mathcal{B}$ which is false.
Am I missing something? If I am, how would you actually prove it?
Best Answer
I think the claim about $\sigma(\mathcal{I})$ can be proved as follows with no mention of the 'interesting direction'.
We have that $X_i^{-1}(\mathcal{S})\subseteq \mathcal{I},\,\forall i \in I$, as for all $J=\{i\}$ we have $X^{-1}_i(A)\in \mathcal{I},\forall A \in \mathcal{S}$. Therefore, $\cup_{i \in I}X^{-1}_i(\mathcal{S})\subseteq \mathcal{I}$.
Now for arbitrary $X_i\in (X_i)_{i \in I},A_i\in \mathcal{S}$ we have $X_i^{-1}(A_i)\in \sigma(X_i,i\in I)$. By the properties of $\sigma$-algebras, any finite intersection $\cap_{j \in J}X_j^{-1}(A_j)\in \sigma(X_i,i\in I)$.
It follows that
$$\bigcup_{i \in I}X_i^{-1}(\mathcal{S})\subseteq \mathcal{I}\subseteq \sigma(X_i,i \in I)$$
and that $\sigma(\mathcal{I})=\sigma(X_i,i \in I)$.