Given a probability space $\Omega$, a measurable space $S$ and a set $T$, a stochastic process $X: \Omega \times T \to S$ is defined as a $T$-indexed family of random variables $\{X_t: \Omega \to S, t \in T\}$.
It can be viewed as a functional-valued mapping $\Omega \to S^T$ with the product sigma algebra on $S^T$.
I was wondering how to show that this functional-valued mapping is measurable wrt the product sigma algebra on $S^T$?
Thanks and regards!
Best Answer
The product sigma-algebra on $S^T$ is generated by the projections
$$\Pi_t: S^T \to S, f \mapsto f(t) \qquad (t \in T)$$
i.e. it's the smallest sigma-algebra such that projections $\Pi_t$ ($t \in T$) are measurable. This means that a mapping $X: \Omega \to S^T$ is measurable (with respect to the product sigma-algebra on $S^T$) iff the mappings
$$\Pi_t \circ X: \Omega \to S, \omega \mapsto X(\omega)(t) = X_t(\omega)$$
are measurable for all $t \in T$.
(Here: $Y:=S^T$, $I:=T$, $Y_i := S$, $f_i := \Pi_i$ for $i \in T$.)