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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!

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1 Answer 1

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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 $t \in T$.

Let $Y$ a set, $(Y_i,\mathcal{A}_i)$ measure spaces ($i \in I$) and $f_i: Y \to Y_i$ arbritary mappings ($i \in I$). Denote by $\sigma(f_i,i \in I)$ the $\sigma$-algebra generated by the mappings $f_i$. Moreover, let $(\Omega,\mathcal{A})$ a measure space and $g: \Omega \to Y$ a mapping. Then the following statements are equivalent.

  1. $g$ is $\mathcal{A}/\sigma(f_i,i \in I)$-measurable
  2. $\forall i \in I: f_i \circ g$ is $\mathcal{A}/\mathcal{A}_i$-measurable

(Here: $Y:=S^T$, $I:=T$, $Y_i := S$, $f_i := \Pi_i$ for $i \in T$.)

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