How to show that a stochastic process is a measurable function-valued mapping?

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!

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 $$i \in T$$.)