Why does $\sigma (X_t) \subset \sigma (X)$ hold? We have a random process $X=\{X_t\;,t\in T\}$, where $X_t:(\Omega,\mathcal{A})\to(S_t,\mathcal{S}_t)$ are random variables. I am confused as to why does
$$\sigma (X_t) \subset \sigma (X)$$
hold $\forall t\in T$. 
More background:  Intuitively, this makes sense, rigorously, I am at loss. I suspect this might have something to do with my poor understanding of $\sigma(\text{anything})$ (perhaps later spawning a subsequent question), however, this is my thinking so far:
$\sigma(X_t)=\{[X_t\in B_t],B_t\in \mathcal{S}_t\}$ and 
$\sigma(X)=\{[X\in B],B\in \mathcal{S}\}$ where $\mathcal{S}=\bigotimes S_t$ (the product $\sigma$-algebra, i.e. the smallest $\sigma$-algebra on $\prod S_t$ containg measurable cylinders)
Now, perhaps I could write $[X\in B]=[\{X_t\}\in B]$ and perhaps somehow $B=\prod B_t$, and so maybe $[X\in B]$ really equals to something like $\bigcap[ X_t\in B_t]$ which looks much more like what I want, but a) this is almost just wishful thinking with my poor understanding of the topic and b) even then I am not sure how to proceed.
Hope this makes sense (doesn't to me so far!) and thanks for any help.
 A: By definition, $\sigma(X_t)$ is the smallest $\sigma$-algebra $\mathcal{F} \subseteq \mathcal{A}$ such that $X_t:(\Omega,\mathcal{F}) \to (S_t,\mathcal{S}_t)$ is measurable. Therefore, it suffices to show that $X_t$ is measurable with respect to $\sigma(X)$. To this end, we associate to $B_t \in \mathcal{S}_t$ the cylinder set
$$B := \{(s_r)_{r \in T} \in \prod_{r \in T} S_r; s_t \in B_t\}.$$
It follows from the very definition of $\sigma(X)$ and $B$ that
$$\begin{align*} X_t^{-1}(B_t)  &= X^{-1}(B) \in \sigma(X). \end{align*}$$
This already proves that $X_t$ is measurable with respect to $\sigma(X)$.

Answers to your questions:


*

*Measurable as a mapping $X_t: (\Omega,\mathcal{F}) \to (S_t,\mathcal{S}_t)$, i.e. $X_t^{-1}(A) \in \mathcal{F}$ for all $A \in \mathcal{S}_t$.

*If $X_t$ is measurable with respect to $\sigma(X)$, i.e. $X_t: (\Omega,\sigma(X)) \to (S_t,\mathcal{S}_t)$ is measurable, then $\sigma(X_t) \subseteq \sigma(X)$ since $\sigma(X_t)$ is the smallest $\sigma$-algebra such that $X_t$ is measurable (in the sense of 1.).

*$(s_r)_{r \in T}$ is simply an element of the product space $\prod_{r \in T} S_r$, i.e. for each $r \in T$, we have $s_r \in S_r$.

