Generator of a product $\sigma$-algebra Let

*

*$I\subseteq\mathbb R$;

*$(E,\mathcal E)$ be a measurable space;

*$\pi_J$ denote the projection from $E^I$ onto $E^J$ for $J\subseteq I$ and $\pi_i:=\pi_\{i\}$ for $i\in I$.

By definition, the product $\sigma$-algebra $\mathcal E^{\otimes I}$ on $E^I$ is given by $$\mathcal E^{\otimes I}:=\sigma(\pi_i,i\in I)$$ and the $\sigma$-algebras $\mathcal E^{\otimes J}$ on $E^J$, for any $J\subseteq I$, are defined accordingly.

Question 1: If $J\subseteq I$, are we able to show that $\sigma(\pi_J)=\sigma(\pi_j,j\in J)$?

This is claim seems to be so trivial that I'm not able to prove it. By definition, $\sigma(\pi_J)=\pi_J^{-1}(\mathcal E^{\otimes J})$ and it's clear to me that $\pi_J$ is $(\mathcal E^{\otimes I},\mathcal E^{\otimes J})$-measurable.

Now let $(\Omega,\mathcal A)$ be a measurable space and $(X_i)_{i\in I}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$. If $J\subseteq I$, we may consider the random variable $$X_J:=(X_j)_{j\in J}:\Omega\to E^J.$$

Question 2: Let $\mathcal F^Y$ denote the filtration generated by any process $Y$. Are we able to show that $\mathcal F^X_t=X_I^{-1}(\mathcal F^\pi_t)$ for all $t\in I$?

Let $t\in I$ and $J:=\{s\in I:s\le t\}$. By definition, $$\mathcal F^\pi_t=\sigma(\pi_s,s\in J)=\mathcal E^{\otimes J}\tag1.$$ On the other hand, $$\mathcal F^X_t=\sigma(X_s,s\in J)\tag2$$ and if the claim in Question 1 is correct, then $$\sigma(X_J)=X_J^{-1}(\mathcal E^{\otimes J})=X_I^{-1}(\pi_J^{-1}(\mathcal E^{\otimes J}))=X_I^{-1}(\mathcal F^\pi_t)\tag3.$$ So, if that's correct, all that remains to show would be $$\sigma(X_s,s\in J)=\sigma(X_J)\tag4,$$ which seems again to be the same question as in Question 1.
 A: You are right when you say that the results you need for Question 1 and Question 2 are rather obvious (or trivial). However, we need some care when proving them. Here are detailed proofs.
Let us start with a lemma:
Lemma: Let $(Y, \Sigma)$ be a measure space and let $S \subseteq \Sigma$ be such that $\Sigma =\sigma(S)$.  Let $f : X \rightarrow Y$ be a function. Then,
$$\sigma(f)= \sigma (\{f^{-1}(D) : D\in S\})$$
Proof: Since  $S \subseteq \Sigma$, we clearly have
$$\sigma (\{f^{-1}(D) : D\in S\}) \subseteq \sigma (\{f^{-1}(C) : C\in \Sigma\})= \sigma(f) \tag{1}$$
Now, let
$ \Gamma =\{B \subseteq Y : f^{-1}(B) \in \sigma (\{f^{-1}(D) : D\in S\})$.
Note that

*

*$\emptyset \in \Gamma$


*if $B \in \Gamma$ then $f^{-1}(B) \in \sigma (\{f^{-1}(D) : D\in S\})$. So,
$$f^{-1}(B^c) = (f^{-1}(B))^c \in \sigma (\{f^{-1}(D) : D\in S\}) $$
so $B^c \in \Gamma$.


*If, for all $i \in \Bbb N$,  $B_i \in \Gamma$, we have $f^{-1}(B_i) \in \sigma (\{f^{-1}(D) : D\in S\})$. So
$$f^{-1} \left(\bigcup_i B_i \right)= \bigcup_i f^{-1}(B_i) \in \sigma (\{f^{-1}(D) : D\in S\}) $$
so $\bigcup_i B_i \in \Gamma$.
So $\Gamma$ is a $\sigma$-algebra. Clearly $S \subseteq \Gamma$. So, $\Sigma = \sigma(S) \subseteq \Gamma$. But that means that, for all $C \in \Sigma$, $f^{-1}(C)  \in \sigma (\{f^{-1}(D) : D\in S\})$. So,
$$\{f^{-1}(C) : C\in \Sigma\} \subseteq \sigma (\{f^{-1}(D) : D\in S\})$$
So
$$ \sigma(f) = \sigma (\{f^{-1}(C) : C\in \Sigma\}) \subseteq \sigma (\{f^{-1}(D) : D\in S\}) \tag{2}$$
From $(1)$ and $(2)$, we have $\sigma(f)= \sigma (\{f^{-1}(D) : D\in S\})$. This completes the proof of the lemma. $\square$
Now,

*

*Let $I\subseteq \Bbb R$ and $J\subseteq I$ be two fixed sets;


*Let $(E,\mathcal E)$ be a measurable space;


*For all $i \in I$, let $\pi_i: E^I \rightarrow E$ be the projection from $E^I$ onto $E$ corresponding to $i$.


*We define $\mathcal E^{\otimes I}= \sigma(\pi_i: i \in I) = \sigma(\{\pi_i^{-1}(A): A \in \mathcal E \text{ and } i \in I\})$


*For all $j \in J$, let $\rho_j: E^J \rightarrow E$ be the projection from $E^J$ onto $E$ corresponding to $j$.


*We define $\mathcal E^{\otimes J}= \sigma(\rho_j: i \in J) = \sigma(\{\rho_j^{-1}(A): A \in \mathcal E \text{ and } j \in J\})$
Let  $\pi_J : E^I \rightarrow E^J$ be the projection from  $E^I$ onto $E^J$. Note that, for all $j \in J$, we have $\pi_j= \rho_j \circ \pi_J$
Let us prove

Question 1: $\sigma(\pi_J)=\sigma(\pi_j,j\in J)$

Proof:
We have
\begin{align*}
\sigma(\pi_J) & = \sigma(\{\pi_J^{-1}(B): B \in \mathcal E^{\otimes J} ) &\\
&= \sigma(\{\pi_J^{-1}(\rho_j^{-1}(A)): A \in \mathcal E \text{ and } j \in J ) & \text{ by the lemma} \\
&= \sigma(\{(\rho_j \circ \pi_J)^{-1}(A): A \in \mathcal E \text{ and } j \in J ) & \\
&= \sigma(\{\pi_j^{-1}(A): A \in \mathcal E \text{ and } j \in J ) & \\
&= \sigma(\pi_j: j \in J)
\end{align*}
This completes the proof. $\square$
Remark: Regarding Question 2, your argument is correct. For the last step you need

Let $(\Omega,\mathcal A)$ be a measurable space and $(X_i)_{i\in I}$ be an $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$. If $J\subseteq I$, we may consider the random variable $$X_J:=(X_j)_{j\in J}:\Omega\to E^J.$$
Then $\sigma(X_J)= \sigma(X_j,j\in J)$

Proof It is similar to proof of Question 1.
Note that, for all $j \in J$, we have $X_j= \rho_j \circ X_J$.
We have
\begin{align*}
\sigma(X_J) & = \sigma(\{X_J^{-1}(B): B \in \mathcal E^{\otimes J} ) &\\
&= \sigma(\{X_J^{-1}(\rho_j^{-1}(A)): A \in \mathcal E \text{ and } j \in J ) & \text{ by the lemma} \\
&= \sigma(\{(\rho_j \circ X_J)^{-1}(A): A \in \mathcal E \text{ and } j \in J ) & \\
&= \sigma(\{X_j^{-1}(A): A \in \mathcal E \text{ and } j \in J ) & \\
&= \sigma(X_j: j \in J)
\end{align*}
