Let $I$ be an uncountable set and consider the product sigma algebra on $\mathcal{B}(\mathbb R)^{I}$
I want to show that $\mathcal{B}(\mathbb R)^{I}=\{ \{f \in \mathbb R^{I}\lvert (f(t_{1}),f(t_{2}),...)\in A\}\lvert (t_{n})_{n\in \mathbb N}\subseteq I,\; A \in \mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}\}=: Q$
I have already shown the inclusion $\mathcal{B}(\mathbb R)^{I}\subseteq Q$ by using the fact that $Q$ is a sigma-algebra and the preimages of projections of measurable sets indeed lie in $Q$
Apparently the direction $\mathcal{B}(\mathbb R)^{I}\supseteq Q$ is the easier one but I am lost on how to go about it.
Attempt: Assuming that I have $M:=\{f \in \mathbb R^{I}\lvert (f(t_{1}),f(t_{2}),...)\in A\}$ such that $(t_{n})_{n\in \mathbb N}\subseteq I,\; A \in \mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}$. Then if $A$ is of the form $A=\prod\limits_{n=1}^{\infty}A_{t_{i}}$ for $A_{t_{i}}\in \mathcal{B}(\mathbb R)$ clearly:
$M:=\mathbb R^{I\setminus (t_{n})_{n \in \mathbb N}}\times A= \bigcap\limits_{n=1}^{\infty}p_{t_{j}}^{-1}(A_{t_{j}})\in \mathcal{B}(\mathbb R)^{I}$. But my issue is that not every $A \in \mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}$ is of the form $A=\prod\limits_{n=1}^{\infty}A_{t_{i}}$. But I know that the set
$$\mathcal{C}:=\{ \prod\limits_{n=1}^{\infty}A_{t_{i}}\lvert A_{t_{i}}\in \mathcal{B}(\mathbb R)\; \forall i \in \mathbb N\} $$ is a generator of $\mathcal{B}(\mathbb R)^{(t_{n})_{n \in \mathbb N}}$. Is there any way I could use this generator to extend the result, like one usually does with the monotone class theorem?
If my understanding of your notation is correct then $M$ is a Borel subset of $\Bbb R^I$, being a preimage of a Borel set $A$ by the (continuous) projection map from $\Bbb R^I$ to $\Bbb R^{\{t_n\}}$.