Show identity between product-$\sigma$-algebra and a set

Let $T$ be any index set and $(\Omega_i,\mathcal{A}_i)_{i\in T}$ a family of measurable spaces and $\mathcal{A}:=\bigotimes_{i\in T}\mathcal{A}_i$. Show that
$$
\mathcal{A}=\left\{A\subset \times_{i\in T}\Omega_i | \exists R\subset T\text{ countable} : A\in\pi_R^{-1}\left(\bigotimes_{i\in R}\mathcal{A}_i\right)\right\}=:\mathcal{B},
$$
whereat
$$
\pi_R\colon \times_{i\in T}\Omega_i\to\times_{i\in R}\Omega_i, (\omega_i)_{i\in T}\mapsto (\omega_i)_{i\in R}.
$$

(Sorry, I do not know how to write the big times here so I used the small \times.)

Hello!

$\subseteq$:

My assumption is that $\mathcal{B}$ is a $\sigma$-algebra, is that right? ( Iwas not able to show it yet.)

Let $\mathcal{F}(T)$ be the set of the finite subsets of $T$. Then in our lecture we had, that
$\mathcal{A}$ is generated by
$$
\mathcal{Z}=\mathcal{Z}(\mathcal{A}_t: t\in T):=\bigcup_{S\in\mathcal{F}(T)}\mathcal{Z}_S,~~~\mathcal{Z}_S:=\pi_S^{-1}\left(\bigotimes_{t\in S}\mathcal{A}_t\right),
$$
so $\sigma(\mathcal{Z})=\mathcal{A}$.

When I see it right, then $\mathcal{Z}\subset \mathcal{B}$. If my assumption that $\mathcal{B}$ is a $\sigma$-Algebra is right, then it follows $\mathcal{A}=\sigma(\mathcal{Z})\subset \mathcal{B}$.

$\supseteq$:

Consider $A\in \mathcal{B}$. Then there exists a countable $R\subset T$ so that $A\in\pi_R^{-1}\left(\bigotimes_{i\in R}\mathcal{A}_i\right)$.

Now I think one has to distinguish (i) $R$ is finite and (ii) $R$ is countably infinite.

Case (i): Then $A\in\mathcal{Z}_{R}\subset\mathcal{Z}\subset\sigma(\mathcal{Z})=\mathcal{A}$.

I am not able to handle case (ii).

Would be great if you could help me.

Solutions Collecting From Web of "Show identity between product-$\sigma$-algebra and a set"

Let $A$ be a set in the product $\sigma$-algebra. We want to show $A\in\mathcal{B}$. It is enough to find a countable set $R$ such that $A=\pi_R^{-1}(A)$. This is certainly the case if $A=\emptyset$. Also, it is the case if $\pi_t(A)=\Omega_t$ for all but countably many $t$. In that case, we can let $R=\{t:\pi_t(A)\neq\Omega_t\}$. So it is enough to prove the following lemma:

Lemma: Let $\mathcal{W}$ be the family of all product measurable sets $A$ that are empty or such that $\pi_t(A)=\Omega_t$ for all but countably many $t$. Then $\mathcal{A}=\mathcal{W}$.

Proof: $\prod_t\Omega_t$ is clearly in $\mathcal{W}$. Let $A$ be a nonempty set with nonempty complement in $\mathcal{W}$ such that $\pi_t(A)=\Omega_t$ for all $t$ outside the countable set $R$. Then $$A^C=\Big\{x\in\prod_t\Omega_t:x\notin A\Big\}=\Big\{x\in\prod_t\Omega_t:x_t\notin \pi_t(A)\text{ for some }t\Big\}$$ $$=\Big\{x\in\prod_t\Omega_t:x_t\notin \pi_t(A)\text{ for some }t\in R\Big\}.$$ So $A^C$ is a product measurable set such that $\pi_t(A^C)=\Omega_t$ for all $t\notin R$. So $\mathcal{W}$ is closed under complements.

Now let $(A_n)$ be a sequence of sets in $\mathcal{W}$. We can asumme wlog that all sets in the sequence are nonempty and have nonempty complement. For each $n$, let $R_n$ be such that $\pi_t(A_n)=\Omega_t$ for all $t\notin R_n$. Let $R=\bigcap_n R_n$.Now let $t\notin R$. Then there is $m$ such that $t\notin R_m$ and hence $\pi_t(A_m)=\Omega_t$. Now $$A_m\subseteq \bigcup_n A_n\text{ and therefore }\Omega_t=\pi_t(A_m)\subseteq\pi_t\bigg(\bigcup_n A_n\bigg),$$ which shows that the latter set equals $\Omega_t$. So $\mathcal{W}$ is closed under countable unions. Since it is also closed under complements, it is therefore also a $\sigma$-algebra. Moreover, we clearly have $\pi_t^{-1}(A_t)\in\mathcal{W}$ for all $A_t\in\mathcal{A}_t$. So $\mathcal{W}=\mathcal{A}$. $ \square$

For the other direction, let $R$ be countable, and $A$ be product measurable and nonempty. We want to show that $\pi_R^{-1}(A)$ is product measurable. Now $$\pi_R^{-1}(A)=\{x\in\prod_t\Omega_t: x_t\in\pi_t^{-1}(A)\text{ for all }t\in R\}=\bigcap_{t\in R}\pi_t^{-1}(A)\in\mathcal{A},$$ which is measurable as the countable intersection of measurable sets.