If $\mathcal E_i=\mathcal E_0\times\{i\}$ is $\sigma\left(\bigcup_{i\in\mathbb N}\mathcal E_i\right)=\mathcal E_0\times\mathbb N$? Let $(E_0,\mathcal E_0)$ be a measurable space, \begin{align}E_i&:=E_0\times\{i\};\\\mathcal E_i&:=\mathcal E_0\times\{i\}\end{align} for $i\in\mathbb N$ and \begin{align}E&:=\bigcup_{i\in\mathbb N}E_i;\\\mathcal E&:=\sigma\left(\bigcup_{i\in\mathbb N}\mathcal E_i\right).\end{align}
Note that $\mathcal E_i$ is actually a $\sigma$-algebra on $E_i$ for all $i\in\mathbb N$ and $$E=E_0\times\mathbb N.$$
$\mathcal E$ is the smallest $\sigma$-algebra $\mathcal S$ on $E$ such that $\mathcal E_i\subseteq\mathcal S$ for all $i\in\mathbb N$.

Is $\mathcal E$ equal to $\mathcal E_0\times\mathbb N$ or is $\mathcal E=\mathcal E_0\otimes2^{\mathbb N}$?

 A: We can write every subset of $A\subseteq E=E_0\times\mathbb N^+$ as:$$A=\bigcup_{i=1}^{\infty}A_i\times\{i\}$$where $A_i=\{a\in E_0\mid (a,i)\in A\}\subseteq E_0$ for every positive integer $i$.
The set $A$ determines the sets $A_i$ and vice versa.
Further we find:

*

*$\left(A^c\right)_i=A_i^c$

*$\left(\bigcup_{n=1}^{\infty} A^{(n)}\right)_i=\bigcup_{n=1}^{\infty}A^{(n)}_i$
On base of this define $\mathcal E\subseteq\mathcal P(E)$ by stating:
$$A\in\mathcal E\text{ if }A_i\in\mathcal E_0\text{ for every positive integer }i\tag1$$
It can be proved now on a basical way that $\mathcal E$ is a $\sigma$-algebra on $E$ with $\mathcal E_i\subseteq\mathcal E$ for every positive integer $i$.
Moreover it is evident that every $\sigma$-algebra $\mathcal S$ on $E$ that contains the $\mathcal E_i$ as subcollections must satisfy the condition posed on $\mathcal E$ in $(1)$, so that we are allowed to conclude that:$$\mathcal E=\sigma\left(\bigcup_{i=1}^{\infty}\mathcal E_i\right)$$

In short: the elements of $\mathcal E$ are sets of shape:$$\bigcup_{i=1}^{\infty}A_i\times\{i\}$$where $A_i\in\mathcal E_0$.
They correspond one to one with sequences $(A_1,A_2,\dots)$ in $\mathcal E_0$.
These sequences correspond also one-to-one with functions $\mathbb N\to\mathcal E_0$ so that $\mathcal E$ can also be looked at as the set $\mathcal E_0^{\mathbb N}$.
An element of $\mathcal E_0\times\mathbb N$ is a set of shape $A\times\{i\}$ where $A\in\mathcal E_0$ and $i$ is a positive integer.
So $\mathcal E$ and $\mathcal E_0\times\mathbb N$ are essentially different.
