Problem in understanding sigma algebra in infinite product space $\textbf{(1)}$ Okay, So I understand the concept of product measure when we have two measure spaces, let's suppose $(\Omega_1, F_1,\mu_1)$ and $(\Omega_2,F_2,\mu_2)$ then we defined $(\Omega,F,\mu)$ where $\Omega=\Omega_1\times\Omega_2$, $F=F_1\times F_2$ and $\mu=\mu_1\times \mu_2$.
what we did here is that we proved that $F_1\times F_2$ is a semi-algebra and then we defined a set function $\mu: F_1\times F_2\to[0,\infty]$ and then we proved that it is countably additive hence it is a measure and then using caratheodory extension theorem we extended this to sigma-algebra generated by $F_1\times F_2$ which is $\sigma(F_1\times F_2)$.
But I am stuck in infinite product space 
$\textbf{(2)}$Let us have $(\Omega_i, F_i,\mu_i),i\in I$ be a collection of measure space.If I follow the same steps as above in the case of two measure spaces then $\Omega=\Pi_{i\in I}\Omega_i$, My problem is in the case of two measure space I was able to visualize that with the help of rectangles but in the infinite case, I am finding it difficult to visualize it. The book I am reading they have defined the sigma-algebra on $\Omega$ as follows 
Let $\pi_{i}:\Omega\to \Omega_{i}$ be a projection map then the sigma-algebra $F$ is the smallest sigma-algebra such that $\pi_{i}$ is measurable for all $i\in I$ and $F$=$\sigma\{\pi_{i}^{-1}(E_i):E_i \in F_i,i\in I\}$
My question: Can someone explain to me why we have defined the sigma-algebra like this in the infinite case and how can I connect this with the sigma-algebra generated in the case when we had two measure space as we have defined in $\textbf{(2)}$, also if possible suggest some books where I can find this topic to understand it in a better way.Thanks
 A: Let's take a closer look at your definition.
$$
F = \sigma\{\pi_{i}^{-1}(E_i):E_i \in F_i,i\in I\}
$$
Fix $i \in I$, and $E_i \in F_i$. Now we have that
$$ \pi_i^{-1}(E_i) = \prod_{j \in I }E_j, \quad E_j =\Omega_j \text{ for } j \neq i$$
so it is product with only one set being not whole $\Omega_j$. Clearly $\sigma$-algebra generated by those sorts of sets is the same
as one generated by
$$ \left\{ \prod_{i\in I} E_i : (\forall i\in I)(E_i\in F_i)\land (E_i=\Omega_i \text{ for all but finitely many } i )\right\}$$
because such sets are finite intersections of $\pi^{-1}_j(E_j)$.
Those sets are called cylinder sets, and appear in defining product topology as well. Here we could actually take countable intersections, but let's stick to cylinder sets for now.
Note that the case of finite product is included in this case: we allow finitely many elements $E_j$ to not be whole $\Omega_j$, and so our family of cylinder sets (which due to general definition are told to generate product $\sigma$-algebra) is precisely
$$
\left\{\prod_{i = 1}^n E_i : (\forall i)E_i \in F_i \right\}
$$- it is the same as generator in definition for finite product. This justifies our general definition as generalization of finite one.

You could ask why don't we make different generalization, for example define product $\sigma$-algebra to be one generated by $\{\prod_{i\in I} E_i : E_i \in F_i\}$. This actually would make bigger $\sigma$-algebra, and while projections still would be measurable, it would no longer be "smallest $\sigma$-algebra in which projections are measurable".
Take $F_x = \mathcal P(\{0, 1\})$. You can prove that product $\sigma$-algebra on $\prod_{x \in \mathbb R}\{0, 1\}$ has only $\mathfrak c$ elements, while our new definition would give $2^{\mathfrak c}$ sets. This "new" approach is very similar to box topology (again, not a standard way to define topology on a product).
In finite case, definitions by products on every axis and by measurability of projections are equivalent.
But in infinite, may not be. So as a standard, we pick one which seems more useful, and that happens to be one defined via projections.

Edit. Answering comment here, since it adds nice insight.
Suppose that $F_j$ were Borel $\sigma$-algebras; that is let $\tau_i, i\in I$ be topologies that satisfy $\sigma(\tau_i) = F_i$. Then
if each $\tau_i$ has countable basis, our
definition of product $\sigma$-algebra ensures that it is Borel $\sigma$-algebra of product topology on $\prod_{i\in I}\Omega_i$.
Product topology is (similarly) defined as smallest one in which
projections are continuous. This implies, that for this topology
the family of cylinder sets spanned by axis topologies forms a basis.
$$
\mathcal B_{\text{prod}} = 
\left\{\prod_{i \in I} U_i : (\forall i)(U_i \in \tau_i)
\land (U_i=\Omega_i \text{ for all but finitely many } i ) \right\}
$$
This family must be contained in product topology by continuity of projections, and the fact that topology is closed under finite intersections. Taking arbitrary sums of sets of this form gives
product topology, but assumption that each $\tau_i$ has countable
basis ensures that only countable sums are enough, and therefore
product topology is contained in product $\sigma$-algebra.
You can easily prove that in the process
of generating $\sigma$-algebra out of open cylinders
any measurable cylinder can be obtained, and therefore generated $\sigma$-algebras are the same.
A: Not an answer, just an alternative formulation to the product of measure space that looks more like the first one you mention.
This is the definition used in 'Linear Operators Part I : General Theory' by Dunford and Schwartz.
If instead of using the product of the $\sigma$-algebra as a semi algebra you use the following :
\begin{align*}
F=\left\{ \prod_{i\in I} E_i : \forall i\in I,E_i\in F_i, E_i\neq \Omega_i \text{ for finitely many } i \right\}
\end{align*}
Then you get exactly the same $\sigma$-algebra when you extend using Caratheodory.
I don't fully understand why it is like this but for sure this is to avoid degenerate cases when an arbitrary number of the $E_i$ can be strictly smaller than $\Omega_i$, some weird things could happen in those cases.
