inclusion of $\sigma$-algebra generated by random variables Consider the following random variables
$$X:\Omega\to\mathbb{R}\quad\text{and}\quad Y:\Omega\to \mathbb{R}$$
and
$$Z:=XY$$.
One may interpret it as follows, i.e.
$$Z(\omega) = X(\omega)Y(\omega).$$
In general, we cannot say much about the relations amongst $\sigma(X),\sigma(Y)$ and $\sigma(Z)$, as discussed in the answer below.

However, if we consider $X_i:\Omega \to \mathbb{R}$ for $i\in(\infty,\infty)$.
Why do we have $\sigma(\cdots,X_{n-1}X_n)\subsetneq \sigma(\cdots,X_n)$?
 A: The first is valid. The second implies $\omega_1 = \omega_2$. It's valid too, but a waste of indices. You can imagine as if Tyche would choose a random point of  $\Omega$. This point is then mapped according to $X$ and $Y$ to two real numbers. Nothing particular follows from this for the relationship between $\sigma(X)$ and $\sigma(Y)$. You can construct both r.v., such that $\sigma(X) \subset \sigma(X)$ or identical or even $\sigma(Y) \cap \sigma(X) = \{\emptyset,\Omega\}$.
Concerning subset-relations between these $\sigma$-algebras, one can not say anything in general. Usually $\sigma(Z)$ is finer than $\sigma(X),\sigma(Y)$, but if e.g. $X(A)=0, Y(A^C)=0$, then $\sigma(Z)=\{\emptyset,\Omega\}$ and $\sigma(Z)$ is even coarser than the other algebras. It is possible, that $\sigma(Z) = \sigma(X \times Y) = \sigma(X,Y)$ 
Generally not possible is $Z:\Omega^2\rightarrow \mathbb{R}$ with your definition $Z(\omega)=X(\omega)Y(\omega)$. Your definition implies that the same chosen $\omega$ is mapped by $X$ and $Y$. The latter implies that Tyche may choose a tupel $(\omega_1, \omega_2) \in \Omega^2$, so $X$ and $Y$ may map two different points.
