Sigma algebra's generated by random variables Let $\Omega = \{0,1\}^\mathbb{N}$ and denote $\omega = (\omega_1,\omega_2,\ldots)$ as an element of $\Omega$. Now let $X_n: \Omega \rightarrow \mathbb{R}$ be defined by $X_n(\omega)= \omega_n$ and we put $\mathcal{F}_n = \sigma(X_1,\ldots,X_n)$.
My question now is how does $\mathcal{F}_1$ and $\mathcal{F}_2$ explicitly look like? I know that $\mathcal{F}_1 = \sigma(X_1) = \{X_1^{-1}(A)|A \in \mathcal{B}(\mathbb{R})\}$ but I don't know how to proceed further neither for $\mathcal{F}_2 = \sigma(X_1,X_2) = \sigma(X_1 \lor X_2)$. Any help is greatly appreciated!
 A: $X_1$ can only take two values: $0$ or $1$, so $\sigma(X_1)$ is generated by $$X_1^{-1}(\{0\}) = A_1 = \{\omega: \omega_1 = 0\}$$ and $$X_1^{-1}(\{1\}) = B_1 = \{\omega: \omega_1 = 1\}= A_1^c$$
So $\mathcal{F}_1 = \{\Omega, \emptyset, A_1, A_1^c\}$
Similarly, $\sigma(X_2)$ is generated by 
$$X_2^{-1}(\{0\}) = A_2 = \{\omega: \omega_2 = 0\}$$ and $$X_2^{-1}(\{1\}) = B_2 = \{\omega: \omega_2 = 1\}= A_2^c$$
Then $\sigma(X_2) = \{\Omega, \emptyset, A_2, A_2^c\}$
So $\mathcal{F}_2 = \sigma(X_1, X_2)$ is generated by the $\sigma(X_1)$ and $\sigma(X_2)$ and is equal to  
$$\{\Omega, \emptyset, A_1, A_1^c, A_2, A_2^c, A_1\cap A_2,A_1\cap A_2^c, A_1^c\cap A_2, A_1^c\cap A_2^c, A_1\cup A_2,A_1\cup A_2^c, A_1^c\cup A_2, A_1^c\cup A_2^c\}$$
A: $\mathcal{F}_n$ consists of cylinder sets on $\Omega$ formed by the permutation group of $\{\omega_1,...,\omega_n\}:=\pi(\omega_1,...,\omega_n)$ as the cylinder bases. I.e., you are forming a sub-$\sigma$-algebra on the restriction of $\Omega$ to the finite vectors $\omega'=(\omega_1,...,\omega_n)$ (a finite sample space, $\Omega'$). The sub-$\sigma$-algebra will be  the Cartesian product  of $2^{\Omega'}$ and $(\Omega')^c$. Therefore:
$$\mathcal{F}_n = 2^{\{\omega \in \pi(\omega_1,...,\omega_n)\} } \otimes \{\Omega \setminus \{\omega \in \pi(\omega_1,...,\omega_n)\}\}$$
Basically, $\mathcal{F}_n$ represents the $\sigma$-algebra on $\Omega$ than can be formed from only knowing the first $n$ elements of $\omega$
A: Recall that $\mathcal F_1$ is the smallest sub-$\sigma$-algebra of $\mathcal F$ making $X_1$ measurable. Since $X_1$ can only take the values $0$ and $1$, it is enough to contain $\{X_1=0\}$ and $\{X_1=1\}$.
