TITLE: Sigma algebra's generated by random variables
QUESTION [6 upvotes]: 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!

REPLY [2 votes]: 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 $\{X_1=1\}$.