Suppose one has a sequence ${X_1, X_2, \dots}$ of random variables, for each natural number ${n}$, we can define the ${\sigma}$-algebras ${\sigma( X_i: i >n )}$, as the smallest ${\sigma}$ algebra that makes all of the ${X_i}$ for ${i >n}$ measurable.
(Kolmogorov zero-one law) Let ${X_1,X_2,\dots}$ be a sequence of jointly independent random variables. Then every event ${E}$ in the tail ${\sigma}$-algebra ${{\mathcal T}}$ has probability equal to either ${0}$ or ${1}$.
Proof: Since ${X_1,X_2,\dots}$ are jointly independent, the ${\sigma}$-algebra ${\sigma( X_i: i > n )}$ is independent of ${\sigma(X_1,\dots,X_n)}$ for any ${n}$. In particular, ${{\mathcal T}}$ is independent of ${\sigma(X_1,\dots,X_n)}$. Since the ${\sigma}$-algebra ${\sigma(X_i: i \geq 1)}$ is generated by the ${\sigma(X_1,\dots,X_n)}$ for ${n=1,2,3,\dots}$, a simple application of the monotone class lemma then shows that ${{\mathcal T}}$ is also independent of ${\sigma(X_i: i \geq 1)}$. But ${\sigma(X_i: i \geq 1)}$ contains ${{\mathcal T}}$, hence ${{\mathcal T}}$ is independent of itself. But the only events ${E}$ that are independent of themselves have probability ${0}$ or ${1}$, and the claim follows. $\Box$
Question: By the statement "a simple application of the monotone class lemma then shows that ${{\mathcal T}}$ is also independent of ${\sigma(X_i: i \geq 1)}$", it seems that the author is saying that $\mathcal T$ is independent of a monotone class containing the generators $\sigma(X_1, \dots, X_n), n=1, 2, \dots$ of $\sigma(X_i: i>n)$, hence by the said lemma $\mathcal T$ is independent of the smaller class $\sigma(X_i: i > n)$. But I can't see immediately what this larger class is.
