# Understanding a proof of the Kolmogorov zero-one law

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.

Let $$A \in \mathcal T$$. Consider $$\{B\in \sigma(X_1,X_2,...): P(A\cap B)=P(A)P( B)\}$$. Check that this is a monotone class and it contains $$\bigcup_n \sigma(X_1,X_2,...,X_n)$$. Then verify that $$\bigcup_n \sigma(X_1,X_2,...,X_n)$$ is an algebra. Monotone class lemma now shows that every set in the $$\sigma-$$ algebra generated by $$\bigcup_n \sigma(X_1,X_2,...,X_n)$$ belongs to $$\{B\in \sigma(X_1,X_2,...): P(A\cap B)=P(A)P( B)\}$$. This is precisely what we want to prove since the $$\sigma-$$ algebra generated by $$\bigcup_n \sigma(X_1,X_2,...,X_n)$$ coincides with $$\sigma(X_1,X_2,...)$$