I am having a little trouble understanding what exactly is the tail $\sigma$-algebra.

Just so we are all on the same page, my book defined the tail $\sigma$-algebra like this:

Let $X_n$ be a sequence of random variable defined on $(\Omega, \mathcal{A}, P)$. Define $\mathcal{B}_n = \sigma(X_n)$, $\mathcal{C}_n = \sigma(\cup_{p \ge n} \mathcal B_p)$, $\mathcal C_\infty = \cap_{n=1}^\infty \mathcal C_n$. $\mathcal C_\infty$ is called the tail $\sigma$-algebra.

Suppose now that $X_n$ are independent. I am having trouble understanding why the following are true (we are talking about the consequences of Kolmogorov's Zero-One Law): $$\{ \omega: \lim_{n \to \infty} X_n(\omega) \text{ exists} \} \in \mathcal C_\infty$$ $$\limsup_{n \to \infty} X_n \ \ \liminf_{x \to \infty} X_n \ \ \limsup_{n \to \infty} \frac 1n \sum_{p \le n} X_p \ \ \liminf_{n \to \infty} \frac 1n \sum_{p \le n} X_p \text{ are $\mathcal C_\infty$ measurable} $$

I don't have any intuition about why these are true nor what exactly is a tail $\sigma$-algebra apart from the definition, which does not inspire me that much.

Any help is appreciated!

  • $\begingroup$ Which one of these events depend on $X_1$? $\endgroup$
    – Did
    Commented Dec 1, 2014 at 23:06
  • $\begingroup$ @Did The first event does not, I guess. The others are r.v. which do depend on $X_1$.. Maybe I am misinterpreting what you said. To be honest I'm a little confused (if that wasn't clear already!) Anyhow I'd appreciate a comment about the downvote and the vote to close, so I can improve something :) $\endgroup$
    – Ant
    Commented Dec 1, 2014 at 23:10
  • $\begingroup$ Is $\limsup X_n$ depending on $X_1$? $\endgroup$
    – Did
    Commented Dec 1, 2014 at 23:16
  • 3
    $\begingroup$ For example, let $A_k=\{\lim\sum_{n\ge k}X_k\text{ exists}\}$. Then $A_1=A_2=A_k\in\sigma(X_k,X_{k+1},...)$ (because the series converges iff its tail converges). So, $A_1\in\bigcap_{k\ge 1}\sigma(X_k,X_{k+1},...)\equiv \mathcal C_\infty$ $\endgroup$
    – user140541
    Commented Dec 2, 2014 at 6:51
  • 1
    $\begingroup$ @d.k.o. I think it is a little clearer now. I would be very grateful if you could make an answer and expand a little over what you just said! :) $\endgroup$
    – Ant
    Commented Dec 4, 2014 at 21:42

1 Answer 1


The answer follows the idea of what @d.k.o. pointed out. For example, for the statement

$$\{ \omega: \exists \lim_{n \to \infty} X_n(\omega) \} \in C_\infty$$

we can argue as follows:

For each n, $X_{m}$ with $m \ge n$ is measurable for the $\sigma$-field ${C}_n = \sigma(\cup_{p \ge n} B_{p})$. Now we use the following fact:

The set of convergence of a sequence of measurable functions is measurable

Therefore the set of convergence of the sequence ${X_{m}}$ with $m \ge n$ is measurable in $C_{n}$. As this set of convergence does not change with n, it follows it belongs to $C_\infty$.

When you understand this proof the rest is easier. Let's prove that:

$\limsup_{n \to \infty} X_n$ is a measurable function in $C_\infty$.

$[\lim sup_{n \to \infty} X_{n} < c]=[\lim sup_{k \to \infty, k \ge n} X_{n} < c]$ which belongs to $C_{n}$ and the statement follows.


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