Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top
  1. Suppose $\{A_n, n \in \mathbb{N}\}$ is a sequence of subsets of $\Omega$. Each $A_n$ generates a $\sigma$-algebra as $\mathcal{A}_n:=\{ A_n, A_n^c, \emptyset, \Omega \}$. I was wondering if we can simplify $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$, i.e., the tail $\sigma$-algebra of the sequence of $\sigma$-algebras $\{ \mathcal{A}_n, n \in \mathbb{N} \}$?

    I guess $\limsup_{n \rightarrow \infty} A_n:= \cap_{i=1}^{\infty} \cup_{j=i}^{\infty} A_n$ does belong to $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$, and I don't know if $\liminf_{n \rightarrow \infty} A_n:= \cup_{i=1}^{\infty} \cap_{j=i}^{\infty} A_n$ belongs to $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$?

  2. If $\{A_n, n \in \mathbb{N}\}$ are independent events on probability space $(\Omega, \mathcal{F}, P)$, can we further simplify the tail $\sigma$-algebra $\cap_{i=1}^{\infty} \sigma(\cup_{j=i}^{\infty} \mathcal{A}_j)$? The reason I asked this is because I was wondering why the tail σ-algebra is said to be trivial in this independent case and how it looks like to be trivial?

Thanks and regards!

share|cite|improve this question
up vote 3 down vote accepted

The tail sigma-algebra is the sigma-algebra of sets $B$ such that, for every integer $N$ one can build $B$ from the sets $A_n$ with $n\ge N$ only. For example the limsup/liminf of $(A_n)_n$ is also the limsup/liminf of $(A_{n+N})_n$ hence the limsup/liminf is in the tail sigma-algebra. There is no measure involved here.

In the independent case with respect to a probability $P$, the tail sigma-algebra is trivial in the sense that it contains only sets of $P$-probability zero or one. For example the limsup/liminf of any sequence which is independent for $P$ has $P$-probability zero or one. This is a property of $P$ in relation with the sigma-algebras considered but definitely not a property of the sigma-algebras alone.

share|cite|improve this answer
Thanks! For the tail of a sequence of general $\sigma$-algebras $\{\mathcal{F}_n\}$, (1) can I cay a set $B$ belongs to the tail if and only if $B$ can be built from $\{ \mathcal{F}_n: n \geq N \}$ for any integer $N$? (2) what does "$B$ can be built from $\{ \mathcal{F}_n: n \geq N \}$ " mean? Is it that there exists some mapping $f$, such that $B=f(\{ \mathcal{F}_n: n \geq N \})$ ? – Tim May 10 '11 at 14:20
That $B$ belongs to the sigma-algebra generated by the union over $n\ge N$ of the sigma-algebras $F_n$. – Did May 10 '11 at 17:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.