[Math] Tail $\sigma$-algebra for a sequence of sets

measure-theoryprobability theory

  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!

Best Answer

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.

Related Question