On a set $\Omega$, there is a sequence of sigma algebras $(\mathcal{F}_n)_{n \in \mathbb{N}}$. The tail sigma algebra of $(\mathcal{F}_n)$ is defined to be $\cap_{n=1}^{\infty} \sigma(\cup_{m=n}^\infty \mathcal{F}_m )$. I was wondering if the following two statements are true:
- $\forall B$ in the tail sigma algebra of $(\mathcal{F}_n)$, there exist $ A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$, such that $\limsup_n A_n = B$.
$\forall A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$, $\limsup_n A_n$ is in the tail sigma algebra of $(\mathcal{F}_n)$.
Actually @Did has somehow explained it in one earlier question of mine:
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.
But I don't quite understand it well in the hindsight:
- for any subset $B$ in the tail sigma algebra and any integer $N$, how does one build $B$ from the sets $A_n \in \mathcal{F}_n$ with $n\ge N$?
- how does the above explain that $\limsup_n A_n$ is in the tail sigma algebra of $(\mathcal{F}_n)$?
Similar questions to the above two for relations:
between $\sigma(\cup_{n=1}^{\infty} \cap_{m=n}^\infty \mathcal{F}_m )$ and $\liminf_n A_n$ for $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$?
between $\cap_{n=1}^{\infty} \sigma(\cup_{m=n}^\infty \mathcal{F}_m )$ and $\liminf_n A_n$ for $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$?
between $\sigma(\cup_{n=1}^{\infty} \cap_{m=n}^\infty \mathcal{F}_m )$ and $\limsup_n A_n$ for $A_n \in \mathcal{F}_n, \forall n \in \mathbb{N}$?
Thanks and regards!