Tail sigma algebra and $\limsup$ of a sequence of subsets 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!
 A: (2).  $\limsup_n A_n$ is in the tail sigma-algebra.  
It suffices to show: for any $p$, we have $\limsup_n A_n \in \sigma(\bigcup_{m=p}^\infty \mathcal F_m)$.  And, as Didier said, this is because we can write $\limsup_n A_n$ as $\limsup_n A_{n+p}$.  Indeed, each $A_{n+p}$ is in $\sigma(\bigcup_{m=p}^\infty \mathcal F_m)$, so their countable combinations (such as limsup) are also in $\sigma(\bigcup_{m=p}^\infty \mathcal F_m)$.
(1).  is not right.  We could take $\mathcal F_n$ finite sigma-algebras (consisting of finite unions of intervals) such that $\bigcup_{n=1}^\infty \mathcal F_n$ contains all intervals with rational endpoints, so $\sigma(\bigcup_{n=p}^\infty \mathcal F_n)$ is the Borel sigma algebra $\mathcal B$, so $\bigcap_p \sigma(\bigcup_{n=p}^\infty \mathcal F_n)) = \mathcal B$.  But there are elements in $\mathcal B$ that are more complicated than limsups of finite unions of intervals.
But in a certain sense (1) is almost right.  If there is a finite measure here, each element of the tail agrees with such a limsup up to a set of measure zero.
