Let $\mathcal{F}\_i^j$ be a collection of sigma-algebras for 
$i \in \mathbb{N}$ and 
$j \in \{1,\ldots,n\}$ such that $\mathcal{F}\_{i+1}^j \subseteq \mathcal{F}\_i^j$ for all $i,j$. I would like to show that


$$\sigma\left(\cup\_j\cap\_i\mathcal{F}\_i^j\right) = \cap\_i\sigma\left(\cup\_j\mathcal{F}\_i^j\right)$$


but I'm not sure how, or whether it is true.


Thanks!



Finite union of tail sigma-algebras
$\newcommand{\cF}{\mathcal{F}}$
What you want to show is false. Let $x^j\_i$ and be a set of random variables, each taking values in $\{0,1\}$, where $j=1,2$ and $i=1,2,\ldots$. Let $\cF^j\_i=\sigma(x^j\_i,x^j\_{i+1},\ldots)$.


Consider the event $A=$ "$x^1\_i=x^2\_i$ for all but finitely many $i$".


This event clearly belongs to the right hand side. To see that it does not belong to the left hand side, define the following probability distribution: let the $x^1\_i$ be i.i.d uniformly in $\{0,1\}$. Then either let $x^2\_i=x^1\_i$ for all $i$ or let $x^2\_i\ne x^1\_i$ for all $i$, with probability $\frac12$ for either option.


Now, we see that the probability of $A$ is $\frac12$. However, the probability of any event in the lhs must be either 0 or 1, due to Kolmogorov's 0-1 law. Hence, $A$ does not belong to the lhs.


This question is related to a [question](https://mathoverflow.net/questions/8166/measurable-sets-not-depending-on-even-coordinates) of mine. I guess it is high time that I post the answer to that question too.


