Trying to solve exercise 1.1.18 in D.W. Stroock, Probability Theory, I somehow don't see how to get the hint in that exercise.

Given a set $\Omega$, a tail $\sigma$-algebra $\tau$ generated by $\sigma$-algebras $\cal F_n$, where each $\cal F_n$ is again generated by a set $A_n$, that is ${\cal F}_n = \{\emptyset, \Omega, A_n, \Omega \setminus A_n\}$. The sets $A_n$ are independent.

An atom $C\in \tau$ is a non-empty set which has no non-empty subset in $\tau$: If $\emptyset \neq B \subset C$ and $B\in \tau$, then $B=C$.

Now, if $C$ is an atom in $\tau$ then it can be written as a $\liminf$, more precisely: $C$ is an atom only if one can write $$ C = \liminf_n C_n = \bigcup_n \bigcap_{k\geq n} C_k \quad \mbox{ where } C_n \in \{A_n, \Omega\setminus A_n\} $$

Is that implication simple? I simply don't see it.

Thanks for any help.


You do this in two steps:

  1. The atoms of the $\sigma$-algebra generated by all $A_k$ are of the form $\bigcap_k C_k$. You can verify that nonempty sets of this form are exactly the equivalence classes of points that cannot be separated by some $A_k$. One verifies that this property is preserved under complementation and countable unions.

  2. Now you want to construct the atoms of the tail sets. For this, you want to partition the points into sets that cannot be separated by tail sets. By the reasoning from 1., we know that the atoms of the $\sigma$-algebra generated by the $A_k$ with $k\geq n$ are of the form $\bigcap_{k\geq n} C_k$. So where does the union come in? If there is some $n$ and points $x,y$ that cannot be separated by the $A_k$ with $k\geq n$ for some $n$, they have to be in the same atom. So we have to pool the atoms from all tails together, and this is accomplished by the union.

This is of course only a sketch. A great reference for this kind of questions is the booklet Borel Spaces (1981) by Rao and Rao.

  • $\begingroup$ Thanks! A sketch is perfectly fine for me. That atoms should come from $\bigcap_k C_k$ came vaguely to my mind but I wasn't aware of the characterisation as equivalence classes. Thanks again. $\endgroup$ – Eckhard Giere Sep 6 '12 at 11:04

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