Question:
Consider a probability space (Ω, F, P), and let A be an event(element of F).
Let g be the set of all events(the events denote by $g_{i}$ ) that are independent from A.
How to show that g need not be a σ-field?
I know that a σ-field f has following properties:
∅∈f
If A∈f then $A^{C}$∈f
If $A_{i}$∈f for every i∈$\mathbb{N}$,then $\bigcup_{i=1}^{\infty }$$A_{i}$∈f
And I know that for a field $f_{0}$, the third property is
- If A∈$f_{0}$ and B∈$f_{0}$,then A$\bigcup$B∈$f_{0}$.
For $g_{i}$, the properties 1. and 2. are easy to varify. But I don't know how to answer the question,show g need not be a σ-field(And sholud I use the property 3.?).
Hint: It is true that if $\{A_i\}_{i\in \Bbb N}$ is a sequence of disjoint evets all of which are independent from $E$, then $\bigcup_{i\in\Bbb N}A_i$ is independent from $E$. Let's then call $(3^-)$ the property of a family of subsets of being closed under union of countable sequences of disjoint elements.
Let's also call $(3^{--})$ the property of a family of subsets of being closed under finite intersections; equivalently, of being closed under intersections of pairs of elements. Some would say that the pair version is slightly weaker, because they adopt the convention that $\bigcap\emptyset$ is the whole space; however $(1)\land (2)\land (3^{--})$ is the same condition no matter which choice of $(3^{--})$ you make.
Finally, notice that a family which satisfies $(1)\land (2)\land (3^-)\land (3^{--})$ is a $\sigma$-algebra, via the trick $\bigcup_{i\in \Bbb N}A_i=\bigcup_{i\in \Bbb N}\left(A_i\cap \bigcap_{j=0}^{i-1}A_j^C\right)$.
Therefore your only hope is to find three events $A$, $B$, $C$ in some probability space such that $\{A,B\}$ are independent, $\{B,C\}$ are independent and $\{A\cap C,B\}$ are not independent.