TITLE: Show that if the sum of an diverges, no discrete probability space can contain independent events
QUESTION [2 upvotes]: Suppose that $0\leq p_n\leq 1$, and put $a_n= \min \{p_n, 1-p_n\}$. Show that if 
$\sum a_n$ diverges, then no discrete probability space can contain independent events $A_1, A_2, \ldots$ such that $A_n$ has probability $p_n$.

REPLY [0 votes]: Let's do not consider the $P_n=0$ cases, because if the empty set is independent with any sets.
So , consider the discrete probability space, contains at most countable many points with each point assigned with a positive probability.
All the probabilities sum up to 1.
Now $\sum a_n$ diverge,
suppose $A_n$ are the events,
they are independent, therefore $P(\bigcap_n A_n)=\prod_n a_n=0$, here because $p_n$ are at most $\frac{1}{2}$,
then $\bigcap_n A_n$ is simply empty in our discrete setting.
(*) indeed, if we choose any infinite $A_n$ their intersection will be empty.
Use Borel-Cantelli lemma, $\sum a_n$ diverge then $\limsup_{n\rightarrow\infty} A_n$ will have measure $1$, in our discrete setting this means $\limsup_{n\rightarrow\infty} A_n$ is the whole set.
This would imply that any point in our set will be contained in infinitely many $A_n's$
Contradict with (*)
The whole point is that in discrete probability space, measure 0 set is empty and measure 1 set is the whole set. (of course, the points with 0 probability we can simply delete them without changing anything).
Hope it will help. :D
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