1. Oct 20, 2009

### dottidot

I have been trying to figure out the proof to this problem for the past couple of days and still don't have an answer. The question is as follows:

Let (Q,F,P) be a probability triple such that Q is countable. Prove that it is impossible for there to exist a sequence A1,A2,A3,... E F which is independent, such that P(Ai) = 1/2 for eash i. [Hint: First prove that for each w E Q, and each n E N, we have P({w}) <= 1/(2^n). Then derive a contradiction.

Note that Q, represents Omega, F the sigma-algebra/sigma-fiels and E is "an element of or member of"

Would appreciate any help with this.

2. Oct 21, 2009

### g_edgar

Add this to the hint. If $Q$ is countable, then there is $\omega \in Q$ with $P(\{\omega\}) > 0$. So, for large $n$ we have $P(\{\omega\}) > 1/2^n$.