# Existence of independent events

Let $$(\Omega, \mathcal {F},\mathbb {P})$$ be a probability space such that $$\Omega$$ is countable, and $$\mathcal {F}=2^{\Omega}$$. I want to show that it is impossible to exist a countable collection of events $$A_{1},A_{2},\cdots\in \mathcal {F}$$ which are independent, such that $$\mathbb {P}(A_{i})=\frac {1}{2}$$ for each $$i$$. I think showing $$\mathbb {P}(\omega)\leq \frac {1}{2^n}$$ for $$\omega\in\Omega$$ and $$n\in \mathbb {N}$$ might help?

It suffices to show, that the conditions imply that $$P(\{\omega\}) = 0$$ for all $$\omega \in \Omega$$, and thus that $$P(\Omega)=0$$, contradicting the fact that $$P$$ should be a probability measure. So from now on consider a fixed $$\omega \in \Omega$$. Define sets $$B_1,B_2,\dots$$ in the following way $$B_i := \begin{cases} A_i &, \text{ if \omega \in A_i} \\ \Omega \setminus A_i &, \text{ otherwise} \end{cases}.$$ Now given that $$A_1,A_2,\dots$$ are independent with $$P(A_i)=\frac12$$ it follows that $$B_1,B_2,\dots$$ are independent with $$P(B_i)= \frac12$$. Now by construction we know that $$\omega \in \bigcap_{i=1}^\infty B_i$$ and thus we get that $$P(\{\omega \})\leq P(\bigcap_{i=1}^\infty B_i) =\lim_{n \rightarrow \infty} \frac{1}{2^n} = 0.$$

Hints:

Let $$A_\infty$$ be the set of $$\omega$$ which are contained in infinitely many $$A_i$$.

First, show that each $$\omega\in A_\infty$$ satisfies $$\Bbb P(\{\omega\})=0$$.

Then, use a version of the Borel-Cantelli lemma (one where independence or pairwise independence plays an important role), to show that $$\Bbb P(A)=1$$ holds.

Since $$\Omega$$ is countable, this will result in a contradiction.