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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?

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2 Answers 2

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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 $$P(\{\omega \})\leq P(\bigcap_{i=1}^\infty B_i) =\lim_{n \rightarrow \infty} \frac{1}{2^n} = 0.$$

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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.

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    $\begingroup$ Nice answer :-) (+1) $\endgroup$
    – Surb
    Oct 26, 2020 at 10:00

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