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Let $(\Omega, \mathscr{F}, \mathbb{P})$ be any probability triple where $\Omega$ is countable. Prove that it is impossible that there exists a sequence $A_1, A_2, \ldots \in \mathscr{F}$ of events which are independent and such that $\mathbb{P}(A_i) = \frac{1}{2}$ for all $i \in \mathbb{R}$.

We have a theorem:
Theorem 2.4. let $A_1, A_2, \ldots \in F$ for some $\sigma$-algebra $F$.
If $\sum_{n=1}^{\infty} \mathbb{P}(A_n) = \infty$ and $ \{A_n; n \geq 1 \}$ are independent then $\mathbb{P}(A^*) = 1$, with $A^* = \limsup A_n$

I tried the following but I don't know if its correct.
We have that $\sum_{n=1}^{\infty} \mathbb{P}(A_n) = \sum_{n=1}^{\infty} \frac{1}{2} = \infty$, and also the $A_n$ are independent. However, $\mathbb{P}(A^*) = \mathbb{P}(\limsup_{n \to \infty} A_n) = \frac{1}{2} \neq 1. Contradiction.$

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  • $\begingroup$ Have you attempted something so far ? Do you have any idea of where to start ? $\endgroup$ – Gâteau-Gallois Feb 19 '19 at 19:05
  • $\begingroup$ As a first attempt, try to work out what happens if such a sequence exist. If such a sequence exist, what can you do with it under the assumptions? $\endgroup$ – Stan Tendijck Feb 19 '19 at 19:11
  • $\begingroup$ @Gâteau-Gallois i did $\endgroup$ – Jasper Feb 20 '19 at 16:52

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