# Prove that there exists no sequence of independent events $A_1, A_2, \ldots$ such that $\mathbb{P}(A_i) = \frac{1}{2}$ for all \$i [duplicate]

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, $$\mathbb{P}(A^*) = \mathbb{P}(\limsup_{n \to \infty} A_n) = \frac{1}{2} \neq 1. Contradiction.$$

• Have you attempted something so far ? Do you have any idea of where to start ? – Gâteau-Gallois Feb 19 '19 at 19:05
• 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? – Stan Tendijck Feb 19 '19 at 19:11
• @Gâteau-Gallois i did – Jasper Feb 20 '19 at 16:52