# About probability and countable $\Omega$

I have the following problem:

Let $\Omega$ be a finite or countable set, $\mathfrak{F}$ a $\sigma$-algebra and $\mathbb{P}$ a probability measure. Prove there doesn't exist a collection $(A_i)_{i\in\mathbb{N}}$ of independents events in $(\Omega, \mathfrak{F},\mathbb{P})$, with $\mathbb{P}(A_i)=1/2$ for all $i\in \mathbb{N}.$

First I suppose that the collection exists and then I conclude $$\mathbb{P}\left(\bigcap_{n\geq1}A_i\right)=0$$ and

$$\mathbb{P}\left(\bigcup_{n\geq1}A_i\right)=1,$$ but then I don't know how to continue. Also I don't know if this is useful.

Can someone give me an idea or help?

Try proving it first under the assumption that $\mathcal{F} = 2^\Omega$. Use countable additivity to show that there exists a singleton $\omega$ with $P(\{\omega\}) > 0$. Also note that for any $n$, you can use $A_1, A_2, \dots, A_n$ to partition $\Omega$ into $2^n$ sets, each with probability $1/2^n$. Take $n$ sufficiently large to get a contradiction.
More generally, you can prove that a countable $\Omega$ must contain an atom: a measurable set $B \in \mathcal{F}$ with $P(B) > 0$ and such that $B$ contains no proper nonempty measurable subset. Then you can get a contradiction just as before.
As yet another approach, set $X = \sum_{n=1}^\infty \frac{1}{2^n} 1_{A_n}$. Prove that for any real number $x$ we have $P(X=x) = 0$. But since $\Omega$ is countable, the range of $X$ is a countable set $\{x_1, x_2, \dots\}$. Then $\Omega = \bigcup_n \{X = x_n\}$, meaning $P(\Omega) \le \sum_n P(X = x_n) = 0$ which is absurd.