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Let $(\Omega, M, P)$ be a probability measure space and assume there is an infinite sequence $A_1, A_2, \ldots$ of elements of $M$ which are all independent from each other and such that $P(A_i) = \frac{1}{2}$ for all $i \in \mathbb{N}$. How does one prove the claim that $\Omega$ is uncountable?

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Assume that $\Omega$ is countable, and call $\omega_i$ its elements.

Then, we define $I_i= \{n| \omega _i \in A_n \}$. From Borel-Cantelli each $I_i$ is infinite.

Now, observe that $$\Omega = \bigcup _{ i \geq 1 } \bigcap_{n\in I_i }A_n $$

From independence it is straightforward that $$\mathbb{P}( \bigcap_{n\in I_i }A_n )=0$$ Now, $\Omega$ is a countable union of null sets therefore from subadditivity $$\mathbb{P}(\Omega)=0.$$ A contradiction.

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  • $\begingroup$ I believe the fact that $I_i$ is infinite only happens with probability $1$ by B-C. So maybe we should write $\Omega$ as a union of intersections of $A_n$, together with a union of $\omega_i$ for which $I_i$ is finite. Since the latter has probability $0$, I think your conclusion goes through anyway. $\endgroup$ – Woett Aug 20 '17 at 8:11
  • $\begingroup$ Or maybe it could be just $$\{ \omega_i \} \subseteq \bigcap_{n \in I_i} A_n \cap \bigcap_{n \notin I_i} A_n^c$$ so the intersection is always countable. Therefore $\mu( \{ \omega_i \} ) = 0$ and the contradiction follows. A very nice proof anyway! $\endgroup$ – Adayah Aug 20 '17 at 8:33
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Suppose $(\Omega, \mathcal{M}, \mu)$ is a probability measure space where $\Omega$ is countable.

Lemma 1. There is a decomposition $\displaystyle \Omega = \bigcup_{n < \alpha} \Omega_n$ into $\alpha$ disjoint measurable subsets $\Omega_n \subseteq \Omega$, where $\alpha \leqslant \omega$, such that $(\forall \varnothing \subsetneq A \subsetneq \Omega_n) \, A$ is not measurable.

Proof. Define an equivalence relation $\sim$ on $\Omega$ by

$$a \sim b \iff (\forall E \in \mathcal{M}) \, \big[ a \in E \Leftrightarrow b \in E \big].$$

Let $\{ \Omega_n : n < \alpha \}$ be the set of it's equivalence classes, which is at most countable, so $\alpha \leqslant \omega$. Let $a \in \Omega_n$ for some $n < \alpha$. For each $b \notin \Omega_n$ we find $E_b \in \mathcal{M}$ such that $a \in E_b$ but $b \notin E_b$. Then $\displaystyle \Omega_n = \bigcap_{b \in \Omega} E_b$ which is a countable intersection, so $\Omega_n$ are measurable.
Now suppose $\varnothing \subsetneq A \subsetneq \Omega_n$ and let $a \in A, b \in \Omega_n \setminus A$. If $A$ were measurable, by definition $a \not \sim b$ although $a, b$ belong to a common equivalence class $\Omega_n$, which is impossible. $\ \square$


Lemma 2. $(\mathcal{M}, \mu)$ is isomorphic to $(\mathcal{P}(\mathbb{N}), \nu)$ for some probability measure $\nu$ on $\mathcal{P}(\mathbb{N})$.

Proof. Let $\displaystyle \Omega = \bigcup_{n < \alpha} \Omega_n$ be a decomposition as in Lemma 1. Let $\varphi : \mathcal{P}(\alpha) \to \mathcal{M}$ be given as $\displaystyle \varphi(A) = \bigcup_{n \in A} \Omega_n$. One easily proves that this is an isomorphism of algebras.

Now if $\alpha$ were finite, so would be $\mathcal{M}$, but since $A_i$'s from the question are independent, they are infinitely many distinct sets in $\mathcal{M}$. So $\alpha = \omega$ and $\mathcal{P}(\alpha) = \mathcal{P}(\mathbb{N})$ and it suffices to transport $\mu$ to $\mathcal{P}(\mathbb{N})$ through $\varphi$. $\ \square$


By Lemma 2, without loss of generality we can assume that $\Omega = \mathbb{N}$ and each $A \subseteq \Omega$ is measurable. There is $N \in \mathbb{N}$ such that $\mu( \{ 1, 2, \ldots, N \} ) \geqslant \frac{5}{6}$. Since there are infinitely many $A_n$'s, there are $n < m$ such that $A_n \cap \{ 1, 2, \ldots, N \} = A_m \cap \{ 1, 2, \ldots, N \}$. But given that $\mu(A_n) = \mu(A_m) = \frac{1}{2}$, we have $\mu(A_n \cap \{ 1, 2, \ldots, N \}) = \mu(A_m \cap \{ 1, 2, \ldots, N \}) \geqslant \frac{1}{3}$, so $\mu(A_n \cap A_m) \geqslant \frac{1}{3} > \frac{1}{4}$ and they are not independent.

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  • $\begingroup$ Maybe someone could get confused why we can assume that $\{1,2, \cdots,N \}$ are measurable. $\endgroup$ – clark Aug 20 '17 at 2:15
  • $\begingroup$ @clark That was an essential flaw in my argument that I have now supplemented by adding two lemmas. Thank you! That made the proof a bit lengthy by the way and I don't like it anymore. :-( $\endgroup$ – Adayah Aug 20 '17 at 8:43
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Define $X:= \sum_{n=1}^{\infty} 2^{-n}\cdot 1_{A_n}$.

Using this question, we see that $X$ is uniformly distributed over $[0,1]$.

Now if $\Omega$ was countable, then $$\sum_{\omega \in \Omega} P(X=X(\omega)) = 1$$But since $X$ is uniformly distributed, it follows that $P(X=X(\omega))=0$ for all $\omega$, which contradicts the above expression.

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