In Probability with Martingales (Williams) I came across the following proposition enter image description here

and then they give the following contradictory example enter image description here

Could someone please explain how it can be so? Also why is that in the truth set they use $\rightarrow \frac{1}{2}$ and not $= \frac{1}{2}$ when comparing each outcome?

Thank you.

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    $\begingroup$ The proposition applies to countable collections - what is the cardinality of $\mathcal{A}$? $\endgroup$ – Anthony Carapetis Aug 22 '13 at 3:30
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    $\begingroup$ @MLT $mathcal A$ is not countable! The proposition is stated for sequences. $\endgroup$ – azarel Aug 22 '13 at 3:30
  • $\begingroup$ @AnthonyCarapetis Thank you. How do I find the cardinality of $\mathcal{A}$ $\endgroup$ – triomphe Aug 22 '13 at 3:39
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    $\begingroup$ @MLT: Hint: increasing sequences of integers are in correspondence with the infinite subsets of $\mathbb{N}$. $\endgroup$ – Anthony Carapetis Aug 22 '13 at 3:54
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    $\begingroup$ @MLT: The power set of $\mathbb{N}$ is uncountable... $\endgroup$ – Anthony Carapetis Aug 22 '13 at 4:22

This statement says that for every $\omega$ there exists some $\alpha$ in $\mathcal A$ such that $\omega\notin F_\alpha$. To see that this holds, consider $N_\omega=\{n\in\mathbb N\mid\omega_n=H\}$.

  • If $N_\omega$ is infinite, order it as $N_\omega=\{\nu_\omega(k)\mid k\in\mathbb N\}$ and consider $\alpha$ such that $\alpha(k)=\nu_\omega(k)$ for every $k$. Then $\omega_{\alpha(k)}=H$ for every $k$ hence $\frac1n\#\{k\leqslant n\mid\omega_{\alpha(k)}=H\}=1$ for every $n$, in particular $\omega\notin F_\alpha$.

  • If $N_\omega$ is finite, consider $\alpha$ such that $\alpha(k)=k$ for every $k$. Then $\omega_{\alpha(k)}=H$ for finitely many $k$ hence $\frac1n\#\{k\leqslant n\mid\omega_{\alpha(k)}=H\}\to0$ when $n\to\infty$, in particular $\omega\notin F_\alpha$.

One sees that $\alpha$ such that $\omega\notin F_\alpha$ depends on $\omega$.

And this does not contradict the proposition in the book because $\mathcal A$ is uncountable. To see this, note that $\mathcal A$ is in bijection with $\mathbb N^\mathbb N$ through the application $\alpha\mapsto\beta$ defined by $\beta(1)=\alpha(1)$ and $\beta(n+1)=\alpha(n+1)-\alpha(n)$ for every $n$ in $\mathbb N$. Since $\mathbb N^\mathbb N$ is uncountable, this proves the claim

  • $\begingroup$ Thank you. I have one question. Why do you consider the $N_\omega$ finite case. I was thinking the space $\Omega$ is $\{H,T\}^{\infty}$ (I mean infinite tosses) and therefore there is a sequence with $\omega_n=H$ (and $\omega_n=T$) for each $n \in \mathbb{N}$ $\endgroup$ – triomphe Aug 22 '13 at 14:42
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    $\begingroup$ Consider the sequence $\omega=TTTTTTTTT\ldots$ defined by $\omega_n=T$ for every $n$. Then $N_\omega=\varnothing$ is finite. $\endgroup$ – Did Aug 22 '13 at 15:02

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