Noob Question : Need help to understand : Probability with Martingales : page 25 I was asked to post the question from here
I am trying to learn measure theoretic probability.
The book I am trying to learn it from is Probability With Martingales
and, I am really not understanding much. Almost every page am missing something.
So I thought that I would seek help.
Here is the current blocker.
At page 25 the following exercise is given:-
Let $\alpha \in A$ be a st of functions $\alpha : \mathbb{Z} \to \mathbb{Z} $ 
 such that $\alpha(1) < \alpha(2) < \alpha(3) <\cdots $
Let 
$$
F_\alpha = \left \{   \omega :  \frac{\#( k \le n : \omega_{\alpha(k)} = H) }{n} \to \frac{1}{2}              \right \}
$$ 
For a fair coin tossing with 
$$
\omega = ( \omega_1 , \omega_2 , \omega_3 , ... ) \; \; \omega_n \in \{ H , T \}  
$$
It says it is "The truth set of strong law for the subsequence $\alpha$".
Now I do not get it what it means.
Also I don't get it why 
$$
P(F_{\alpha}) = 1 \; \forall \alpha \in A 
$$ 
And then comes the shocker:-
Asked to show that:-
$$
\bigcap F_\alpha = \varnothing
$$
How?
I just don't get it at all.
Please be easy with me, I am no mathematician - not even a student - just an enthusiast.
 A: The fraction $$\frac{\#( k \le n : \omega_{\alpha(k)} = H) }{n}$$ represents the average number of heads in the first $n$ terms of the sequence $$\omega_{\alpha(1)}, \omega_{\alpha(2)}, \omega_{\alpha(3)},\ldots,$$
which is a subsequence of $$\omega_1,\omega_2,\ldots$$
Using such a strictly increasing function $\alpha:\mathbb{N}\to\mathbb{N}$ is only done here to define the notion of a subsequence rigorously. Now the subsequence chosen depends on $\omega=(\omega_1,\omega_2,\ldots)$ and for some such $\omega$ the average number of heads in the first $n$ terms goes to $1/2$.  
Now if you fix which terms you select for your subsequence, which amounts to fixing $\alpha$, the classical strong law of large numbers tells you that $F_\alpha$ has probability $1$, the terms of the subsequence are still independent fair coin flips. 
Now if for some sequence $\omega$, the average number of heads in the first $n$ terms goes to $1/2$, there must be infinitely many heads occuring. So for each $\omega$, there is a subsequence that consists only of heads and there the average number of heads in the first $n$ terms goes to $1$ trivially. If there are not infinitely many heads, we can take the original sequence to be a subsequence of itself (by letting $a(n)=n$ for all $n$), and for this subsequence, the average number of heads in the first $n$ terms goes to $0$. Either way, for each $\omega$, there is $\alpha$ such that $$\omega_{\alpha(1)}, \omega_{\alpha(2)}, \omega_{\alpha(3)},\ldots,$$ 
does not satisfy that the average number of heads in the first $n$ terms going to $1/2$. Hence, for this $\alpha$, we have $\omega\notin F_\alpha$. So no $\omega$ is in all $F_\alpha$, and the set of $\omega$'s which is in all $F_\alpha$ is exactly $$\bigcap_{\alpha\in\mathcal{A}}F_\alpha.$$
That such an intersection of probability $1$ events is empty is of course only possible since there are uncountably many $\alpha$.
