I am currently reading through Steven Shreve's Stochastic Calculus for Finance II; specifically the creation of $\sigma$-algebras and the application of probability measures of their elements. I am asking a question from a provided example which can be found on pages 6-9 (inclusive) of the book.
First, let $\Omega_{\infty}$ be defined as the set of all outcomes from a coin being tossed infinitely many times. That is
$$\Omega_{\infty}:= \{\omega:\omega=w_{1}w_{2}\cdots\},$$
where $w_{i}$ is either heads or tails. Let $\mathcal{F}_{n}$ represent a $\sigma$-algebra of subsets of $\Omega_{\infty}$ such that every $A \in \mathcal{F}_{n}$ represents a permutation of heads and tails of the first $n$ tosses. As an example, one such $A$ could be all $\omega$ where the first $n$ tosses (of an infinite amount) are heads. Another element in $\mathcal{F}_{n}$ could be the set of all $w$ where the first $n$ tosses alternate between heads and tail with $w_{1}$ being heads.
$\mathcal{F}_{\infty}$ is then defined to be (as per Shreve's book on page 6):
We create a $\sigma$-algebra, called $\mathcal{F}_{\infty}$ by putting in every set that can be described in terms of finitely many coin tosses and then adding all other sets required in order to have a $\sigma$-algebra.
I believe this will eventually generalise the idea behind $\mathcal{F}_{n}$ through the unions and intersections of sets in the $\sigma$-algebra.
An example of a set in $\mathcal{F}_\infty$ is then studied to demonstrate a tricky/interesting construction. That is, let $A$ be a set of elements in $\Omega_\infty$ defined as follows:
$$ A:= \left\{\omega: \lim_{n \to \infty} \frac{H_{n}(\omega)}{n} = \frac{1}{2}\right\},$$
where $H_{n}(\omega)$ is the number of heads in the first $n$ tosses of an infinite number of tosses. The definition for $A$ basically says $\omega \in A$ if the average number of heads, as the number of first tosses considered goes to infinity, will be $\frac{1}{2}$.
Next, it is shown that this set $A$ is in $\mathcal{F}_{\infty}$. Since only a finite number of tosses are considered to define $\mathcal{F}_{\infty}$, the proof is performed by defining the following set $A_{n,m}$ as
$$A_{n,m}:= \left\{\omega: \left\vert \frac{H_{n}(\omega)}{n} - \frac{1}{2} \right\vert \leq \frac{1}{m} \right\}$$
Since $A_{n,m}$ is defined by a finite number of $n$ (first) tosses it is in $\mathcal{F}_{\infty}$. By applying the definition of the limit that for all $m > 0$ there exists an $N$ such that for all $n \geq N$, $\omega \in A_{n,m}$, it is shown that
$$ A = \bigcap_{m=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty}A_{n,m}.$$
Since $A$ is constructed by the unions and intersections of $A_{n,m} \in \mathcal{F}_{\infty}$, it is concluded that $A \in \mathcal{F}_{\infty}$.
Finally to my question. I want to understand how the probability measure is applied to $A$. The probability measure on sets in $\mathcal{F}_{n}$ is defined as:
$$\mathbb{P}(A) = p^{n_{H}}q^{n-n_{H}},$$
where $n_{H}$ is the number of heads in the first $n$ tosses in $\omega \in A$.
So back to the complicated $A \in \mathcal{F}_{\infty}$, Shreve on page 7 says:
...the Strong Law of Large Numbers asserts that $\mathbb{P}(A) = 1$ if $p = \frac{1}{2}$ and $\mathbb{P}(A) = 0$ if $p \neq \frac{1}{2}$
Can someone please explain how the Strong Law of Large Numbers is applied to determine the probability measure of $A$?
Thanks
