Rigorous Understanding of Law of Large Numbers for Coin Toss The law of large numbers states:

If $(\Omega,\mathcal{F},P)$ is a probability space, and $X_i:\Omega\rightarrow\mathbb{R}$ with $i=1,2,3,...$ form an infinite sequence of independent, identically distributed random variables with common, finite mean $\mu$, then the sequence $\{\bar{X}_n\}$ with $\bar{X}_n:=(1/n)\sum_{i=1}^n X_i$ converges (in some precise sense) to $\mu$.

This law is commonly said to explain the phenomenon that if you flip a fair coin and record the proportion of heads, flipping many times results in recorded proportions nearing the expected number of 50%.  
What I would like to understand is what precisely is the connection? Specifically, in this application of the law of large numbers to flipping a coin:
(1) What is $\Omega$?
(2) If not obvious from the answer to (1), what is $P$?
(3) What is the sequence $X_1,X_2,X_3,...$ of real-valued functions on $\Omega$ that are independent and identically distributed of mean $0.5$? 
To be clear, I am interested in a precise mathematical answer, so please do not answer "$X_i$ is the ith coin flip." I want to know what it is as a function on a specified domain. Thanks!
 A: Actually there's a cheap trick you can use to construct iid sequences without worrying about the technicalities of infinite product spaces. I think infinite product spaces give the "right" construction, because it's intuitively clear why they should work, but anyway, if you don't want to worry about what's a measurable set, etc, do this:
Let $\Omega=[0,1]$, with $P$ equal to Lebesgue measure. For $n\ge 1$ let $$A_n=\bigcup{k=0}^{2^{n-1}-1}[(2k+1)/2^n,2(k+1)/2^n].$$Now if $$B_n=\Bbb 1_{A_n}$$then the $B_n$ are in fact iid random variables, uniformly distributed on $\{0,1\}$.
So there's a mathematically precise sample space for a sequence of coin tosses; the strong LLN says just that $$\frac{B_1(x)+\dots+B_n(x)}n\to\frac12$$for almost every $x\in[0,1]$.
But there's more. Define $$U(x)=\sum_{n=1}^\infty B_n(x)2^{-n}.$$Then $U$ is uniformly distributed on $[0,1]$, because in fact $U(x)=x$ (except maybe for countably many $x$).
Now say $S_1,S_2,\dots$ is an infinite sequence of pairwise disjoint infinite subsets of $\Bbb N$. Say $$S_k=\{n_{k,1},n_{k,2},\dots\}.$$Define $$U_k=\sum_{j=1}^\infty B_{n_{k,j}}2^{-j}$$
Then $(U_k)$ is an iid sequence of random variables uniformly distributed on $[0,1]$.
And now you can get an iid sequence $(X_k)$ with any distribution you want by setting $X_k=f(U_k)$ for a suitable function $f$.


Theorem  $[0,1]$ is the only sample space you ever need.


Not that throwing out all the other sample spaces is a good idea, but it seems interesting.
