Infinite sequence of random variables

Question

Suppose that we have an infinite sequence of i.i.d. binary random variables $(X_n)_n$ with $X_n\in\{0,1\}$ for all $n$ (with both options having probability 0.5). Now the outcome of all random variables together will be an element $x\in\{0,1\}^{\mathbb N}$. My intuition says that we can make certain statements about $x$. For example, define $S$ as the set of $x\in\{0,1\}^{\mathbb N}$ such that $\lim_{n\rightarrow\infty}\frac{x(1)+\dots+x(n)}{n}=\frac12$. Then the probability that $(X_n)_n$ will be in $S$ should be $1$. We can also make other statements about $x$.

Now I would like to contruct two things:

1. A $\sigma$-algebra $A\subset P(\{0,1\}^{\mathbb N})$. This $\sigma$-algebra should contain all sets $V\in A$ such that we know what the probability is that $(X_n)_n\in V$. For example, we should have $S\in A$.
2. A probability measure $f$ on $(\{0,1\}^{\mathbb N},A)$ which assigns to each $V\in A$ the probabillity $f(V)$ that $(X_n)_n\in V$. For example, we should have $f(S)=1$.

Is there a standard way in which this is done?

Answer 1

As Nicolas Agote mentioned, we can view a sequence $(x(1),x(2),\dots)$ as a real number in $[0,1]$ by using the correspondence $$f(x(1),x(2),\dots):= \sum_{n=1}^{\infty}x(n)2^{-n}.$$ Now when $(X_n)_n$ is i.i.d. randomly distributed with $X_n=0$ and $X_n=1$ having both probability $\frac12$, it follows that $f(x(1),x(2),\dots)$ is uniformly distributed in $[0,1]$. For the uniform distribution on $[0,1]$, we have a nice well-known probability measure: the Lebesgue measure $\lambda$. The associated $\sigma$-algebra $L$ does not seem to have a name. Now we can pull back this $\sigma$-algebra: we can define $A:=f^{-1}(L)$. For any $V\in A$, the probability that $(X_n)_n\in V$ equals $\lambda(f(V))$.

It remains to prove that $\lambda(f(S))=1$ for the set $S$ defined in the question. The proof will appear at this question.

However, interpreting this can be tricky. For every binary sequence $a_1,\dots$, the Lebesgue measure of the associated set with one element is zero. Therefore, the probability that $(X_n)_n=(a_n)_n$ is 0. But this does not necessarily mean that $(a_n)_n$ can not possibly the result. After all, there is some sequence that will be the outcome of the infinite series.

A solution of this is:

For all Turing machines that try to predict outcomes of $X_n$, the total asymptotic success rate over the entire infinite time will be exactly 0.5.

The corresponding subset of $[0,1]$ of elements that satisfy the above constraint, can be proven to have Lebesgue measure 1. Those are exactly the possible outcomes of the infinite series.