identify the interval $[0, 1]$ with the Lebesgue measure to the probability space for tossing a fair coin I want to identify the interval $[0, 1]$ with the Lebesgue measure to the probability space for infinite tossing a fair coin. 
I know we can define a probability on cylinder sets (the sets can be represented by the outcomes of finite tosses) and then use Caratheodory extension theorem to extend the probability to the $\sigma$-algebra $\mathcal{F}$ generated by all the cylinder sets.
But I am not sure how to identify this probability measure with the Lebesuge measure on $[0,1]$. In particular, I don't know what $\mathcal{F}$ looks like and why it can be identified as $\sigma$-algebra where Lebesgue measure is defined. I just have a brief thought about identify $\omega\in[0,1]$ with a binary representation.
Could you explain how to identify these two measure space ? It will also be helpful if you can provide a reference about this topic.
 A: Say $S=\{0,1\}^{\Bbb N}$ is the space of sequences $s=(s_1,\dots)$ of zeroes and ones. Define a map $f:S\to[0,1]$ by $$f(s)=\sum_{n=1}^\infty s_n2^{-n}.$$Then $f$ is "almost a bijection"; in fact it's a bijection except for countably many $x\in[0,1]$ that have two inverse images.
And $f$ is measure-preserving. For example, if $C$ is the cylinder in $S$ defined by $s_1=0$, $s_2=1$ then $f(C)=[1/4,1/2]$; both $C$ and $f(C)$ have measure $1/4$.

Or to do the same thing from the other direction: Regard $[0,1]$ with Lebesgue measure as a probability space. It's enough to show that this space supports a sequence $X_1,X_2,\dots$ of iid random variables with $P(X_j=0)=P(X_j=1)=1/2$. The Rademacher functions are such a sequence (except that they take the values $1,-1$ instead of $0,1$).

Ok, a detailed proof that $f$ is a bijection except for countably many points. Say $f(s)=f(t)$ but $s\ne t$. Choose $n$ so $s_j=t_j$ for all $j<n$ but $s_n\ne t_n$. WLOG $s_n=1$, $t_n=0$. So we have $$2^{-n}+\sum_{j=n+1}^\infty s_j2^{-j}=\sum_{j=n+1}^\infty t_j2^{-j}.$$
So $$\sum_{j=n+1}^\infty t_j2^{-j}\ge 2^{-n}.$$But$$\sum_{j=n+1}^\infty 2^{-j}=2^{-n}.$$So we must have $$t_j=1\quad(j\ge n+1);$$
now this shows that $$\sum_{j=n+1}^\infty s_j2^{-j}=0,$$so $$s_j=0\quad(j\ge n+1).$$
So both sequences $s$ and $t$ are eventually constant; there are only countably many such sequences. (The above also shows that there cannot be a third sequence with the same sum...)
