Let $\{a_i\}$ be a countable sequence of binary values representing the results of repeated independent fair coin tosses with $a_i= 0$ indicating tail and $a_i = 1$ indicating head.
Let $\Omega$ be the sample space consisting of all possible $\{a_i\}$. The $\sigma$-algebra is constructed in the following way.
Let $\cal F_n, n\in\Bbb N$ be the set of sequences that can be determined by $a_1,a_2,…,a_n$ union $\{\emptyset, \Omega\}$.
For example, let $A_0$ denote the set of sequences with $a_1=0$, $A_1$ denote the set of sequences with $a_1=1$, $A_{00}$ denote the set of sequences with $a_1=0,a_2=0$, $A_{10}$ denote the set of sequences with $a_1=1, a_2=0$, etc. Then $\cal F_1 = \{\emptyset, \Omega, A_0, A_1\}$, $\cal F_2 = \{\emptyset, \Omega, A_0, A_1,A_{00},A_{01},A_{10},A_{11}\}$, and so on.
It is not hard to verify that $\cal F_n$ is increasing, i.e. $\cal F_n \subseteq \cal F_{n+1}$. Define $\cal F_\infty = \bigcup_{i=1}^\infty \cal F_i$, and we state a result that $\cal F_\infty$ is not a $\sigma$-algebra. However, the smallest $\sigma$-algebra generated by $\cal F_\infty$, denoted as $\sigma(\cal F_\infty)$ is what we need, and we can define a sensible probability measure on the measurable space $（\Omega, \sigma(\cal F_\infty))$.
My question is: is $\sigma(\cal F_\infty)=2^\Omega$? Or in other words why the above construction is necessary?
Thank you!
Best Answer
The answer is negative. In fact, through the map $$ (a_1,a_2,\dots) \mapsto \sum_{n=1}^\infty 2^{-n} a_n, $$ your coin tossing probability space is isomorphic (ignoring some set of probability zero) to $([0,1],\mathcal B[0,1],\lambda)$, in particular, there are some non-measurable sets (see also a relevant discussion here).