# Question about the $\sigma$-algebra for infinite coin toss

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!

• $\sigma(\cal F_\infty)$ is the countable product of $\sigma$-algebras $2^{\{0,1\}}$. It is contained in $2^{\Omega}$, but it is not equal to $2^{\Omega}$. – Ramiro Sep 30 '15 at 13:57
• One way to see that $\sigma(\cal F_\infty) \neq 2^\Omega$ is a cardinality argument. Note that $\Omega =\{0,1\}^{\mathbb{N}}$, so the cardinality of $\Omega$ is $\aleph_1$ (cardinality of the continum). So $2^\Omega$ has cardinality $\aleph_2$ (cardinality of the set of parts of the continum). On the other hand, $\cal F_\infty$ is countable, so the cardinality of $\sigma(\cal F_\infty)$ is at most $\aleph_1$. (see Halmos $\S 5$ exercise 9). – Ramiro Sep 30 '15 at 23:26

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).