Let $\Omega = \{ \omega = (\omega_1, \omega_2, \ldots) : \omega_j = 1 \text{ or } 0 \}$. For each positive integer $n$, let $\Omega_n = \{ \omega = (\omega_1, \ldots, \omega_n) : \omega_j = 1 \text{ or } 0 \}$. We can consider $\Omega_n$ as a probability space with $\sigma$-algebra $ 2^{\Omega_n}$ and probability induced by $\mathbb{P}_n(\omega) = 2^{-n}$.
We define $F_n$ to be the collection of all subsets $A$ of $\Omega$ such that there is an
$E \in 2^{\Omega_n}$ with \begin{equation}
A = \{(\omega_1, \omega_2, \ldots) : (\omega_1, \ldots, \omega_n) \in E\}. \tag 1
\end{equation}
$F_n$ is a finite $\sigma$-algebra (containing $2^{2^n}$ subsets) and $F_1 \subset F_2 \subset F_3 \subset \cdots$ (i.e., an ascending sequence of $\sigma$-algebras). If $A$ is of the form $(1),$ we let $\mathbb{P}(A) = \mathbb{P}_n(E_n)$. This gives a function $\mathbb{P}$ on on \begin{equation}
F^{0} = \bigcup_{j=1}^{\infty} F_j
\end{equation}
This is followed by the proposition that $F^{0}$ is an algebra but not a $\sigma$-algebra.
All of the above and most of the proof makes sense to me. But in the proof for the proposition I just mentioned states: $ \Omega \in F_0$ since $\Omega \in F^1$. I get a bit confused because $F_1$ should just consist of the first flip being tails and the first flip being heads. Does this mean that $F_1$ consists of each as the first element in two infinite sequences where such as $(1, \omega_2, \ldots)$ and $(0, \omega_2, \ldots)$, where $\omega_j, j > 1$ are all just not given? But then doesn't $F_1$ trivially contain all possible sequences? Maybe it's because I'm a bit rusty on measure theory or I'm missing something but I'm a bit confused.
Confusion about probability space associated with infinite coin flips
measure-theoryprobability theory
Best Answer
Note that $F_n$ is not a subset of $\Omega$, it is a set of subsets of $\Omega$. So it doesn't make sense to ask about $F_1$ containing all possible infinite sequences.
By definition, $F_1$ consists of four subsets of $\Omega$ (corresponding to the four subsets of $\Omega_1$). These four subsets are:
$\Omega$ (corresponding to $E=\Omega_1$)
$\emptyset$ (corresponding to $E=\emptyset$)
all sequences in $\Omega$ that start with $0$ (corresponding to $E=\{(0)\}$)
all sequences in $\Omega$ that start with $1$ (corresponding to $E=\{(1)\}$)
In general, $\Omega_n$ has size $2^n$ hence has $2^{2^n}$ subsets. These $2^{2^n}$ subsets of $\Omega_n$ correspond to $2^{2^n}$ subsets of $\Omega$, which together make up the collection $F_n$. The correspondence is: start with a subset $E$ of $\Omega_n$ (so this is some collection of sequences of $0$'s and $1$'s of length $n$). Then define the subset $A_E$ of $\Omega$ to consist of all infinite sequences whose first $n$ entries are a sequence in $E$. Now $F_n$ is the collection $\{A_E:E\subseteq \Omega_n\}$.