Why is it that $\mathscr{F} \ne 2^{\Omega}$?

From Williams' Probability with Martingales:

2.3. Examples of $(\Omega, \mathcal{F})$ pairs
We leave the question of assigning probabilities until later.

(a) Experiment: Toss coin twice. We can take $$ \Omega = \{HH, HT, TH, TT\}, \quad \mathcal{F} = \mathcal{P}(\Omega) := \text{set of all subsets of $\Omega$}. $$ In this module, the intuitive event ‘At least one head is obtained’ is described by the mathematical event (element of $\mathcal{F}$) $\{HH, HT, TH\}$.

(b) Experiment: Toin coss infinitely often. We can take $$ \Omega = \{H,T\}^{\mathbb{N}} $$ so that a typical point $\omega$ of $\Omega$ is a sequence $$ \omega = (\omega_1, \omega_2, \dotsc), \quad \omega_n \in \{H,T\}. $$ We certainly wish to speak of the intuitive event ‘$\omega_n = W$’ where $W \in \{H,T\}$, and it is natural to choose $$ \color{red}{ \mathcal{F} = \sigma( \{w \in \Omega : \omega_n = W\} : n \in \mathbb{N}, W \in \{H,T\} ) }. $$ Although $\color{red}{\mathcal{F} \neq \mathcal{P}(\Omega)}$ (accept this!), it turns out that $\mathcal{F}$ is big enough; […]

Why is it that $\mathscr{F} \ne 2^{\Omega}$ ?

What are some elements of $2^{\Omega}$ that are not in $\mathscr{F}$?

Solution 1:

The $\sigma$-algebra generated by the events $\{\omega \in \Omega: \omega_n = W \}$ is the so-called Borel $\sigma$-algebra on $\Omega = \{H,T\}^\mathbb{N}$.

One can show, by transfinite induction (so you need some set-theory background) that there are at most $|\mathbb{R}| = 2^{\aleph_0}$ many Borel sets, while the power set of $\Omega$ has $2^{|\Omega|} = 2^{|\mathbb{R}|}$ many subsets, which is more by Cantor's theorem. So $\mathcal{F}$ is much smaller than the power set of all subsets of $\Omega$. But you only need the sets in $\mathcal{F}$. One can also use the Axiom of Choice to find non-Borel sets (the characteristic function of an ultrafilter, e.g.).

Filling in all the details requires some theory the author presumably did not want to assume the reader to know about.

Solution 2:

Blackwell & Diaconis ["A non-measurable tail set", in Statistics, probability and game theory, pp. 1–5, IMS Lecture Notes Monograph Series, vol. 30, 1996] give an example of a subset of $2^{\Bbb N}$ that is not an element of $\mathscr F$. Their construction uses a free ultrafilter $\mathscr U$ on $\Bbb N$. Let $E\subset 2^{\Bbb N}$ consist of those $a=(a_1,a_2,\ldots)\in 2^{\Bbb N}$ (thus each $a_i$ is either $0$ or $1$) such that $N_a:=\{i\in\Bbb N:a_i=1\}\in\mathscr U$. Then $E\notin\mathscr F$.