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 $\mathscr{F}$?
 A: 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. 
A: 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 $E\notin\mathscr F$. 
