Probability of events in an infinite, independent coin-toss space I am studying Steven E. Shreve's Stochastic Calculus book. Example 1.1.4 (p.4-6) constructs a probability measure on the space of infinely many coin tosses $\Omega_\infty$. In the example the $\sigma$-algebras for $n$ coin tosses are defined like this $\mathcal{F}_0 = \{\emptyset, \Omega\}$, $\mathcal{F}_1 = \{\emptyset, \Omega , A_{H}, A_{T}\}$, where $A_H$ is the set of all sequences beginning with head. For three coin tosses we get $\mathcal{F}_3 = \{\emptyset, \Omega ,A_H, A_T, A_{HH}, A_{HT}, \ldots\}$
Now the authors state

By continuing this process, we can define the probability of every set that can be described in terms of finitely many tosses.

And later:

We create a $\sigma$-algebra, called $\mathcal{F}_\infty$ by putting in every set that can be described in terms of finitely many coin tosses and then adding all other sets required in order to have a $\sigma$-algebra. It turns out that once we specify the probability of every set that can be described in terms of finitely many coin tosses, the probability of every set in $\mathcal{F}_\infty$ is determined.

I find this puzzling! Why do finite descriptions suffice? For example I don't understand how the probability of the event "infinitely many heads" is determined. I would guess it has probability 1 but how can I conclude this from the finite cases? How is this done for general elements $A\in\mathcal{F}_\infty$?

Edit 2: Can I argue like this: for $A\in (\mathcal{F}_\infty\setminus (\bigcup_{n=1}^\infty F_n))$, the complement $A^C$ is in some $\mathcal{F}_m$ and therefore $\mathbb{P}(A) = 1-\mathbb{P}(A^C)$?
 A: No real answer but too much for a comment:
$$\left\{ \text{number of heads infinite}\right\} =\bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}\left\{ \text{head at  }k\text{-th toss}\right\} $$
showing that this event can be described by means of sets in $\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}$. 
I am not sure, but suspect that $\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}$ is an algebra and that a probability 'premeasure' on it determines uniquely a probability measure on $\sigma$-algebra $\sigma\left(\bigcup_{n\in\mathbb{N}}\mathcal{F}_{n}\right)$.
A: First, let us be clear about the setup: We have an countably infinite number of coins; call them coin $1,2,3,\dots$. Now, for any finite set of coins (say coins $7,8,9,$ and $10$), we know the $\sigma$-algebra of events for those coin tosses (and we know the probabilities assigned).
Now, to get a $\sigma$-algebra over the entire infinite sequence (and to get a probability measure), we just need to recall the axioms of a $\sigma$-algebra and a measure space.
A $\sigma$-algebra is closed under countable union and intersection. So after starting with all of the finite measurable sets (those corresponding to the outcomes of any finite number of coins), we need to take all of the complements of these sets and all of their possible unions and add these to our system. For example, for each $j=1,2,3,\dots$, we have the finite event "coin $j$ is heads", so by taking the union over all $j$ of these events, we have the event "at least one coin is heads". We must then continue in this fashion until we have all the sets that can be obtained by countable union or complement. For instance, the complement of "at least one coin is heads" is the event "all coins are tails", so we must add this set. After continuing in all possible ways, we have our $\sigma$-algebra. 
Meanwhile, a measure is always countably additive. So the author is saying that we can also use this rule to deduce what the probabilities of all these new events will be, if we know the probabilities for every finite set of coins.
For instance, how do we determine the probability of the event "infinitely many heads"? Well, it is the complement of the event "finitely many heads". This event is the countable union of all outcomes that have finitely many heads. (Note that I am asserting here that there is only a countably infinite number of outcomes where there are finitely many heads.) Each of these has zero probability, because any given single outcome of the infinite sequence has zero probability (I think you can argue this in different ways, like a contradiction). Since a probability measure is countably additive, the probability of finitely many heads is the sum of countably many zeroes, which is still just zero.
This seems like a surprisingly too-easy proof, but it is true: Any countable-size set of outcomes will have probability zero.
Disclaimer: I haven't read the book in question and am just going off of the quotes.
