# Event Occurs Almost Surely

Consider an uncountably infinite space, an infinite coin-tossing.

Let $$(\Omega,\mathcal{F},\mathbb{P})$$ be the probability space. If a set $$A\in\mathcal{F}$$ satisfies $$\mathbb{P(A)=1},$$ then we say that event A occurs almost surely.

My Question:

1. Why is it the case every individual coin-toss sequence has probability zero in this uncountable probability space? In Shreve's explanation, he defines two sets: $$A_H=\{\omega\in\Omega_\infty:\omega_1=H\}\\A_T=\{\omega\in\Omega_\infty:\omega_1=T\}.$$

He sets $$\mathbb{P}(A_H)=p,\mathbb{P}(A_T)=1-p=q.$$ Hence, instead of assigning probability measure to any single sequence, he looks at a set of sequences where it kicks off with either H or T. Why does this set-up makes sense compared to how we deal with coin-tossing in a finite sample space. Usually, when we deal with two coins, we are given an assumption that the coin is fair, so $$P(H)=\frac{1}{2}$$. However, in the uncountably infinite coin-tossing experiment, any single infinity sequence gets measure $$0$$ while the set of sequences with certain characteristics (e.g. first flip $$H$$) gets a strictly positive probability.

1. What exactly does $$\mathbb{P}(A)=1$$ almost surely mean? According to Shreve:

"Whenever an event is said to be almost sure, we mean it has probability one, even though it may not include every possible outcome. The outcome or set of outocme not included, taken all together, has probability zero."

Does this mean the event that a coin toss gets at least one tail will happen with certainty? Conversely, when we say $$\mathbb{P}(\{\omega\in\Omega_\infty:(\omega_i)_i=H\space\forall\space i\in\mathbb{N}\}=0,$$ does this mean something with probability zero CANNOT happen, so it is impossible to obtain an infinite coin-toss sequence with all Heads?

Reference:
Shreve, Steven E. $$\textit{Stochastic Calculus for Finance II : Continuous-Time Models}$$. Springer, 2008.

• Hi: I could be wrong ( not a mathematician ) but I believe that, in order to address your confusion, you have to deal with the lebesgue integral and the concept of zero measure. I couldn't do it justice here ( or anywhere else ) but I think that the level of thorough understanding you seek requires that level of knowledge. Sep 25, 2019 at 7:45
• Just to touch on the concept here, note that zero measure does not imply that the event can't happen. It just implies that the event that it can happen has zero measure, given the sigma field and the probability measure defined on that sigma field. Sep 25, 2019 at 7:48
• @DrewSaunders here the sample space is the set of all possible infinite coin tosses which can be put into correspondence with the reals in $(0,1)$. For each coin toss encoded as a sequence e.g. $(0,1,1,0,0,0,1...)$ corresponds to the binary expansion of a real in $(0,1)$ etc Sep 25, 2019 at 17:03
• @NapD.Lover So the space is uncountable. Thank you for the correction. I have deleted my incorrect comment.
– Drew
Sep 26, 2019 at 11:43

Consider sequences of $$n$$ flips with $$p= \frac 12$$ for simplicity. The sets defined based only on the first flip obviously divide the space in half, so each has positive probability ($$\frac 12$$) no matter what $$n$$ is. But consider the outcome A (a singleton set) defined by seeing $$n$$ heads. If $$n=2$$, $$P(A)=.25$$. More generally, $$P(A)=0.5^n$$, which goes to zero as $$n$$ goes to infinity. Although this isn't a proof and infinity can be tricky mathematically, in this case, the intuition gives the correct result for the limit: the infinite sequence consisting of all heads has zero probability, and similarly for any one infinite sequence you care to specify.
Don't confuse $$P(A)=1$$ with "certainty"; certainty isn't really a useful concept here, unless maybe you want to apply it to the set of all possible outcomes. I guess it would be easy enough to show that the probability of the set of outcomes that each include at least one head is one, but I'd stay away from the word "certainty". Similarly, to say that an event "cannot happen" or is "impossible" would be wrong (or perhaps not well-defined probabilistically).