Problems with definition of almost surely Let $(\Omega,\mathcal F,\mathbb P)$ be a probability space. In probability theory one says that and event $F\in\mathcal F$ happens $\mathbb P$-almost surely, if $\mathbb P(E)=1.$ Intuitively, as a beginner one thinks that this means that there exists an event $N\in\mathcal F$ with $\mathbb P(N)=0$ and such that the event $E\setminus N$ happens surely? Is this intuition generally true? Counterexample?
 A: In many probability spaces there are events $E \subsetneq \Omega$ and $\varnothing \neq N \subset \Omega$ such that $P(E)=1$ and $P(N)=0$. 
Take for example the following scenario: 
Suppose you flip a fair coin infinitely many times. The sample space $\Omega$ will be all sequences which look like $(x_1, x_2, x_3, …)$ where $x_i = 0$ if you flip a tail and $x_i= 1$ if you flip a heads. For instance, if the first two flips were tails and the third a heads, you have a sequence starting with $(0,0,1,x_4, …)$. Now, let's ask: What is the probability that the coin flipped all heads? The event we want to look at is $N = \{(1,1,1,1,1,1,…)\}$ and 
$$P(N) = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdots = \lim_{n \to \infty}\left(\frac{1}{2}\right)^n = 0.$$
This intuitively makes sense because how likely are you to actually flip all heads with a fair coin and infinitely many flips? But, you can use this same strategy and see that the probability of any one fixed sequence of coin flips must be $0$. But again, this intuitively makes sense because how likely are you to flip exactly the same faces of the coin in exactly the same order as the fixed sequence for infinitely many flips? 
Now, if I ask what is the probability that you flip a heads on your first flip, lets call this event $H$, the answer is sensibly $P(H)=\frac{1}{2}$. So how can we merge a positive probability of events like $H$ when we have that each single outcome has probability $0$? This is the challenge that measure theory has come to answer. If $P$ is that measure (the measure of probability) and now you ask what is the probability of the event $E$ that you don't flip all heads? Well, the probability of flipping all heads is $0$, so  $P(E) =1$. But $E \neq \Omega$ since $E = \Omega \backslash \{(1, 1, 1, 1, …)\}$. Therefore, can we say that if you flip a coin infinitely many times, then you will surely have an outcome that lands in $E$? No! It could so happen that you do flip all heads. Although, since the probability of that happening is $0$ (and just based on our intuition of how likely that is), you will almost surely flip a sequence that lands in $E$.
