Example of probability space s.t. $p(A)=0$, $p(B)=1$ and $A≠∅, B≠Ω$ I had a question of finding a probabilty space $(Ω,β,p)$ s.t. $A,B∈β$ ,  $p(A)=0$, $p(B)=1$ and $A≠∅, B≠Ω$.                               
This is what I thought for solution: I thought about the problem of throwing a dice with $3$ faces- $1$, $2$, $3$. The probability to get $1$ is $0.4$, the probability to get $2$ is $0.6$ and the probability to get $3$ is $0$ (sort of unfair dice).
So I can take the following probability space: $Ω=\{1,2,3\}$, $β=2^Ω$, and 
 $p:β\to [0,1]$ will be defined according to the probabilities I wrote above. We can see that this is  a valid probability space, and can take $A=\{3\}$ and $B=\{1,2\}$ to get what we want.                     
However, something in my solution doesn't seem right to me. For example, I suspect that if $p(3)=0$, then $3$ should not be included in $Ω$.
Is my solution correct? 
 A: 
For example, if I suspect that if $P(3) = 0$, then $3$ should not be included in $\Omega$.

Mathematically speaking, there is nothing wrong with including $3$ in your $\Omega$.  In a certain setting, it may be sensible to exclude it, but there are many reasons to leave events of probability zero in your space.
Here's another solution along these lines: we flip a fair coin and define
$$
\Omega = \{H,T,Z\} 
$$
$H$ denotes the event in which the coin lands "heads".  $T$ denotes the event in which the coin lands "tails".  $Z$ denotes the event in which the coin hovers midair, becomes sentient, learns to speak, and teaches you probability.  We may suppose that $Z$ has probability zero.

As for why one might include such an event in $\Omega$, one reason is for convenient comparisons of distributions.  For example, we might have two random variables $X,Y$ with
$$
P(X = 1) = P(X=2) = 1/2; \quad P(X = 3) = 0\\
P(Y = 1) = 0; \quad P(Y = 2) = P(Y = 3) = 0
$$
Rather than take $\Omega_X = \{1,2\}$ and $\Omega_Y = \{2,3\}$, it is often convenient to use a common sample space such as $\Omega = \{1,2,3\}$ or even $\Omega = \Bbb N$.
