Probability theory says that if an event $E$ is certain to happen, then $P(E)=1$ which makes sense. Similarly, an impossible event has probability $0$.
What surprised me is the fact that you can still find mathematical texts (notice that this paper comes from a renowned American university) that say the converse are also true, namely:
$P(E)=1 \implies E\quad$ is certain to happen
and
$P(E)=0 \implies E\quad$ can't happen.
Now let's consider the second case. Let's say I'm choosing a point randomly from the interval $[0,2]$. Even though it's possible for every particular point to be chosen (I can easily choose $2$ or $0.5$), the calculated probability for randomly choosing that particular point is $0$. But I have chosen a point, right? Thus it is can happen.
In this case, the probability should be considered as a limiting value. When $P($the randomly chosen number equals $1$$)=0$, it should be understood as the limit of the number of times I've chosen $1$ divided by the number of trials. As the number of trials increases, this fraction approaches $0$ – but it doesn't have to be $0$ at any point during that process.
Similarly, in the first case, I might consider of an event that the randomly selected point from $[0,1]$ is from interval $[0,1)$. The measures of those sets are identical, so the probability equals $1$. Does it mean I will certainly select a point from $[0,1)$? Of course not, because $1$ can be chosen. Thus the event is not certain to happen.
Is there anything wrong with reasoning above? Why are so many people convinced the two implications are true?
Best Answer
It is not true in general that probabilities of $0$ and $1$ necessarily mean that the event is impossible or certain.
Your example with a point chosen from $[0,2]$ shows clearly that such a claim can't be upheld if we want to speak about continuous distributions at all.
As Lulu notes in a comment, the text you're linking to contradicts itself: On page 1 it wrongly claims that
whereas on page 3 it contradicts this with the correct