Question on the consequences of the Kolmogorov axioms I am reading the (german) Applied Statistics and on page 140 as a consequence of the Kolmogorov axioms it is stated that if $P(A)=0$ one cannot conclude that $A=\emptyset$ . Similarly if $P(A)=1$ one also cannot conclude that $A=S$. Why is that?
Also, if $P(A)=0$ this means that event A is almost never possible and if $P(A)=1$ will almost surely occur.
I am having a bit of a trouble intuitively understanding the need for the above statements (almost surely or almost never) and why if $P(A)=1$ one cannot conclude that $A=S$. 
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 A: Let X be a standard normal random variable with $S=(-\infty,\infty)$. Here, $P(X=1)=0$ but $\{1\}\neq \emptyset$. 
To show that $P(A)=1$ does not imply that $A=S$, consider the following. You flip a coin an infinite number of times. The event of getting all heads $\{H,H,H,H,...\}$ is in the sample space because it is physically possible that tails never appears. Now, let $A = \{\text{flip at least one heads in the infinite flips}\}\neq S$. However, $P(A) = 1 - P(\text{all tails in the infinite flips}) = 1 - (.5)^{\infty} = 1$.
A: Take the uniform distribution on $S=[0,1]$. Now, $P[S]=1$, but also $P[S\setminus \{1\}]=1$ and $P[S \setminus \{1,1/2,1/3,\ldots\}] = 1$ as well even though both of the latter two are strict subsets of $S$. (Here the notation $A \setminus B$ means all elements belonging to set $A$ and not belonging to $B$.)
To answer the almost surely part of your question:

if P(A)=0 this means that event A is almost never possible and if
  P(A)=1 will almost surely occur.

I think you have got it the wrong way round: "almost surely" is defined to be $P(A)=1$.
To fully understand the intricacies and subtleties of the definitions, some familiarity with measure theory is helpful. See, e.g. the Cantor set, which has measure ["length"] zero but contains a uncountably infinite number of points, compared to any interval on the real line $(a,b)$ which has again uncountably infinite number of points but nonzero measure $b-a$.
Given that you are working in Applied statistics, these complicated sets are probably irrelevant to you, so I would not worry about it ( its just something authors like to put in!).
A: Following the arguments from @TrynnaDoStat:
Let $X$ be a standard normal random distribution, and therefore $X$ has support on whole $\mathbb{R}$. $P(X=1) = 0$ but $\{1\} \neq \emptyset$
Then using the principle that
$P(\Omega \backslash E) = 1 - P(E)$
$P(X \in \mathbb{R} \backslash ${1}$) = 1 - 0 = 1$, but $\mathbb{R} \backslash \{1\} \neq \mathbb{R}$
