# 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 $A=S$.

[deleted edit]

• You have to dive into measure theory and understand the concept of an event that is not Lebesgue-measurable. Such events, although totally non-intuitive, do exist, as a rather annoying consequence of the Axiom of Choice. The predominant consensus is that we can live with non-Lebesgue measurable sets (adding all the time "almost surely" and "almost never" to various results) much better that we would go by if we didn't accept the Axiom of Choice. Dec 2, 2013 at 0:02
• Dear @Alecos: I don't see how this question has much of anything to do with nonmeasurable sets (beyond the fact that we can't always construct a space where all subsets are measurable). Indeed, all of the sets referenced are implicitly measurable (since we can apply $P(\cdot)$ to them). Provided the probability space in question is sufficiently rich (e.g., $([0,1], \mathcal B[0,1], \lambda)$), there are, of course, plenty of distinct measurable sets with probability zero (and, hence, equivalently, plenty with probability one as well). Dec 2, 2013 at 1:30
• The proof using $P(X)$ for a $X$ continuously distributed (@TrynnaDoStat, @wonghang, @seanv507), gets support from the notion that $P(X=1)=0$, which is for all practical means true, but logically it shouldn't be zero (inexistence) but it should be equal to some infinitesmall quanntity$\neq 0$. Then the proofs collapse and for $P(A)=1, A=S$ .
– ECII
Dec 2, 2013 at 10:46

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, $P(A) = 1 - P(\text{all tails in the infinite flips}) = 1 - (.5)^{\infty} = 1$.

• In your example if the sample space is defined as the set of all possible outcomes then $0\notin S$ and $A=S$ because a single dice coming up zero is not an outcome. Dec 1, 2013 at 23:44
• There is no restriction that S is defined as the set of all possible outcomes as far as I can tell. Dec 1, 2013 at 23:46
• According to wikipedia, the sample space is defined as the set of all possible outcomes and so my first example doesn't work. I will edit my answer accordingly to give you a more satisfying example. Dec 2, 2013 at 0:05

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 $\mathbb{R} \backslash \{1\} \neq \mathbb{R}$

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 $B$.)

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 $b-a$.