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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  • $\begingroup$ 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. $\endgroup$ Dec 2, 2013 at 0:02
  • 2
    $\begingroup$ 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). $\endgroup$
    – cardinal
    Dec 2, 2013 at 1:30
  • $\begingroup$ 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 $P(A)=1, A=S$ . $\endgroup$
    – ECII
    Dec 2, 2013 at 10:46

3 Answers 3


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

  • $\begingroup$ 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. $\endgroup$
    – nostock
    Dec 1, 2013 at 23:44
  • $\begingroup$ There is no restriction that S is defined as the set of all possible outcomes as far as I can tell. $\endgroup$ Dec 1, 2013 at 23:46
  • $\begingroup$ 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. $\endgroup$ Dec 2, 2013 at 0:05

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


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!).

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    $\begingroup$ I wasn't sure exactly what to make of your last line, so I didn't edit that. Maybe you could edit to clarify. That said, at a glance, I'm not sure the reason behind the downvote, either. Cheers. $\endgroup$
    – cardinal
    Dec 2, 2013 at 2:41

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