# Does probability theory require infinitesimals to work?

I'm taking an intro to probability class, and my book lists this as an axiom:

$$P(\bigcup_{k=1}^{\infty}A_k) = \sum_{k=1}^{\infty}P(A_k)$$

where $A_k$ is an event and $P(A_k)$ is the probability of $A_k$.

Now consider the question of picking a random number in an interval. For instance, a random number is picked between 0 and 1. What's the probability of picking 0.5?

1) 0: If it's 0, then you get weird results. For instance, say you pick a random number between 0 and 1. What's the probability of the number being less than 0.5? Well using the axiom at the very top, the probability of this is the sum of the probabilities of picking each number less than 0.5. Since they are all 0, the probability of picking a number less than 0.5 is 0. Which is obviously not true.

2) Infinitesimals. This would give you a result I'm assuming would make sense

So the way I see it, probability for continuous sample spaces cannot exist without infinitesimals.

Am I misunderstanding something or can probability for continuous sample spaces not exist without infinitesimals?

• The axiom is for th union of a countable number of (pairwise disjoint) sets. You are misunderstanding the axiom if you interpret it as applying to unions of an uncountable number of sets. There has been some work done on probability from the viewpoint of non-standard analysis. But non-standard analysis is definitely not needed. – André Nicolas Sep 14 '15 at 0:36
• @AndréNicolas Hmm, my book doesn't say anything about the set having to be countable. It just lists the theorem. – pdfgdfg Sep 14 '15 at 0:40
• @AndréNicolas Also, how can all the samples in a sample space have 0 probability? – pdfgdfg Sep 14 '15 at 0:43
• The individual $A_k$ need not be countable. However, the indices $k$ run over the natural numbers. So there are only countably many $A_k$. – André Nicolas Sep 14 '15 at 0:44
• For certain probability distributions, over some sample spaces, the probability assigned to any individual element is indeed $0$. This is the case for all the familiar continuous distributions. Probability $0$ does not mean impossible. – André Nicolas Sep 14 '15 at 0:47

• @pdfgdfg The key is the countability. While each point separately has probability zero, $[0,1]$ is not a countable union of single points (i.e. it is uncountable). So there is no difficulty. One way of viewing the zero probability aspect is in a frequentist description: typically, whatever your (say) first value in an infinite sequence of random numbers is, you will only see it that one time. Thus the fraction of times that it appeared will converge to zero. To a frequentist, this is the meaning of "probability". – Ian Sep 14 '15 at 0:44