Is there a measure on $\mathbb{N}$? (What is the chance that a random integer from $\mathbb{N}$ is even?)

Question

The more I study, I realize I really know nothing. sigh. Now, here is my newest problem. Picking up an integer at random from integer set $\mathbb{N}$. What is the chance that the integer would be even? What is the chance that the integer would be odd?

Ok, I understand it - we can have a bijection and all - and show it is $\frac{1}{2}$ - in both the cases - but how? Can we introduce some measure for $\mathbb{N}$ - which shows that the set of even numbers $\mathbb{E}$ and set of odd numbers $\mathbb{O}$ are, sharing it 50-50? Thanks to bijection we have all of them $\aleph_0$.

As usual, I am completely at lost here, please help.

So we do not have anything formal done for 1/2 to select a random even number from number set? :( — Noga Tailcutter, Jun 05 '14 at 05:40
Not if (as usual) we ask for measures to be countably additive. — André Nicolas, Jun 05 '14 at 05:41
So there is NO other way than primitive counting? So half is --- simply intuition? Can you please add it as a solution so that I can accept it - and upvote it? — Noga Tailcutter, Jun 05 '14 at 05:44
The "half" is not simply intuition, it is a precise (easy) technical result. Just happens not to use a measure. — André Nicolas, Jun 05 '14 at 05:45
But if you want only finitely additive (not countable additive) measure, there are such measures assigning a value to each subset of $\mathbb N$, see this question for details. — Martin Sleziak, Jun 05 '14 at 05:53

Michael Albanese · Answer 1 · 2014-06-05T11:38:34.440

11

As André Nicolas points out there is no countably additive probability measure on $\mathbb{N}$ which assigns equal measure to singletons.

The way people usually deal with such questions is by considering natural density.

A subset $A \subseteq \mathbb{N}$ is said to have natural density (or asymptotic density) $\alpha \in [0, 1]$ if $$\lim_{n\to\infty}\frac{|A\cap\{1, \dots, n\}|}{n} = \alpha.$$

As André Nicolas also pointed out, the set of positive even integers has natural density $\frac{1}{2}$.

edited Jun 05 '14 at 11:38

answered Jun 05 '14 at 05:45

Michael Albanese

99,526

Why can I not accept two answers? Both are awesome? :( – Noga Tailcutter Jun 05 '14 at 05:47
Because the accepted answer is supposed to be the best answer according to the OP (i.e. you). – Michael Albanese Jun 05 '14 at 05:49
That is so not cool (and formal)! Arbitrary axiomatic assignment! – Noga Tailcutter Jun 05 '14 at 05:50
Well you can upvote answers you find useful, and I suggest you do (always, not just on your own questions), but accepting an answer is supposed to indicate that no other answers are required because the OP's question has been answered (by the accepted answer). To be fair, most of what I said was said by André Nicolas in the comments, so accepting his answer is totally OK with me. – Michael Albanese Jun 05 '14 at 05:54
Thanks and I always do it. It is just that given limited resource justice to individuals can not be done. :( – Noga Tailcutter Jun 05 '14 at 06:02

André Nicolas · Accepted Answer · 2014-06-05T06:41:56.540

6

There is no (countably additive) probability measure on $\mathbb{N}$ that assigns equal weight to all points.

But we can easily get the following density result. Let $E(n)$ be the number of even natural numbers up to $n$. Then $\lim_{n\to\infty}\frac{E(n)}{n}=\frac{1}{2}$.

edited Jun 05 '14 at 06:41

answered Jun 05 '14 at 05:46

André Nicolas

507,029

Man, seriously I do not know anything. But whats the point. – Noga Tailcutter Jun 05 '14 at 05:46
Umm... The counting measure is countably additive measure on $\mathbb{N}$ that assigns the measure $1$ to all points. – Eric Towers Jun 05 '14 at 06:34
@EricTowers, the counting measure is not a probability measure. – Aditya Jun 06 '14 at 04:12
@Aditya: The OP does not request a probability measure. – Eric Towers Jun 06 '14 at 06:25
Let $\cal{A}(N)$ be sum of two ideals $I(O)$ and $I(E)$, where Let $I(O)$ be an algebra of subsets of the set $O(N)$ of odd natural numbers which are finite or whose complement in $O(N)$ is finite; – George Jun 06 '14 at 16:58

score 3 · Answer 3 · answered Jun 05 '14 at 05:46

You could look at the counting measure. This usually leads to density argument:

Let $n \in \mathbb{N}$ and use the counting function $o(n) = \text{the number of odd numbers in [0,n]}$. See that $\lim_{n \ \rightarrow \infty} \frac{o(n)}{n} = 1/2$. Alternatively, see that $o(n) = n/2 + O(1)$.

For similar arguments (typically having a number theoretic or computer science flavor, and littered with big O notation), see the prime number theorem, and the de Bruijn function (density of smooth integers).

Is there a measure on $\mathbb{N}$? (What is the chance that a random integer from $\mathbb{N}$ is even?)

3 Answers3

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