Is there a measure on $\mathbb{N}$? (What is the chance that a random integer from $\mathbb{N}$ is even?) 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.
 A: 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 $\lim_{n\to\infty}\frac{E(n)}{n}=\frac{1}{2}$.
A: 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).
A: 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 $\frac{1}{2}$.
