Why isn't there a uniform probability distribution over the positive real numbers?

If a uniform probability distribution did exist, then for any integer $n$ the probability a real number $x$ satisfies $n \leq x < n+1$ would have to be the same for all $n$. Call this probability $p$.

One of the rules that a probability distribution has to satisfy is that if $\{E_n\}$ are a countably infinite collection of disjoint events then $P(\cup_n E_n) = \sum_n P(E_n)$. Let $E_n$ be the event "the real number $x$ satisfies $n \leq x < n+1$". Since every real number is between some $n$ and $n + 1$, $P(\cup_n E_n) = 1$. On the other hand, $\sum_n P(E_n) = p + p + p + ....$. If $p > 0$, this gives infinity. If $p = 0$ it gives zero. In either case, you'll never add up to $1$. Hence you cannot have a uniform probability distribution over the reals.

A uniform probability over the positive real numbers would not satisfy all three axioms of probability. Specifically, there is a conflict between the second axiom ($P(\Omega) = 1$) and the third axiom (countable additivity).

Check out the wikipedia entry http://en.wikipedia.org/wiki/Axioms_of_probability. I think it should be clear how a uniform probability over the positive reals would cause problems.

For every probability density $f$, $\liminf_{x\to \pm \infty} f(x) = 0$ must hold.

It is clear that for a uniform distribution, the density has to be constant on the considered interval.

Combining both requirements, only $g(x) = 0$ remains as a choice for a uniform density on $(-\infty, \infty)$. But $\int_{-\infty}^\infty g(x)dx = 0 \neq 1$, that is $g$ is no proper probability density.