How are the Real Numbers distributed?

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HyperbolicMan
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Question: What is the probability that a random variable X with domain all real numbers will take a value in the closed interval [a,b]?

It seems to me that in order to answer this question you have to know how the real numbers are distributed. Given the appropriate distribution function, you can integrate it from a to b to find the probability.

Common sense says that the real numbers should have a constant distribution (e.g. P(x)=c for all x). However, the integral of any constant function from -[tex]\infty[/tex] to [tex]\infty[/tex] is not 1.

So how exactly are the real numbers distributed?
 
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I am pretty sure that for any finite numbers a and b, the chance that x is in [a,b] is zero.

For any finite number a, the chance that x is in [a,infinity) is 50%. The same with (-infinity,a]
 
HyperbolicMan said:
So how exactly are the real numbers distributed?
They are distributed how you want them to be. A set does not have an intrinsic distribution. For instance you may define a probability density function p(t) by p(t) = 1/2 for t in [1,3] and p(t)=0 otherwise.

If what you want is a uniform distribution, then no such thing can exist on the real numbers for the reason you mentioned (that if p(x) =c for all x, then [itex]\int_{-\infty}^\infty p(x) dx[/itex] is 0 or [itex]\infty[/itex] depending on whether c =0 or c>0, but never 1).