Is there a uniform distribution over the real line?

For every interval $[a,b]$, there exists a uniform probability density over this interval, which is the constant function $f(x)=\frac{1}{|a-b|}$ for $a < x < b$, and $f(x)=0$ for all other $x$.

Now it's clear that there is no ordinary probability density which is constant on all of $\mathbb{R}$ which also has integral $1$. (The height of such a function would have to be infinitesemally small.)

However, it seems to me as if the Dirac-Delta $\delta_x$ has very similar problems: It is conceptually a function that is infinitely thin but still has integral $1$.

In the Dirac-Delta case, one could solve the issue by generalizing the definition of a function and defining $\delta$ not as a function, but as a Schwartz distribution.

So my question is: Is there some way, for example by generalizing some definition somewhere, to get a function that acts like a uniform probability density over the real numbers, i.e. when multiplied with any other density $f$ the resulting density will be equal to $f$ (up to a constant factor)?

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Well, you could relax the condition that the total probability measure must be finite :-) – Henning Makholm Nov 27 '11 at 4:04
+1 for a nice question. And the correct spelling of Schwartz. =) – Srivatsan Nov 27 '11 at 4:29
Related question: Is it possible or useful to define a uniform distribution on the integers or the real line, by permitting infinitesimal probabilities/probability densities? – Mechanical snail Nov 28 '12 at 9:56

(1) Google the term "improper prior".

(2) Look at the works of Edwin Jaynes.

(3) Maybe even search for "improper prior" in the works of Edwin Jaynes.

My suspicion (really only a suspicion) is that something broader than probability is involved. If you restrict yourself to probability as conceived by Kolmogorov's system of axioms then of course the answer is "there is no ordinary probability density which is constant on all of $\mathbb{R}$ which also has integral 1." But Kolmogorov's axioms are not the last word. Countable, as opposed to finite, additivity has been questioned by de Finetti, maybe Renyi, and others. There's a book called Theory of Charges (by K. P. S. Bhaskara Rao and M. Bhaskara Rao) that deals with finitely additive measures that are not necessarily countably additive, that may be worth checking out. But quite aside from the question of whether Kolmogorov is God's last prophet in the field of probability, I think one might wonder whether probability itself is the last word.

(Admittedly, Jaynes is annoying at times. And sometimes he's immensely entertaining. You can also find his writings on music, including stuff on the thermodynamics of muscles in human fingers as applied to piano technique.)

Later note: I don't know why, but I forgot to mention Sir Harold Jeffreys' book on probability theory. His Ph.D. was in mathematics. For a time he was professor of astronomy. He wrote a lot of stuff about seismology. And perhaps only because of an appendix in one edition of one of his somewhat philosophical books do we know today about Mary Cartwright's proof of the irrationality of $\pi$. Anyway, you'll find a fair amount of stuff on improper priors in that book, and maybe a certain amount of disdain for logical rigor as understood by mathematicians.

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Michael, +1 Nice answer. I took the liberty to add a link to the book and the authors' names. Hope it's ok. Regards, – Srivatsan Nov 27 '11 at 16:28
Thank you, Srivatsan. – Michael Hardy Nov 27 '11 at 17:06

Dirac delta can be defined as $\lim_{t\rightarrow 0} (x^2+t^2)^{-1}t/\pi$.

The measure $\lim_{t\rightarrow \infty} (x^2+t^2)^{-1}t/\pi$ has uniform density over the real line and integral 1.

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You should say in which sense the convergence is in. A delta is surely not a point function. – user12014 Nov 27 '11 at 9:11
@PZZ What do you mean by 'sense' and 'point function'? This is definition from mathworld.wolfram.com/DeltaFunction.html – TROLLKILLER Nov 27 '11 at 9:14
@ cosine666is1 Well, no function $f: \mathbb{R} \to \mathbb{R}$ can satisfy $f(x) = 0$ for $x \neq 0$ and $\int_{\mathbb{R}} f(x) dx = 1$. Therefore, the limit cannot be interpreted as a pointwise limit, since this would define a function with that property. I'm assuming that the definition should be interpreted as an approximate identity for the Banach algebra $L^1$ (en.wikipedia.org/wiki/Approximate_identity) or in a distributional sense – user12014 Nov 27 '11 at 9:35
@cosine, calling on a website's authority should not prevent us to think, right? In the present case the expression The measure $\lim_{t\to\infty}(x^2+t^2)^{-1}t/\pi$ is awfully confusing. What you write after the limit sign is a number, its limit when $t\to\infty$ is zero, for every fixed $x$, and there is no measure there. If only for these reasons you should explain more precisely what you mean. – Did Nov 27 '11 at 16:39
That's exactly the point. If you just write some limit, explicitly leaving the variables, then the natural interpretation is pointwise and this is not what is meant in this case. And BTW, most serious mathematical literature tends to define the delta as a linear functional on $C_0^{\infty}$. – user12014 Nov 28 '11 at 22:36