Comparing Divergent Integrals

I have a (vector) space of step functions from the unit interval to the reals. Each function is discontinuous at a countable, possibly infinite, number of points. Since the discontinuities have measure zero, I can integrate the functions. I want a strict weak order in which one is less than another if its integral is less. However, some of the integrals diverge. What is the most natural extension of this ordering to these cases (or is there a reason why I should not expect any one to be more natural than another)?

I'm not really sure what field this is in. I've read about distributions, vector spaces, and measure theory, but I can't find anything that really applies. If you know of any related idea, I can try to make the question more precise.

flag	asked Feb 10 2011 at 3:49 Dan 1

	The usual trick for infinite metrics d(x,y) is to use d(x,y)/(1+d(x,y)), which preserves ordering if you replace d(x,y) with whatever integral norm you like. Sadly it is not a norm – Alex R. Feb 10 2011 at 3:59
	You can get a partial order by truncation: $f \geq g$ if for all sufficiently large $N>0$, the integral of $f$ over the region where $f<N$ is greater than the integral of $g$ over the region where $g<N$. I don't know of a canonical extension to a strict order. – S. Carnahan♦ Feb 10 2011 at 4:26
	Can you regularize the integrals? en.wikipedia.org/wiki/… – Steve Huntsman Feb 10 2011 at 12:49