If is a signed measure then we know that the total variation is a measure. It then makes sense to discuss whether or not a measurable function is integrable with respect to . In this case, will be integrable with respect to both and . Indeed, since this is obviously true for simple , and general integrable functions are limits of simple integrable functions.
This being the case, we can define both integrals
and, since , it makes sense to define
This integral shares some properties with “positive” integrals. For instance, it’s clearly linear:
Unfortunately, it doesn’t play well with order. Indeed, if is a measurable -negative set, then everywhere, but
This throws off most of our basic properties. However, some can be salvaged. It’s no longer necessary that a.e. for the integral to be zero, but it’s sufficient. And, thus, if a.e. then their integrals are equal, although the converse doesn’t hold.
One interesting fact is that for every measurable set we find
where we take the supremum over all measurable functions with everywhere. Indeed, if we take a Hahn decomposition for , then since is measurable so are and . If we take and , then we find
Thus we can actually attain this value. Can we get any larger? No. We can’t achieve anything by adding to the value of outside , since the integral is only taken over anyway. And within we could only increase the positive component of the integral by increasing the value of in , or increase the negative component by decreasing the value of in . Either way, we’d make some , which isn’t allows. Thus the total variation over is indeed this supremum.