## Integrals are Additive Over Regions

Before I state today’s proposition, I need to define what I mean by saying that two sets in are “nonoverlapping”. Intuitively, we might think that this means they have no intersection, but that’s not quite it. We’ll allow some intersection, but only at boundary points. Since the regions we’re interested in for integrals are Jordan measurable, and their boundaries have zero Jordan content, so we know changing things along these boundaries in an integral will make no difference.

Let be a collection of bounded regions in , so that any two of these regions are nonoverlapping. We define their union

and let be a bounded function defined on this union. Then is integrable on if and only if it’s integrable on each , and we find

Indeed, if is some -dimensional interval containing , then it will also contain each of the smaller regions , and we can define

and

We can use Lebesgue’s condition to establish our assertion about integrability. On the one hand, the discontinuities of in must be contained within those of , so if the latter set has measure zero then so must the former. On the other hand, the discontinuities of consist of those within each of the , and maybe some along the boundaries. Since the boundaries have measure zero, and we assume that the discontinuities in each are of measure zero, their (countable) union will also have measure zero. And then so must the set of discontinuities of in have measure zero as well.

The inclusion-exclusion principle tells us that we can rewrite the characteristic function of :

We can put this into our integral and use the fact that integrals are additive with respect to finite sums in the integrand

But we assumed that all the are nonoverlapping, so any intersection of two or more of them must lie only along their boundaries. And since these boundaries all have Jordan content zero the integrals over them must come out to zero. We’re left with only the sum over each subregion, as we wanted.