From our study of functions of bounded variation, we know that can be written as the difference between two increasing functions . Then the linearity of the integral in the integrator tells us that
Or does it? Remember that we have to understand this equation as saying that if two of the integrals exist then the third one does and the equality holds. But we also know that we’ve got a lot of choice in how to carve up , and it’s easily possible to do it in such a way that the integrals on the right don’t exist.
Luckily, we can show that there’s always at least one splitting of into two parts like we want, and further so that all the integrals above exist and the equation holds. And it turns out that the first function is just the variation ! That is, I assert that if is Riemann-Stieltjes integrable with respect to on the interval and there exists some with on , then is Riemann-Stieltjes integrable with respect to on the same interval. And because is increasing, we can prove it by showing that satisfies Riemann’s condition with respect to on .
So, given we need to find a partition so that for any finer partition we have .
We start by picking a partition so that for any finer partition and any choices and of tags for we have
Such a partition exists because the sum is the difference between two Riemann-Stieltjes sums for , and we’re assuming that these sums converge to the value of the integral.
Now let’s refine it a bit. Any refinement will still satisfy the same condition we already have, but let’s make it also satisfy
This we can do because the total variation is the supremum of such sums over all partitions, and so we can find a fine enough partition to come within of it. This new partition will be the one we need to establish Riemann’s condition. Now we must actually prove that it works.
First, notice that , since the variation only grows as fast as the function itself changes. Also, the supremum and infimum in the th subinterval are each less than in absolute value, so . Together, these show us that
by the second property of we established when we refined our partition.
Now we set . We want to pick two sets of tags so that for subintervals where , and so that for subintervals where . Then we find the inequality
by the first property of the partition we chose. Notice that in some of the subintervals I took the absolute value off of the change in by switching which sample of I subtracted from which, introducing a new negative sign.
So, adding these two big inequalities together, we find that
which shows that the upper and lower sums for the partition differ by less than . Therefore indeed satisfies Riemann’s condition, and it thus integrable with respect to on .
What this means is that integrals with respect to integrators of bounded variation can always be reduced to those with respect to increasing integrators, and thus to situations where Riemann’s condition can be brought to bear.