Actually, I did go to an event today, but despite rather than because of the day. Jeffrey Bub was talking up at UMBC, and it gave me the chance to clothesline him and ask about convex sets and ordered linear spaces, which Howard Barnum had said he (Dr. Bub) knew something about the interpretation of as state- and measurement-spaces.
If we want our Riemann-Stieltjes sums to converge to some value, we’d better have our upper and lower sums converge to that value in particular. On the other hand, since the upper and lower sums sandwich in all the others, their convergence is enough for the rest. And their convergence is entirely captured by their lower and upper bounds, respectively — the upper and lower Stieltjes integrals. So we want to know when .
We’ll prove this equality in general by showing that the difference has to be arbitrarily small. That is, for any partition of we have the inequalities
by definition. Subtracting the one from the other we find
So if given an we can find a partition for which the upper and lower sums differ by less than then the difference between the upper and lower integrals must be even less. If we can do this for any , we say that the function satisfies Riemann’s condition with respect to on .
The lead-up to the definition of Riemann’s condition shows us that if satisfies this condition then the lower and upper integrals are equal. Then just like we saw happen with Darboux sums we can squeeze any Riemann-Stieltjes sum between and upper and a lower sum. So if the upper and lower integrals are both equal to some value, then the limit of the Riemann-Stieltjes sums over tagged partitions must exist and equal that value, and thus is Riemann-Stieltjes integrable with respect to on .
Now what if the is Riemann-Stieltjes integrable with respect to on ? We would hope that then satisfies Riemann’s condition with respect to on , and so these three conditions are equivalent. So given we need to find an actual partition of so that .
Since we’re assuming that is Riemann-Stieltjes integrable, we’ll call the value of the integral . Then we can find a tagged partition so that for any finer tagged partitions and we have
Combining these we find that
Now as we pick different and we can make the difference in values of get as large as that between and . So for any we can choose tags so that . In particular, we can consider , which is positive because is increasing.
The difference between the upper and lower sums is
which is then less than
which is then less than .
Thus we establish the equivalence of Riemann’s condition and Riemann-Stieltjes integrability, as long as the integrator is increasing.