Sufficient Conditions for Integrability
Let’s consider some conditions under which we’ll know that a given Riemann-Stieltjes integral will exist. First off, we have a straightforward adaptation of our old result that continuous functions are Riemann integrable. Now I assert that any continuous function on an interval
is Riemann-Stieltjes integrable over that interval with respect to any function
of bounded variation on the same interval. In particular, the function
is clearly of bounded variation, and so we will recover our old result.
In fact, we can even adapt the old proof. The Heine-Cantor theorem says that the function , being continuous on the compact interval
is uniformly compact. As usual, we can assume that
is increasing on
. And now Riemann’s condition tells us to consider the difference
We want this difference to go to zero as we choose finer and finer partitions
By uniform continuity we can pick a small enough (depending only on
) so that when
we’ll have
, where
is the total variation
. Then picking a partition whose subintervals are thinner than
makes it so that
, which we can then pull out of the sum. What remains sums to exactly the total variation
, and so the difference
is below
, and our theorem holds.
Immediately from this result and integration by parts we come up with another set of sufficient conditions. If is continuous and
is of bounded variation on an interval
then
is Riemann-Stieltjes integrable with respect to
over
. Since the integrator
is also continuous, this tells us that any function of bounded variation is Riemann integrable!
Of course, these conditions are just sufficient. That is, if they hold then we know that the integral exists. However, if an integral exists, we can’t use these to conclude anything about either the integrand or the integrator. For that we need necessary conditions.
[…] Conditions for Integrability We’ve talked about sufficient conditions for integrability, which will tell us that a given integral does exist. Now we’ll consider the situation where […]
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[…] the function is continuous and of bounded variation (in fact it’s decreasing), and so it’s integrable with respect to . Then it’s integrable over the subinterval . Why not just start by saying […]
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