In class we proved we can write a function in bounded variation as the difference of two non-decreasing functions. (Jordan's theorem)
My prof then said this suggests that this could be a good source of integrators w.r.t. which continuous functions can be integrated. She wrote:
for some $\alpha \in BV(a,b)$, $\int fd\alpha$ = $\int f d(\gamma-\beta) =? \int f d\gamma – \int f d\beta$ for some non-decreasing $\gamma$ and $\beta$. I am wondering why this doesn't follow necessarily? Is there something I am missing?
Best Answer
The basic definition of a Riemann-Stieltjes integral of a function $f$ with respect to any integrator $\beta$ is similar to the Riemann integral -- through convergence of Riemann-Stieltjes sums or equivalently through upper and lower Darboux integrals if the integrator is non-decreasing.
In the definition, we say that $f$ is Riemann-Stieltjes integrable with respect to $\beta$ if there exists a number $I$ such that for any $\epsilon > 0$ there exists a partition $P_\epsilon$ of $[a,b]$ such that if $P = (x_0,x_1,\ldots, x_n)$ refines $P_\epsilon$, then
$$\left|\sum_{j=1}^n f(t_j) (\beta({x_j}) - \beta(x_{j-1}))- I\right| < \epsilon,$$
where $t_j \in [x_{j-1},x_j]$ are arbitrary intermediate points.
As long as $f$ is continuous and $\gamma$ and $\beta$ are non-decreasing then we have existence of the Riemann-Stieltjes integrals
$$\int_a^b f \, d\gamma, \quad \int_a^b f \, d\beta$$
A Riemann-Stieltjes integral can fail to exist if the integrand and integrator have a common point of discontinuity. The non-decreasing functions could have jump discontinuities, but since $f$ is continuous there is no possibility of shared discontinuities.
It is then straightforward to show that $f$ must be Riemann-Stieltjes integrable with respect to $\alpha$ and
$$\int_a^b f d\alpha =\int_a^b f \, d(\gamma - \beta) =\int_a^b f \, d\gamma - \int_a^b f \, d\beta$$
The proof uses integration by parts and the linearity property:
$$\int_a^b f \, d(\gamma - \beta) = \left.(\gamma- \beta)f\right|_a^b - \int_a^b (\gamma - \beta) \, df = \left.(\gamma- \beta)f\right|_a^b-\int_a^b \gamma \, df + \int_a^b \beta \, df \\ = \int_a^b f \, d\gamma - \int_a^b f \, d\beta $$
Aside
An integrator $\alpha$ of bounded variation has no unique Jordan decomposition $\alpha = \gamma - \beta$. If $f$ is not continuous but is Riemann-Stieltjes integrable with respect to $\alpha$, then there is no guarantee that integrals $\int_a^b f \, d\gamma$ and $\int_a^b f \, d\beta$ exist for any given decomposition. However there will always exist at least one pair $(\gamma, \beta)$ such that the integrals exist.