# Jordan's theorem implication

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?

• I'm guessing that what your professor meant was that it would make good sense to take $\ \int fd\gamma-\int fd\beta\$ as being the definition of $\ \int fd\alpha\$. To say that the equation "follows necessarily" must mean that you already have some other definition of $\ \int fd\alpha\$ in mind. To the best of my knowledge, however, the definition apparently suggested by your professor is the usual one, and I'm not aware of any other. Mar 11, 2019 at 3:52
• It is not just a definition. The definition is more basic, and the Jordan decomposition can be used to compute the integral when the integrator is not a simple monotonic function, but rather of bounded variation.
– RRL
Mar 11, 2019 at 4:23

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.

• Thanks a lot. That's a great answer Mar 11, 2019 at 5:24
• +1 I think you can give a reference (regarding last part of your answer) that if $f$ is integrable with respect to $\alpha$ then it is also integrable with respect to variation function $V$ of $\alpha$ so that we can take $\beta=V, \gamma =V-\alpha$. Mar 11, 2019 at 15:37
• @ParamanandSingh: Thanks. One reference is my answer here which proves integrability with respect to the variation function $V$. Then integrability with respect to $V - \alpha$ obviously follows.
– RRL
Mar 11, 2019 at 16:56