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?

  • $\begingroup$ 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. $\endgroup$ Mar 11, 2019 at 3:52
  • 1
    $\begingroup$ 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. $\endgroup$
    – RRL
    Mar 11, 2019 at 4:23

1 Answer 1


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:

$$\int_a^b f \, d(\gamma - \beta) = \left.(\gamma- \beta)f\right|_a^b - \int_a^b (\gamma - \beta) \, df = \left.(\gamma- \beta)f\right|_a^b-\int_a^b \gamma \, df + \int_a^b \beta \, df \\ = \int_a^b f \, d\gamma - \int_a^b f \, d\beta $$


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.

  • $\begingroup$ Thanks a lot. That's a great answer $\endgroup$ Mar 11, 2019 at 5:24
  • $\begingroup$ +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 $\beta=V, \gamma =V-\alpha$. $\endgroup$
    – Paramanand Singh
    Mar 11, 2019 at 15:37
  • $\begingroup$ @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. $\endgroup$
    – RRL
    Mar 11, 2019 at 16:56

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