MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can you define a "derivative" operator such that its antiderivative F(x) of f(x) can be used in the sense of F(b)-F(a) to calculate the Riemann–Stieltjes integral of f(x)?

Perhaps it would be related to the $\Delta$-derivative in time scale calculus.

This question was motivated by an edit to that wikipedia article which said that the ideas of unifying sums and integrals go back to the idea of the Riemann–Stieltjes integral. Now I'm not sure it's correct to say the RS-integral is a precursor of time-scale calculus as the starting point of time-scale calculus is the derivative and the unification of difference and differential equations, but integrals on other time-scales such as the q-integral in quantum calculus can be related to the RS integral (page 7 of a paper by Abreu), so maybe definite integrals on all time scales can be written in the form of RS-integrals for some suitable choice of step function depending on the time-scale.

Edit: About the same time, it seems, as I asked this question, a new post appeared on the wikipedia discussion page for "Time scale calculus" which suggests the Hilger derivative (delta-derivative) is the same as the Radon–Nikodym derivative of the Lebesgue–Stieltjes integral.

share|cite|improve this question
If you want the fundamental theorem to have any chance of being true, the obvious definition is lim_{h to 0} ((f(x+h) - f(x))/(g(x+h) - g(x)). – Qiaochu Yuan Oct 27 '10 at 12:43
Follow up question:… – Roy Maclean Oct 27 '10 at 16:58

For a Riemann-Stieltjes integral $$\int_a ^b f(x) d \phi(x)$$ the corresponding derivative should be $$f^{(\phi)}(x)=\lim_{y\to x} \frac{f(y)-f(x)}{\phi(y)-\phi(x)}$$ This works at least if $\phi\in C^1$ and $\phi'$ is never zero, that is, in this case we have $$\int_a ^b f^{(\phi)}(x) d \phi(x)=f(b)-f(a)$$

share|cite|improve this answer
Thanks Florian, and thanks also to Qiaochu Yuan for the similar comment. For step functions though, which would be necessary to represent integrals such as the q-integral, ϕ′ would be zero, and undefined at the jumps. – Roy Maclean Oct 27 '10 at 13:01
You only need $F,\phi$ to be finite variation with $\int\cdot\,dF$ absolutely continuous with respect to $\int\cdot\,d\phi$. Then the Radon-Nikodym theorem defines $dF/d\phi$ ( – George Lowther Oct 27 '10 at 13:02
A differential equation $dF(x)/d\phi(x)=g(F(x),x)$ using the Radon-Nikodym derivative can be written $dF(x)=g(F(x),x)\,d\phi(x)$, interpreted as an integral equation. So, yes, "Radon-Nikodym differential equations" can be defined, but they are the same as integral equations. Stochastic differential equations are defined via the integral form. – George Lowther Oct 27 '10 at 14:03
And on the Wikipedia Ito calculus page - I'm rather wary of the bit about the derivatives (whether it is even new at all). But I see there is some argument about it on the discussion page. – George Lowther Oct 27 '10 at 14:11
The existence of the Radon-Nikodym derivative requires that the measure $dF$ be absolutely continuous with respect to Lebesgue measure which is not satisfied by a pure jump function where the support of the measure is a finite or countable set. – Hany Oct 27 '10 at 16:48

It's a long time since this question was active but I'm going to answer the question anyway. This year I have been working in this issue. I defined a Stieltjes derivative from the beginning because I didn't know that something quite similar already exists. There are a few articles, each of one has its own "Stieltjes" derivative, some of them are the following:

Daniell, P. J.: Differentiation with respect to a function of limited variation. Trans. Amer. Math. Soc. 19 (1918), no. 4, 353–362.

S. Saks: Theory of Integral, Ch. IV, end of §2 and Chapter IX, §5.

M. Gradinaru: On the derivative with respect to a function with applications to Riemann- Stieltjes integral. Collection: Seminar on Mathematical Analysis (Cluj-Napoca, 1989–1990), 21–28, "Babes-Bolyai'' Univ., Cluj-Napoca, 1990.

Aversa, V.; Preiss, D.: Lusin's theorem for derivatives with respect to a continuous function. Proc. Amer. Math. Soc. 127 (1999), no. 11, 3229–3235.

In any case, I have proved the fundamental theorems of calculus for this derivative using not only Lebesgue-Stieltjes integration but also with the Kurzweil-Stieltjes integration.

In addition, I have proved that the delta-derivative (of Time Scales Calcules) is a particular case of the Stieltjes derivative, so every differential equation in time scales could be seen as a differential equation with the Stieltjes derivative. Moreover the impulsive differential equations with fixed times could be seen as a differential equation with the Stieltjes derivative too.

share|cite|improve this answer

Let $G:[0,1]\longrightarrow[0,1]$ be a continuous increasing function that is constant on each subinterval of the complement of the Cantor ternary set $K$ and satisfying $G(0)=0, G(1)=1$.

The Riemann-Stieltjes integral with respect to $G$ cannot be 'differentiated'. Consider a function $F(x)$ and suppose there is some Riemann-Stieltjes integrable function $h$ such that $$ F(b)-F(a)=\int_a^b h(t)dG(t),\quad \forall a,b \in[0,1]$$ Then, as the R.H.S. vanishes on any interval contained in the complement of $K$, $F$ must be constant on each such interval. So $F(x)=x$ for example cannot satisfy the above formula.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.