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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 pre-cursor 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 paper by Abreau), 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.

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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: mathoverflow.net/questions/43807/… –  Roy Maclean Oct 27 '10 at 16:58
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2 Answers

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 $$\int_a ^b f^{(\phi)}(x) d \phi(x)=f(b)-f(a)$$

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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
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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 $dF/d\phi$ (en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem). –  George Lowther Oct 27 '10 at 13:02
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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
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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
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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
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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.

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