# Variation of a Integral function

Let $g$ be a function of finite $q$ Variation and $f$ be a function of finite $p$ Variation, and $\frac{1}{q}+\frac{1}{p}>1$. What can be said about the variation of $H$ with $$H(t):=\int_{a}^{t}g(s)df(s)\text{ ? }$$

Edit #1: The $p$ variation $\operatorname{Var}_{p}(f;[a,b])$ of a Function $f$ over $[a,b]$ is defined as: $$\operatorname{Var}_{p}(f;[a,b])=\sup \left(\sum_{i=1}^{n}\left|f(t_{i})-f(t_{i-1})\right|^{p}\right)^{\frac{1}{p}}$$

where the supremum runs over all partitions of $[a,b]$

Edit #2: If $f$ would be monotone on $[a.b]$ then I could use a mean value theorem for RS Integrals. Is there a similar mean Value theorem for a function $f$ which is not monotone? I think, that then I could conclude: $H$ has finite $p$ variation.

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Do you mind adding the definition of p-variation? The one I have in mind does not imply that $f$ has finite variation in the usual (p=1) sense, which raises the question of how the integral is defined. – user31373 Jul 2 '12 at 22:45
Nobody an idea? I think what I need is something similar to the Mean Value Theorem.Does anyone know such a result for a nonmonotone $f$? – Peter Moor Jul 3 '12 at 15:32
Well, so far it's not even clear that $H$ is defined. The finiteness of $\mathrm{Var}_p(f;[a,b])$ does not guarantee that $f$ is a function of bounded variation, that is, $\mathrm{Var}_1(f;[a,b])<\infty$. How do you define $\int g\,df$ if $f$ is not of bounded variation? Another issue: even if $f$ has bounded variation, $g$ may fail to be integrable with respect to $df$, for example if $f$ and $g$ have a common point of discontinuity. – user31373 Jul 3 '12 at 15:40
According to a Paper of Young the condition $\frac{1}{p}+\frac{1}{q}>1$ ensures that $H$ is well defined. Edit 1: We can assume that $f$ is continuous. – Peter Moor Jul 3 '12 at 15:41
Do you think that I can ask this Question on mathoverflow? Or is it too elementary? – Peter Moor Jul 3 '12 at 19:55