The famous ( ? ) Jordan decomposition Theorem for Bounded variation functions is generally stated as below.

Theorem. If $f:[a,b] \to\mathbb R$ is a function of bounded variation then there is a pair of nondecreasing functions $f^+$ and $f^-$ such that $f= f^+- f^-$ and $TV (f) = f^+(b)-f^+ (a) + f^- (b)- f^- (a)$ where $TV (f)$ denotes the total variation of $f$. The pair is unique up to addition of a constant, i.e. if $g^+$ and $g^-$ is a second pair with the same property, then $g^+-g^-=f^+-f^-\equiv {\rm const}$.

My question is motived by Hahn-Jordan decomposition Theorem to signed measure.

Question 1: It is possible to generalize the theorem of Jordan decomposition for functions of bounded variation defined on intervals [a, b] for functions of bounded variation defined on some mensurable space $(\Omega,\mathcal{F})$ with some partial order $\prec$?

In this case, a function $f$ is nondecreasing if $\omega \prec \eta$ implies $f(\omega)\leq f(\eta)$.

Of course, to generalize this theorem we must define the that becomes the total variation VT $ (f) $ for functions $ f $ defined in $ \Omega $.

In this case I prefer to be flexible and not risk any definition not to restrict the generality of the solution.

Question 2:Suppose that the theorem admits no such generazation. Now if $(\Omega,d) $ is a compact metric space with a partial order $\prec$ and the functions $ f $ are continuous then there is some generalization for this case?

Thank you.

  • $\begingroup$ I think the Riez-Markov theorem gives a direction for a positive answer to this question when $\Omega $ is against a set of measurable functions. $\endgroup$ – MathOverview Nov 9 '12 at 16:25
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    $\begingroup$ You probably also want to have some kind of relation between the partial order and the $\sigma$-algebra, e.g., that the $\sigma$-algebra contains the one generated by the partial order, or equivalently that all sublevel sets are measurable. $\endgroup$ – Lukas Geyer Nov 9 '12 at 16:28
  • $\begingroup$ I think a natural generalization of the concept of total (and similarly, positive and negative) variation would be to take the supremum over $\sum_{k=1}^n |f(b_k) - f(a_k)|$ where $(a_1,b_1), \ldots, (a_n, b_n)$ are mutually disjoint open intervals w.r.t. the partial order. $\endgroup$ – Lukas Geyer Nov 9 '12 at 16:39
  • $\begingroup$ @LukasGeyer I think a generalization like total variation of signed measures. See encyclopediaofmath.org/index.php/Signed_measure $\endgroup$ – MathOverview Nov 9 '12 at 16:49

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